PREDICATE LOGIC Jorma K. Mattila LUT, Department of Mathematics and Physics 1 Basic Concepts In predicate logic the formalism of propositional logic is extended and is made it more finely build than propositional logic. Thus it is possible to present more complicated expressions of natural language and use them in formal inference. Example 1.1. Consider the inference All ravens fly. Peter is a raven. So, Peter flies. In the view of propositional logic, the sentences are totally different atoms, and thus they have no such common parts needed in the analysis of the inference. However, there are clearly similar parts in the sentences. In the first sentence there is spoken about ravens and flying, in the second sentence about ravens and Peter, and in the last sentence there is spoken about Peter and flying. So, points of contacts can be found. Predicate language, as propositional language, too, consists of the following parts: Syntax: Alphabet and rules for formation wff’s Semantics: Questions connecting models: interpretation, determining truth values, i.e. how to translate wff’s for example into natural language, and what are the exact conditions for their truth. Proof theory: The principle of calculus, axioms and inference rules. Theo- rems are deduced from axioms by means of inference rules. (Axioms are 1
Transcript
PREDICATE LOGIC
Jorma K. MattilaLUT, Department of Mathematics and Physics
1 Basic Concepts
In predicate logic the formalism of propositional logic is extended and is madeit more finely build than propositional logic. Thus it is possible to present morecomplicated expressions of natural language and use them in formal inference.
Example 1.1. Consider the inference
All ravens fly. Peter is a raven. So, Peter flies.
In the view of propositional logic, the sentences are totally different atoms,and thus they have no such common parts needed in the analysis of the inference.However, there are clearly similar parts in the sentences. In the first sentencethere is spoken about ravens and flying, in the second sentence about ravens andPeter, and in the last sentence there is spoken about Peter and flying. So, pointsof contacts can be found.
Predicate language, as propositional language, too, consists of the followingparts:
Syntax: Alphabet and rules for formation wff’s
Semantics: Questions connecting models: interpretation, determining truth values,i.e. how to translate wff’s for example into natural language, and what arethe exact conditions for their truth.
Proof theory: The principle of calculus, axioms and inference rules. Theo-rems are deduced from axioms by means of inference rules. (Axioms are
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premises that are taken the basic truths of the system. Thus they must belogically true.)
Let us denote the pure predicate language by the symbol P .
1.1 Syntactic Components of Predicate Logic
Predicate logic contains all the components of propositional logic, includingpropositional variables and constants. In addition, predicate logic contains terms,predicates, and quantifiers. Terms are typically used in place of nouns and pro-nouns. They are combined into sentences by means of predicates. For example,in the sentence "John loves Mary", the nouns are "John" and "Mary", and thepredicate is "loves". The same is true if this sentence is translated into predicatelogic, except that "John" and "Mary" are now called terms. Predicate logic usesquantifiers to indicate if a statement is always true, if it is sometimes true, or itis never true. In this sense, the quantifiers are used to correspond words such as"all", "some", "never", and related expressions.
Definition 1.1. Alphabet of P consists of connectives (familiar from proposi-tional logic) ¬, ∧, ∨, → and ↔and parentheses (, ) and dot ,. Other symbolsbelonging to alphabet are
individual constants: a, b, c, . . .
variables: x, y, z, . . .
predicate symbols P , Q, R, . . .
function symbols f , g, . . .
identity symbol =
quantifiers ∀ and ∃.
Constants are thought to be names of creatures of (real or ideal) world (moreprecisely: interpretation maps constants to some elements of a set A of crea-tures). Variables refer to creatures, too, but not to any certain ones. They corre-spond nearest to pronouns, like he, she, it, . . .. Expressions including so-called
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free variables can have a truth value not until variables have got their constantvalues.
1.1.1 The Universe of Discourse
To explain the main concepts of this section, we use the following logical argu-ment:
1. Jane is Paul’s mother.
2. Jane is Mary’s mother.
3. Any two persons having the same mother are siblings.
4. Paul and Mary are siblings.
The truth of the statement "Jane is Paul’s mother" can only be assessed within acertain context. There are many people named Jane and Paul, and without furtherinformation the statement in question can defer to many different people, whichmakes it ambiguous. To remove such ambiguities, we introduce the concept of auniverse or a domain.
Definition 1.2. The universe of discourse or domain is the collection of all per-sons, ideas, symbols, data structures, and so on, that affect the logical argumentunder consideration. The elements of the universe of discourse are called indi-viduals
In the argument concerning Mary and Paul, the universe of discourse may,for example, consist of the people living in a particular house or on a particularblock.
Many arguments involve numbers and, in this case, one must stipulate whetherthe domain is the set of natural numbers, the set of integers, the set of real num-bers, or even the set of complex numbers. In fact, the truth of a statement maydepend on the domain selected. The statement "There is a smallest number" istrue in the domain of natural numbers, but false in the domain of integers.
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To avoid trivial cases, one stipulates that every universe of discourse mustcontain at least one individual. Hence, the set of all natural numbers less that0 does not constitute a universe because there is no negative natural numbers.Instead of the word individual, one sometimes uses the word object, such as in"the domain must contain at least one object". To refer to a particular individualor object, identifiers are used.These identifiers are called individual constants.Each individual constant must uniquely identify a particular individual and noother one. For example, if the universe of discourse consists of persons, theremust not be two persons with the same name.
1.1.2 Predicates
Generally, predicates make statements about individuals. To illustrate this notion,consider the following statements:
(a) Mary and Paul are siblings.
(b) Jane is the mother of Mary.
(c) Tom is a cat.
(d) The sum of 2 and 3 is 5.
In each of these statements, there is a list of individuals, which is given bythe argument list, together with phrases that describe certain relations among orproperties of the individuals mentioned in the argument list. These properties arereferred to as predicates. For example, in the statement (a), the argument list isgiven by "Mary" and "Paul", in that order, whereas the predicate is described bythe phrase "are siblings". The entries of the argument list are called arguments.In this sense, arguments are terms, i.e. variables or individual constants.
In predicate logic, each predicate is given a name, which followed by the listof arguments. For example, to express "Jane is the mother of Mary" one wouldchoose an identifier, say, "mother, to express the predicate "is mother of", andone would write ’mother(Jane, Mary)’. Very often only single letters are usedfor predicate names and terms. We usually write, for example, M(j,m) instead
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of mother(Jane, Mary). The order of the arguments is important. Clearly, thestatement M(j,m) and M(m, j) have completely different meaning.
Each predicate is associated with arity number. It indicates the number ofelements in the argument list of a predicate.Unary predicates describe propertiesof of objects, for example, "P (x)" indicates that x has the property P . Theinterpretation of a predicate P in a set of objects, A, is the set of those elementsof A that have the property P , i.e.,
{α ∈ A | P (α)}.
A predicate with arity n is often called an n-place predicate. These predicatesindicate relations between objects. For example, if Q is a two-place predicatethen we interpret Q as a binary relation on a universe of discourse A, i.e. as thepairs of elements (α, β) ∈ A×A to denote that α is in the relationQ to β. Hencewe can write
{(α, β) ∈ A× A | Q(α, β)}.
Example 1.2. The predicate "is a cat" is one-place predicate. The predicate "isthe mother of" is two-place predicate, i.e., its arity is 2. The predicate in thestatement "The sum of 2 and 3 is 6" (which is false) contains the three-placepredicate "is the sum of".
The identity ’=’ is a constant two-place predicate always having the sameinterpretation: identity on a set of terms A is
{(α, β) | α, β ∈ A}.
The identity symbol can be inserted only between two terms.
Definition 1.3. In predicate language P , an atomic formula (or an atom) is
(a) a predicate name followed by an argument list,
(a) an identity t1 = t2, where t1 and t2 are terms (i.e. individual constants orvariables).
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Atomic formulas are statements, and they can be combined by logical con-nectives in the same way as atoms in propositional logic. For example, usingabove mentioned natural language counterparts, we can write
M(j,m) → ¬M(m, j)
to mean that "If Jane is Mary’s mother then Mary is not Jane’s mother."
If all arguments of a predicate are individual constants, then the resultingatomic formula must either be true or false. For example, if the universe ofdiscourse consists of Jane, Doug, Mary, and Paul, we have to know for eachordered pair of individuals whether or not the predicate "is the mother of" (or"mother" for short) is true. This can be done in the form of a table. Any methodthat assigns truth-values to all possible combinations of individuals of a predicateis called an assignment of the predicate.For example, the following table is anassignment of the predicate "mother".
Doug Jane Mary PaulDoug F F F FJane F F T TMary F F F FPaul F F F F
In general, if a predicate has two arguments, its assignment can be given bya table in which the lows correspond to the first argument and columns to thesecond.
In a finite universe of discourse, one can represent the assignments of predi-cates with arity n by n-dimensional arrays. For example, properties are assignedby one-dimensional arrays, predicates of arity 2 by two-dimensional arrays, andso on.
Note that mathematical symbols ≥ and ≤ are predicates. However, thesepredicates are normally used in infix notations. By this, we mean that they areplaced between the arguments. For example, we usually write 2 > 1 instead of> (2, 1).
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1.1.3 Variables and Instantiations
Often, one does not want to associate the arguments of an atomic formula witha particular individual. To avoid this, variables are used. Variable names arefrequently chosen from the end of the alphabet: x, y, z etc., with or withoutsubscripts. Examples of expressions containing variables include
cat(x)→ hastail(x),
dog(y) ∧ brown(y),
grade(x)→ (x ≥ 0) ∧ (x ≤ 100).
As in propositional logic, expressions can be given names, i.e. meta-variablesare used. For example, one can give the name A to the expression
A := B(x) → C(x)
which means that when we write A we really mean "B(x) → C(x)".
Example 1.3. Consider a statement "If x is a cat then x has tails." We formalizeit as follows: C(x) := "x is a cat", T (x) := "x has tails", then the formalizationof the whole sentence is
A := C(x) → T (x).
This is an open formula, because there exists a variable x in the formula. Wecannot give a truth-value to it before we know the value of of x. If S is a domainof objects, where x can have values, we can choose some object and replace it toall occurrences of x in the formula. After doing this, we can state the truth-valueof the corresponding closed formula. Suppose that a ∈ S, and a refers to Tom.When we replace a to the instances of x of the formula, we have
C(a) → T (a),
and its interpretation in natural language is "If Tom is a cat then Tom has tails".
Generally, if A is an expression, the expression obtained by replacing alloccurrences of a variable x in A by term t is denoted by Sx
t A. According toExample 1.3, Sx
aA stands for C(a) → T (a).
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Definition 1.4. Let A represent an expression, x represent a variable, and t rep-resent a term. Then Sx
t A represents the expression obtained by replacing alloccurrences of x in A by t. Sx
t A is called an instantiation of A, and t is said tobe an instance of x.
Example 1.4. Let a, b, and c be individual constants, P and Q be predicatesymbols, and x and y be variables. Find
Sxa (P (a) → Q(x)),
Syb (P (y) ∨Q(y)),
SyaQ(a),
Sya(P (x) → Q(x)).
Solution: Sxa (P (a) → Q(x)) is P (a) → Q(a), and Sy
b (P (y) ∨ Q(y)) is P (b) ∨Q(b). Since (Q(a)) does not contain any y, replacing all occurrences of y by aleaves Q(a) unchanged, which means that
SyaQ(a) = Q(a).
Similarly,Sy
a(P (x) → Q(x)) = P (x) → Q(x).
1.1.4 Quantifiers
Consider the following statements:
1. All cats have tails.
2. Some people like their meat raw.
3. Everyone gets a break once in a while.
All these statements indicate how frequently certain things are true. In pred-icate logic, one uses quantifiers in this context. Specially, we will discuss twoquantifiers: the universal quantifier, which indicates that something is true forall individuals, and the existential quantifier, which indicates that a statement istrue for some individuals.
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Definition 1.5. Let A represent an expression, and let x represent a variable. Ifwe want to indicate that A is true for all possible values of x, we write ∀x A.Here ∀x is called the universal quantifier, and A is called the scope of the quan-tifier. The variable x is said to be bound by the quantifier. The symbol ∀ ispronounced "for all".
The quantifier and the bounded variable that follows have to be treated as aunit, and this unit acts somewhat like a unary connective. Statements containingwords like "every", "each", and "everyone" usually indicate universal quantifica-tion. Such statements must typically be reworded such that they start with "forevery x", which is then translated to ∀x.
Example 1.5. Express "Everyone gets a break once in a while" in predicate logic.Solution: We define B to mean "gets a break once in a while". Hence, B(x)
means that x gets a break once in a while. The word "everyone" indicates thatthis is true for all x. This leads to the following translation
∀x B(x).
Example 1.6. Express "All cats have tails" in predicate logic.Solution: We first have to find the scope of the universal quantifier, which is
"If x is a cat, then x has tails". After choosing descriptive predicate symbols, weexpress this by the following compound formula:
cat(x) → hastails(x).
This expression must be universally quantified to yield the required solution:
∀x (cat(x) → hastails(x)).
At last, when we let C stand for "cat" and T stand for "hastails", we have theresult
∀x (C(x) → T (x)).
Under quantification, we are able to use what ever variable we like, for example∀ y (C(y) → T (y)) tells the same thing.
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Definition 1.6. Let A represent an expression, and let x represent a variable. Ifwe want to indicate that A is true for at least one value of x, we write ∃xA. Thisstatement is pronounced "There exists an x such that A". Here ∃x is called theexistential quantifier, and A is called the scope of the existential quantifier. Thevariable x is said to be bound by the quantifier.
Statements containing such phrases as "some" and "at least one" suggest exis-tential quantification. They should be rephrased as "there is an x such that",which is translated by ∃x.
Example 1.7. Let P be the property "like their meat raw". Then ∃xP (x) canbe translated as "There exists people who like their meat raw" or "Some peoplelike their meat raw".
Example 1.8. If the universe of discourse is a collection of things, ∃x blue(x)should be understood as " There exist objects that are blue" or "Some objects areblue".
Quantifiers ∀x and ∃x have to be treated like unary connectives. The quanti-fiers are given a higher precedence than all binary connectives. For example, ifP (x) and Q(x) means that x is living and that x is dead, respectively, then onehas to write ∀x(P (x)∨Q(x)) to indicate that everything is either living or dead.∀xP (x) ∨Q(x) means that everything is living or x is dead.
The variable x in a quantifier is just a placeholder, and it can be replaced byany other variable name not appearing elsewhere in the expression. For example,∀xP (x) and ∀ y P (y) mean the same thing: they are logically equivalent. Theexpression ∀ y P (y) is called a variant of ∀xP (x).
Definition 1.7. An expression is called a variant of ∀xA if it is of the form∀ y Sx
y A, where y is any variable name, and Sxy A is the expression obtained
from A by replacing all instances of x by y. Similarly, ∃xA and ∃ y Sxy A are
variants of one another.
Quantifiers may be nested, as demonstrated by the following examples.
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Example 1.9. Translate the sentence "There is somebody who knows everyone"into the language of predicate logic. To do this, use K(x, y) to express the factthat x knows y.
Solution: The best way to solve this problem is to go in steps. We writeinformally
∃x(x knows everybody).
Here "x knows everybody" is still in English and means that for all y it is truethat x knows y. Hence,
x knows everybody = ∀ y K(x, y).
We now add the existential quantifier and obtain
∃x ∀ y K(x, y).
Example 1.10. Translate "Everybody has somebody who is his or her mother"into predicate logic.
Solution: We define M to be the predicate "mother"; that is, M(x, y) standsfor "x is the mother of y". The statement "Someone is the mother of y" becomes
∃xM(x, y).
To express that this must be true for all y, we add the universal quantifier, whichyields
∀ y ∃xM(x, y).
The statement "Nobody is perfect" also includes a quantifier, "nobody", whichis the absence of an individual with a certain property. In predicate logic, the factthat nobody has property P cannot be expressed directly. To express the fact thatthere is no x for which an expression A is true, one can either use ¬∃xA or∀x¬A. For example, if P represents the property of perfection, both ¬∃xPand ∀x¬P indicate that nobody is perfect. In the first case, we say, in verbaltranslation, "It is not the case that there is somebody who is perfect", whereasin the second case we say "For everyone, it is not the case that he or she isperfect". The two methods to express that "nobody is A" must of course belogically equivalent, i.e.,
¬∃xA ≡ ∀x¬A. (1.1)
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According to Definitions 1.5 and 1.6, the variable appearing in the quantifieris said to be bound. For example, in the expression ∀x (P (x) → Q(x)), xappears three times, and each time x is bound variable. Any variable that is notbound is said to be free. Later, we will see that the same variable can occur bothbound and free in an expression. For this reason, it is important to also indicatethe position of the variable in question.
Example 1.11. Find the bound and free variables in
∀ z (P (z) ∧Q(x)) ∨ ∃ y Q(y).
Solution: Only the variable x is free. All occurrences of z are bound, and soare all occurrences of the variable y.
Note that the status of a variable changes as expressions are divided intosubexpressions. For example, in ∀xP (x) x occurs twice, and it is bound bothtimes. This statement contains P (x) as a subexpression. Nevertheless, in P (x),the variable x is free.
Instantiations only affect free variables. Specifically, if A is an expression,Sx
t A only affects the free occurrence of the variable x. For example, Sxy∀xP (x)
is still ∀xP (x), that is, the variable x is not free. However„ Sxy (Q(x)∧∀xP (x))
yields Q(y) ∧ ∀xP (x). Hence, instantiation treats the variable x differently,depending on whether it is free or bound, even if this variable appears twice inthe same expression. Obviously, two things are only identical if they are treatedidentically. This implies that, if a variable appears both free and bound within thesame expression, we have in fact two different variables that happen to have thesame name. From this it follows that if several quantifiers use the same boundvariable for quantification, then all these variables are local to their scope, andthey are therefore distinct. To illustrate this, consider the statement "y has amother". If M is the predicate name for "is mother of", then this statement trans-lates into ∃xM(x, y). One obviously must not form the variant ∃ yM(y, y),which means that y is her own mother. For similar reasons, there are restric-tions on instantiations. For example, the instantiation Sy
x(∃xM(x, y)) is illegal.because its result is ∃xM(x, x). In such cases, one tampers with the way inwhich a variable is defined, and this undesired side effects.
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If all occurrences of x in an expression A are bound, we say "A does notcontain x free". If A does not contain x free, then the truth-value of A does notchange if x is instantiated to an individual constant. A is independent of x in thissense.
1.1.5 Restrictions of Quantifiers to Certain Groups
Sometimes, quantification is over a subset of the universe of discourse. Suppose,for instance, that animals form the universe of discourse. How can one expresssentences such as "All dogs are mammals" and "Some dogs are brown"? Con-sider first the statement "All dogs are mammals". Since the quantifier shouldbe restricted to dogs, one rephrases the statement as "If x is a dog, then x is amammal". This immediately leads to
∀x(dog(x) → mammal(x)).
Generally, the sentence∀x(P (x) → Q(x))
can be translated as "All individuals with property P also have property Q".Consider now the statement "Some dogs are brown". This statement means
that there are some animals that are dogs and that are brown. Of course, thestatement "x is a dog and x is brown" can be translated as
dog(x) ∧ brown(x).
"There are some brown dogs" can now be translated as
∃x(dog(x) ∧ brown(x))
The statement∃x(P (x) ∧Q(x))
can in general be interpreted as "Some individuals with property P also haveproperty Q". Note that if the universal quantifier is to apply to individuals witha given property we use the conditional to restrict the universe of discourse. Onthe other hand, if we similarly restrict application of the existential quantifier, weuse conjunction.
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Finally, consider statements containing the word "only", such as "only dogsbark". To convert this into predicate logic, this must be reworded as "It barksonly if it is a dog", or, equivalently, "If it barks, then it is a dog". One hastherefore
∀x (barks(x) → dog(x)).
2 Interpretations and Validity
2.1 Introduction
This section gives the semantical approach to predicate logic, and it deals withinterpretations of logical statements and with the soundness of logical arguments.Interpretations are obviously fundamental to predicate logic, and they are there-fore important in their own right. Moreover, interpretations allow one to distin-guish between arguments that are sound and arguments that are not. Soundnessis closely related to validity. Generally, an expression A is valid if A is truefor all interpretations. Valid expressions in predicate logic play the same role astautologies in propositional logic. In particular, logical implications and logicalequivalences are defined as valid implications and valid equivalences, respec-tively.
For consideration of truth of sentences of predicate language, we divide thealphabet of L into two parts, such that predicate symbols, function symbols,and individual constants form non-logical alphabet and other symbols logicalalphabet of L. Non-logical alphabet refer to those things illustrated by the lan-guage, i.e., to objects and their mutual relationships outside the language. Hence,in each case, we can restrict L by giving a list of the non-logical alphabet underconsideration. For example,
L = {P,Q,R; a, b, c, d}
is a predicate language having the predicate symbols P , Q, and R and the indi-vidual constants a, b, c, and d as its non-logical alphabet.
2.2 Interpretations
An interpretation of a logical expression contains the following components:
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1. There must be a universe of discourse.
2. For each individual, there must be an individual constant that exclusivelyrefers to this particular individual, and to no other.
3. Every free variable must be assigned a unique individual constant.
4. There must be an assignment for each predicate used in the expression,including predicates of arity 0, which represents propositions.
The truth status of a sentence can be determined after an interpretation isgiven to the sentence or to the whole L, i.e. we can speak about truth in somemodel.
Let L = {P,Q, . . . ; a, b, . . .} be a predicate language and A a non-empty set.
Definition 2.1. An interpretation of L in a set A is a valuation V that associatesan element of A to every constant symbol and an n-ary relation to every n-arypredicate symbol in L.
Example 2.1. Let L = {P,Q; a, b} be a predicate language where the both pred-icate symbols are binary predicates. Let A = Z+ (the set of positive integers).We can choose an interpretation to L, for example, such that
V (a) := 3, V (b) := 2, V (P (x, y)) := x < y, andV (Q(x, y)) := x ≥ y.
A model of L consists of a non-empty set and an interpretation according tothe following definition.
Definition 2.2. A model of L is an ordered pair M = (A, V ) where A is anon-empty set and V is an interpretation of L in the set A. A is the universe ofdiscourse of M.
A term of predicate logic has an interpretation according to the followingdefinition.
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Definition 2.3. (a) If a term t is a constant symbol then its interpretation in the aA is a certain element V (t) ∈ A.
(b) If a term t is a variable symbol then it can have any element of A as itsvalue in a given interpretation V , i.e., V (t) ∈ A arbitrarily.
Example 2.2. Consider the predicate language in Example 2.1 in the modelM = (Z+, V ) where V is the valuation of Example 2.1. Determine the truth-value of a formula P (c, b) → ¬Q(c, a) in the model M.
We have the interpretation 3 < 2 → 2 6≥ 3 for the formula, and hence it istrue in M.
We consider more exactly the truth of a formula in a model.
Definition 2.4. Let M = (A, V ) be a model, and α a formula of L.
(a) If P is a k-ary predicate symbol and a1, . . . , ak ∈ A are constants thenP (a1, . . . , ak) is true in M, denote M |= P (a1, . . . , ak), iff the k-tuple(a1, . . . , ak) belongs to the relation V (P ).
(b) A formula ¬α is true in a model M iff α is false in M, i.e., M |= ¬α iffM 2 α.
(c) M |= α→ β iff M 2 α or M |= β.
(d) M |= ∀xα iff M |= Sxaα(x) for any a ∈ A.
Metatheorem 2.1. Let M = (A, V ) be a model, α and β formulas of L, and γan open formula of L where x is free. Hence we have
(a) M |= α→ β iff M |= α⇒M |= β.
(b) M |= α ∨ β iff M |= α or M |= β.
(c) M |= α ∧ β iff M |= α and M |= β.
(d) M |= α↔ β iff either M |= α and M |= β or M 2 α and M 2 β.
(e) M |= ∀ x γ(x) iff M |= Sxaγ(x) for any a ∈ A.
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(f) M |= ∃ x γ(x) iff M |= Sxaγ(x) for some a ∈ A.
Proof. The assertions of Metatheorem follow from Definition 2.4 by the mutualdependency of connectives and that of quantifiers.
Note that only closed formulas can have a truth-value. It is nonsense to speakabout the truth status of open formulas. On the other hand, we can speak aboutsatisfaction of open formulas in a sense that in an interpretation variables of anopen formula can have such values belonging to the universe of discourse of thecorrespondent model, and according to the interpretation, it is possible to get atrue expression from the formula. It may also happen that a given expression isnot true and not false in a model. For this reason, we define for every statementα in L a set of relevant models as follows:
Mα = {M | M = (A, V ), such that either M |= α or M 2 α}.
Definition 2.5. A statement α is valid iff M |= α for all M∈Mα.A statement α is satisfiable iff M |= α for some M∈Mα.A statement α is refutable iff M 2 α for some M∈Mα.A statement α is contradiction iff M 2 α for all M∈Mα.
The closure of an open formula α is the closed formula obtained from α bybounding universally all occurrences of the free variables in α. Hence, accordingto Definition 2.5, the validity of an open formula means the consideration of truthof the closure of the formula in all relevant models.
Definition 2.6. Let ∆ be a set of statements of L and M = (A, V ) a model. Mis the model of the set of statements ∆ iff M |= α for all statements α of the set∆, denoted by M |= ∆.
Now the concept of semantic entailment (or logical consequence) in L can bedefined.
Definition 2.7. A statement α is a semantic entailment of a set ∆ of statementsiff α is true in every model of ∆.
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Example 2.3. Show that the statement
∃ y ∀xQ(x, y) → ∀x ∃ y Q(x, y)
is valid.LetM = (A, V ) be any relevant model of the statement. IfM 2 ∃ y ∀xQ(x, y)
then M |= ∃ y ∀xQ(x, y) → ∀x ∃ y Q(x, y) by Definition 2.4 (c). Thus, weconsider the case where M |= ∃ y ∀xQ(x, y). In this case, M |= ∀xQ(x, b)for some element b ∈ A by Metatheorem 2.1 (f). Hence, for some elementb ∈ A it holds that for all elements a ∈ A, M |= Q(a, b) by Definition 2.4(d). Hence, for all elements a ∈ A, there exists an element b ∈ A, such thatM |= Q(a, b). Hence, M |= ∃ y Q(a, y) holds foa all a ∈ A by Metatheorem2.1 (f), and thus M |= ∀x ∃ y Q(x, y) holds by Definition 2.4 (d). Hence, in thiscase, too,
M |= ∃ y ∀xQ(x, y) → ∀x ∃ y Q(x, y)
holds by Definition 2.4 (c). BecauseM was an arbitrarily chosen relevant modelof α, the formula ∃ y ∀xQ(x, y) → ∀x ∃ y Q(x, y) is valid.
Example 2.4. Show that the statement
∀x ∃ y Q(x, y) → ∃ y ∀xQ(x, y)
is not valid.Consider the formula in the model M = (N, V ) (N = {1, 2, . . .} is the set
of natural numbers) where V (Q) = {(a, b) ∈ N × N | a < b}. Hence, M |=∀x ∃ y Q(x, y), because for all natural number, there exists a natural number b,such that M |= Q(a, b), i.e. a < b. On the other hand, M 2 ∃ y ∀xQ(x, y)because there does not exist a natural number b, such that M |= Q(a, b) for allnatural numbers a, because the set N does not have the greatest element. Hence,M is such a model where our statement is not true. From this it follows that thestatement is not valid.
Consider the satisfaction of an arbitrary formula more closely. Let M =(A, V ) be a model of L and t1, . . . , tn are terms in L.
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Definition 2.8. Let P be a unary predicate symbol. Then a formula P (t) issatisfied in a model M = (A, V ) iff in the interpretation V , t has a value in A,such that it has the property V (P ), i.e. V (t) ∈ V (P ).
Example 2.5. Consider a formula P (x), and let V (P (x)) := x is a father, i.e.V (P (x)) = {x | x is a father. Let A be the set of students. Hence the formulaP (x) is satisfied in the model M = (A, V ) iff V gives x a value a ∈ A, suchthat a is a father, i.e. x is a student who is also a father.
Definition 2.9. Let P be a n-ary predicate symbol, n ≥ 2. A formula P (t1, . . . , tn)is satisfied in a modelM = (A, V ) iff in the interpretation V , the terms t1, . . . , tnhave such values V (ti) ∈ A (i = 1, . . . n) which are in the relation V (P ), i.e.(V (t1), . . . , V (tn)) ∈ V (P ).
Example 2.6. Consider a formula P (xc), and let M = (A, V ) be a model,such that A is the set of soldiers of Finnish army, V (P (x, y)) means that x isthe superior of y, and V (c) is a soldier named Karhunen (we use the individualconstant k to refer him/her), i.e. k ∈ A. Hence, the formula P (x, c) is satisfiablein M iff a ∈ A, such that (a, k) ∈ V (P (x, y)), i.e. the formula P (xc) issatisfiable inM iff a is a superior of k, for example, a is a captain and Karhunenis a sergeant.
If we want, we can use the satisfiability in a model of a formula when deter-mining the truth status of a statement in a given model. However, we have todistinguish the case "a statement to be satisfiable" mentioned in Definition 2.5from the case "a formula to be satisfiable in a model" mentioned in Definitions2.5 and 2.9.
Example 2.7. Show that the formula
∀x (P (x) ∨ ¬P (x))
is valid. Let M = (A, V ) be an arbitrary relevant model. Hence
M |= ∀x (P (x) ∨ ¬P (x)) iff M |= P (a) ∨ ¬P (a) for all a ∈ A.
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This is the case iffM |= P (a) orM |= ¬P (a) for all a ∈ A, i.e. M |= P (a)or M 2 P (a) for all a ∈ A. It is obviously clear that P (a) is either true or falsein any model where a is any element of the domain of the model. From this itfollows that ∀x (P (x) ∨ ¬P (x)) is true in any relevant model and hence valid.
Example 2.8. Show that the formula
∃x (P (x) ∧ ¬P (x))
is a contradiction. Let M = (A, V ) be an arbitrary relevant model. Hence
M 2 ∃x (P (x) ∧ ¬P (x)) iff M |= ¬∃x (P (x) ∧ ¬P (x)) iffM |= ∀x¬(P (x) ∧ ¬P (x)) iff M |= ∀x (¬P (x) ∨ P (x)),
which holds by Example 2.7. Hence the formula ∃x (P (x) ∧ ¬P (x)) is a con-tradiction.
Example 2.9. Show that a formula
∃ x (P (x) ∧ ¬Q(x))
is satisfiable. Consider the model M = (Z, V ), such that
V (P (x)) = {x | x > 2} and V (Q(x)) = {x | x < 10}.
M |= ∃ x (P (x) ∧ ¬Q(x)) iff M |= P (a) ∧ ¬Q(a) for some a ∈ Z. This isthe case iff M |= P (a) and M |= ¬Q(a), i.e. M 2 Q(a). For example, thevalue a = 15 satisfies the last condition, because 15 ∈ V (P ) (i.e. 15 > 2) and15 /∈ V (Q) (i.e. 15 ≮ 10). Hence, the formula is satisfiable.
Example 2.10. Examine, whether the formula
∃ x (P (x) ∧ ¬Q(x))
is refutable. Consider the model M = (Z, V ) where
V (P (x)) = {x | x = 2} and V (Q(x)) = {x | x < 10}.
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The interpretation in the model M is as follows:
∃x (x = 2 ∧ x ≮ 10)
which is false for any integers x ∈ Z. Hence, M 2 ∃ x (P (x) ∧ ¬Q(x)) andthus the formula is refutable.
Example 2.11. Examine, whether the set of formulas
∆ = {∀x(P (x) → Q(x)),∃x(¬P (x) ∧Q(x))}
is satisfiable. We refer to the first formula by (1) and to the second formula by(2). Let M = (A, V ) be a model, such that
A = {1, 2, 3, 4}, V (P (x)) = {x | x > 2}, and V (Q(x)) = {x | x > 1}.
M |= ∀x(P (x) → Q(x)) iff M |= P (a) → Q(a) for all a ∈ A. Clearly,x > 2 ⇒ x > 1 holds for all elements of A. Hence, the formula (1) is true inM.M |= ∃x(¬P (x) ∧Q(x)) iff M |= ¬P (a) ∧Q(a) for some a ∈ A. Clearly,
x ≮ 2 ∧ x > 1 holds for example for 2 ∈ A. Hence, (2) is true in M.Because there exists at least one model, such that the formulas (1) and (2) are
true in it, the set ∆ is satisfiable.
Example 2.12. Examine, whether in the following set of formulas the last for-mula follows logically from the two first formulas:
We refer to these formulas from the first to the third by the respective numbers(1), (2), and (3).
We consider the formulas in the modelM = (A, V ) whereA = {1,−2, 3,−4},V (P (x)) = {x | x is odd}, V (Q(x)) = {x | x < 5}, and V (R(x)) = {x |x is positive}.
(1) M |= ∀x(P (x) → Q(x)) iff M |= P (a) → Q(a) for all a ∈ A. All theodd numbers are less that 5 in A. Hence the formula (1) is true in M.
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(2) M |= ∀x(P (x) → R(x)) iff M |= P (a) → R(a) for all a ∈ A. All theodd numbers are positive in A. Hence the formula (2) is true in M.
(3) M |= ∀x(Q(x) → R(x)) iff M |= Q(a) → R(a) for all a ∈ A. Becausethe elements of A which are less than 5, are not positive, the formula (3) isnot true in M.
Because there exists at least one model of formulas (1) and (2), where the for-mula (3) is not true, the formula (3) is not a logical consequence of the formulas(1) and (2).
3 Axiomatization and Proof Theory
In predicate logic, the concepts deduction, proof, and theorem are defined in thesimilar way as in propositional logic.
3.1 Axiom Schemes
Axiom schemes of predicate logic are created by extending those of proposi-tional logic. Hence, we add to the axioms of propositional logic additionalaxioms concerning quantification. Also, we add to the set of inference rules arule concerning the use of universal quantifier. The meta-variables appearing inthe axioms coming from propositional logic refer to formulas of predicate logic.
Definition 3.1 (Axioms). Let α, β, and γ be formulas of predicate logic, then thefollowing formulas are the axiom schemes of predicate logic:
A1 α→ (β → α),
A2 (α→ (β → γ)) → ((α→ β) → (α→ γ)),
A3 (¬β → ¬α) → (α→ β),
A4 ∀xα→ Sxaα,
A5 ∀x(α→ β) → (α→ ∀x β) if x is not free in α.
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Definition 3.2 (Inference Rules). The inference rules of predicate logic are asfollows:
R1 Modus ponens (MP):
αα→ β
β
R2 Universal Generalization (UG):
α
∀xα,
if x does not appear as a free variable in any premises.
According to the rule Universal Generalization, since x becomes bound inthe process, we say that the universal generalization is over x, or that one gen-eralizes over x. We will justify universal generalization later. At the moment, itmust be pointed out that universal generalization is subject to restrictions, If onegeneralizes over x, then x must not appear in any premise as free variable, If xdoes appear free in any premise, then x always refers to the same individual, andit is fixed in this sense. For example, if P (x) appears in a premise, then P (x) isonly true for x and not necessarily true for any other individual. If x is fixed, onecannot generalize over x. Generalizations from one particular individual towardthe entire population are unsound. If, on the other hand, x does not appear in anypremise or if x is bound in all premises, then x is assumed to stand for everyone,and universal generalization may be applied without restriction.
Example 3.1. To demonstrate universal generalization, consider the followingproblem whose domain consists of a group of computer science students. Ofcourse, all computer science students like programming. The derivation mustprove that everyone in the domain likes programming. If P (x) and Q(x) standfor "x is a computer science student" and "x likes programming", respectively,the premises become
∀xP (x), ∀x (P (x) → Q(x)).
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The desired conclusion is∀xQ(x).
Hence, we have to prove that ∀xP (x),∀x (P (x) → Q(x)) ` ∀xQ(x). Thededuction is as follows:
Formula Rule Comment1. ∀xP (x) premise Everyone is a CS major.2. ∀x (P (x) → Q(x)) premise CS majors like programming.3. P (x) Sx
x , 1 x is a CS major.4. P (x) → Q(x) Sx
x , 2 If x is a CS major, x likes programming.5. Q(x) MP, 3,4 x likes programming.6. ∀xQ(x) UG, 5 Everyone likes programming.
In the proof,Q(x) is derived in line 5, which means that x likes programming.This statement is then generalized to ∀xQ(x). This generalization is only pos-sible because all instances of x in the premises are bound. If the premise ∀xP (x)is replaced by P (x), then universal instantiation over x is no longer sound. Thisis the case because x is fixed and universal generalization over fixed variables isunsound.
Example 3.2. As a second example, we derive ∀y ∀xP (x, y) from ∀x ∀y P (x, y).We have the deduction:
Formula Rule Comment1. ∀x ∀y P (x, y) premise2. ∀y P (x, y) Sx
x Drop the first quantifier.3. P (x, y) Sy
y Drop the second quantifier.4. ∀xP (x, y) UG, 3 This is a sound generalization.5. ∀y ∀xP (x, y) UG, 4 Generalize again to obtain the desired conclusion.
On the line 3, we drop the second quantifier to obtain an expression withoutquantifiers. Then we use UG to add the quantifiers back in reverse order. On theline 4, it is sound to generalize, because the premise does not contain x as a freevariable. All occurrences of x in the premises are bound. The situation is similaron the line 5.
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Example 3.3. In the third example of UG, we show that the variable x in auniversal quantifier may be changed to the variable y i.e. we prove ∀xP (x) `∀y P (y):
Formula Rule Comment1. ∀xP (x) premise2. P (y) Sx
y , 1 Instantiate the premise for y.3. ∀y P (y) UG, 2 Generalize P (y) to obtain the conclusion.
In fact, when we use the argument of instantiation Sxt in deductions, we use
the axiom A4 in the following way:
Formula Rule Comment1. ∀xP (x) premise Premise, (or deduced from earlier steps).2. ∀xP (x) → P (t) A4 A4 is used in its original meaning.3. P (t) MP, 1,2 This is the result of Sx
t applied to 1.
Hence, we have proved that ∀xP (x) ` P (t). This justifies the use of instan-tiation in that form used in the previous inferences. Thus, we have a derived therule of universal specification (US):
∀xα(x)
Sxt α(t)
Example 3.4. Consider an additional example about the rule US. We prove thefollowing inference to be correct:
Healthy people live long.Socrates was healthy.Socrates lived long.
To do the derivation, let H(x) indicate that x is a healthy people, L(x) x lives(or lived) long, and s stands for Socrates. We have the deduction:
3.2 Deduction Theorem and Universal Generalization
In the deduction theorem, one assumes B, proves C, using the assumption Blike a premise, and concludes, that B → C. Once this is done, B is discharged.The question now is how to treat free variables occurring in B. First, while B isused as an assumption, that is, as long as B is not discharged, B has to be treatedlike any other premise. In particular, should B contain x as a free variable, thenone must not generalize over x. However, as soon as B is discharged, this is nolonger true. Once B is discharged, it has no effect whatsoever on the status ofany variable. Hence, if x is not free in any other premise, one can universallygeneralize over x even if x appears free in B.
The deduction theorem is now demonstrated by an example.
Example 3.5. Let S(x) stand for "x studied" and P (x) stands for "x passed".The premise is that everyone who studied passed. Prove that everyone who didnot pass did not study.
Solution: The premise "everyone who studied passed" can be translated as∀x(S(x) → P (x)), and the statement "everyone who did not pass did not study"becomes ∀x(¬P (x) → ¬S(x)).
Formula Rule Comment1. ∀x(S(x) → P (x)) premise Everyone who studied passed.2. S(x) → P (x) US, 1 If x studied, x passed.3. ¬P (x) assumption Assume that x did not pass.4. ¬S(x) MT, 2,3 x cannot have studied.5. ¬P (x) → ¬S(x) DT, 3,4 Apply DT and discharge ¬P (x).6. ∀x(¬P (x) → ¬S(x)) UG, 5 Anyone who did not pass cannot have studied.
To obtain the result, the assumption ¬P (x) is introduced in line 3. As longas this assumption is not discharged, no generalization over x is allowed. Toindicate that an assumption is in effect, lines 3 and 4 are indented. However, oncethe deduction theorem is applied, the indentation is removed and the assumption¬P (x) is discharged, and one can generalize over x. This is done in line 6. In allother aspects, the proof is self-documenting.∀x(S(x) → P (x)) is a premise, and ∀x(¬P (x) → ¬S(x)) follows. To arrive
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at the desired conclusion, one uses universal generalization. This can be donebecause x is not free in any premise.
3.3 Dropping the Universal Quantifiers
In mathematics, universal quantifiers are frequently omitted. For example, in thestatement x+ y = y+ x, both x and y are implicitly universally quantified. Thiscauses problems when such statements are used as premises because, accordingto our rules, any variable appearing free in a premise is foxed in the sense thatthroughout the proof it is bound to one and the same individual. To get aroundthis difficulty, we single out certain variables in the premises and explicitly statethat these variables are not fixed. All variables that are not fixed will be calledtrue variables. A variable may be universally generalized if and only if it is atrue variable. If a variable appears in a premise, then it is assumed to be fixed,unless it is explicitly stated that the variable is a true variable.
By using true variables, one can omit many universal quantifiers and this, inturn, simplifies proofs. Moreover, we allow from now on that any true variablecan be instantiated to any term. The same effect can, of course, be achieved byusing universal generalization first, following by universal instantiation. How-ever, direct instantiation is shorter and often clearer.
Until now, instantiations were always represented by the symbols S, such asSx
y . From now on, we will frequently make use of the notation x := y to indicatethat x is replaced by y. In some programming languages, the notation := means"assign to", which is the same as "instantiate to".
Example 3.6. Let P (x, y, z) : x + y = z. Given the premises P (x, 0, x)and P (x, y, z) → P (y, x, z), where x, y, and z are true variables, prove that0 + x = x, i.e., prove P (0, x, x).
Solution: The following derivation is used to prove P (0, x, x). Note that thefirst two lines are premises and that x, y, and z are explicitly declared as truevariables.
We have the deduction:
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1. P (x, y, z) → P (y, x, z) Premise: x+ y = z ⇒ y + x = z, x, y, and z are true.2. P (x, 0, x) Premise: x+ 0 = x, x is a true variable.3. P (x, 0, x) → P (0, x, x) Line 1 with x := x, y := 0, z := x.4. P (0, x, x) MP, 2,3; 0 + x = x.
All true variables are strictly local to the line on which they appear. Hence,if the true variable x appears on two different lines, then these two instances ofx are really two different variables. For example, in the proof of Example 3.6,x in line 1 and x in line 2 are two different variables. When doing the proof,one obviously has to establish some type of connection between the variables,and this connection is through instantiation. Of course, instantiations must not bemade blindly. Instead, one has to do the instantiations in such a way that progressis made toward the desired conclusion. How this is done in detail depends on thegeneral strategy, and in each proof, some type of strategy should be followed.However, there are some general principles that are helpful, and one of them isunification.
Definition 3.3. Two expressions are said to unify if there are legal instantiationsthat make the expressions in question identical. The act of unifying is calledunification. The instantiation that unifies the expressions in question is called aunifier.
Example 3.7. Suppose the expressions Q(a, y, z) and Q(y, b, c) are expressionsappearing on different lines. Show that the two expressions unify, and give aunifier. Here, a, b, and c are fixed, and y and z are true variables.
Solution: Since y in Q(a, y, z) is a different variable than y in Q(y, b, c),rename y in the second expression to become y1. This means that one must unifyQ(a, y, z) withQ(y1, b, c). An instance ofQ(a, y, z) isQ(a, b, c), and an instanceof Q(y1, b, c) is Q(a, b, c). Since these two instances are identical, Q(a, y, z) andQ(y, b, c) unify. The unifier is a = y1, b = y, c = z.
There may be several unifiers. For example, if a and b are constants, thenR(a, x) and R(y, z) have the unifier y = a, z = x, which yields the commoninstance R(a, x). However, there is also the unifier y = a, x = b, z = b, which
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yields the common instance R(a, b). However, R(a, b) is an instance of R(a, x),and the unifier y = a, x = b, z = b is in this sense less general than the unifiery = a, z = x. Of course, we always want to find the most general unifier, if oneexists.
The solution of Example 3.6 involved unification. Specifically, to make useof the modus ponens, line 2 was unified with the antecedent of line 1. Generally,unification is performed in such a way that some rule of inference can be appliedafter unification.
Example 3.8. Clearly, if x is the mother of y and if z is the sister of x, then z isthe aunt of y. Suppose now that Brent’s mother is Jane and Liza is Jane’s sister.Prove that Liza is Brent’s aunt.
Solution: If "mother(x, y)" is the predicate that is true if x is the mother ofy, and if "sister(x, y)" and "aunt(x, y)" are defined in a similar fashion, one canstate the premises as follows:
The conclusion that Liza is Brent’s aunt now follows from 4 and 5 by modusponens.
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3.4 Rules for Existential quantifiers
Existential Generalization
If there is any term t for which P (t) holds, then one can conclude that some xsatisfies P (x). Hence, P (t) logically implies that ∃xP (x). More generally, ∃xαcan be derived from Sx
t α, where t is any term. This leads to the following ruleof inference, existential generalization (EG):
Sxt α
∃xα
Example 3.9. Let c be Aunt Cornelia, and let P (x) stand for "x is over 100 yearsold". Then one has
P (c)
∃xP (x)
The reason is that if one replaces x by c in P (x) then one finds P (c).
The following example demonstrates demonstrates how to use existential gen-eralization within a formal proof. The premises of our derivation are
1. Everybody who has won a million is rich.
2. Mary has won a million.
We want to show that these two statements logically imply that
3. There is somebody who is rich.
If somebody were asked to demonstrate that the conclusion follows from thepremises, he or she would probably argue as follows. If everybody who wins amillion euros is rich, then Mary is rich if she wins a million. Since we know thatMary has won a million, we apply modus ponens and conclude that Mary is rich.There is thus somebody, Mary, who is rich. This argument is now formalized.W (x) means that x has won a million, R(x) means that x is rich, and m standsfor Mary. Hence, we prove that ∀x (W (x) → R(x)), W (m) ` ∃xR(x). Thededuction is as follows:
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1. ∀x (W (x) → R(x)) premise2. W (m) → R(m) US, 1, x := m3. W (m) premise4. R(m) MP, 2,35. ∃xR(x) EG, 4
It was stated earlier that ¬∃xP (x) is logically equivalent to ∀x¬P (x). Wenow prove the first half of this statement by showing that ¬∃xP (x) ` ∀x¬P (x).The deduction is as follows:
We will later prove ∀x¬P (x) ` ¬∃xP (x) The two proofs together establishthe logical equivalence of ¬∃xP (x) and ∀x¬P (x).
Existential Spesification
If ∃xα is true, then there must be some term t that satisfies α; that is, Sxt α must
be true for some t. For example, if P (x) stands for "x does somersaults", then∃xP (x) means that Sx
t P (x) = P (t) must be true for some t. The problem is thatwe do not know whether it is Aunt Eulalia, Uncle Petronius or even somebodyelse who makes somersaults. In a proof, the question must therefore be kept asto who the individual is who makes somersaults. To do this, a new variable, sayb, is selected to denote this unknown individual. This leads to the following ruleof inference, existential specification (ES):
∃xαSx
t α
The variable introduced by existential instantiation must not have appearedearlier as a free variable. For example, when applying ES to the two statements
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"There exists someone who is over 100 years old" and "There exists someonewho makes somersaults", one must not use the same variable b for existentialinstantiation in both cases. Otherwise, one could conclude that b is both over100 and makes somersaults, which certainly does not follow logically. Similarly,one cannot use any variable that appears free in any of the premises. Hence,ES must not introduce any variable that has appeared already as a free variablein the derivation. Moreover, the variable introduced is fixed in the sense thatone cannot use universal generalization over this variable. For example, if bmakes somersaults, then one cannot use UG to conclude that everyone makessomersaults. Moreover, a variable with an unknown value must not appear inthe conclusion, and since any variable introduced ny ES is unknown, it must notappear in the conclusion either.
For the purpose of demonstration, suppose that there is someone who won amillion euros, and we want to prove that there is someone who is rich. Hence,the premises are
1. Someone has won a million euros.
2. Everybody who has won a million is rich.
We want to show that these two statements logically imply
3. There is somebody who is rich.
Hence, we have to prove ∀x (W (x) → R(x)),∃xW (x) ` ∃xR(x). The deduc-tion is as follows:
1. ∃xW (x) premise2. W (b) ES, 1, x := b3. ∀x (W (x) → R(x)) premise4. W (b) → R(b) US, 3, x := b5. R(b) MP, 2,46. ∃xR(x) EG, 5
In this proof, existential specification is used on line 2, where the winner iscalled b. Once this is obtained, the second premise is given on line 3, and thispremise is instantiated with x := b on line 4. Note that one must not derivelines 3 and 4 before lines 1 and 2; that is, one must not apply the universal
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specification before the existential specification. The reason for this is that, onceW (b) → R(b) is obtained, b is no longer a new variable, and it therefore mustnot be used for existential specification. For this reason, it is generally a goodidea to apply existential specification first.
As a second example of ES, we prove ∀x¬P (x) ` ¬∃xP (x). The deductionis as follows:
1. ∃xP (x) assumption2. P (b) ES, 1, x := b3. ∀x¬P (x) premise4. ¬P (b) US, 3, x := b5. P (b) ∧ ¬P (b) CI, 3,46. ¬∃xP (x) RAA, 1, 5
We used indirect proof in this derivation. Since we want to derive ¬∃xP (x),the assumption to be rejected is ∃xP (x). Existential specification allows us toderive P (b) on line 2, which contradicts ¬P (b) derived on line 4 by using uni-versal specification. The resulting contradiction is given on line 5. This lineallows one to reject the assumption ∃xP (x); that is, ¬∃xP (x) must be true.Hence, ∀x¬P (x) implies logically¬∃xP (x), which provides the promised secondhalf of the proof of logical equivalence of these formulas.
3.5 Natural Inference Rules in Predicate Logic
The important basis for natural inference which supports human intuitive deduc-tion is the set of natural inference rules for propositional logic, i.e. Suppes-Genzen type inference rules. To this set, we add rules concerning quantifiers,logical equivalence, and identity. We collect here together all the needed infer-ence rules concerning these things.
Rule of Universal Specification (US): Let ϕ(x) be a formula where x is free,and ϕ(t) = Sx
t ϕ(x), then∀xϕ(x) ` ϕ(t).
Rule of Universal Generalization (HG): Let ϕ(t) be a formula where t is an
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arbitrary term, and ϕ(x) = Stxϕ(t), then
ϕ(t) ` ∀xϕ(x)
iff x does not occur in ϕ(t), and t does not occur in the premises, on which theconclusion ∀xϕ(x) depends.
Rule of Existential Specification (ES): Let t be a term, x is free in ϕ(x), andϕ(t) = Sx
t ϕ(x), then∃xϕ(x) ` ϕ(t).
Remark:
• In a deduction, the rule ES must be applied before applying the rule US, ifpossible.
• Applying ES, t must not occur in the same deduction in earlier steps.
• t must not occur in the conclusion of the deduction.
Rule of Existential Generalization (EG): Let ϕ(t) be a formula where t is aterm, and ϕ(x) = St
xϕ(t), then
ϕ(t) ` ∃xϕ(x).
Quantifier Exchange Rule (Q1): ∀x can be replaced with ¬∃x¬.
Quantifier Exchange Rule (Q2): ∃x can be replaced with ¬∀x¬.
Rule of Logical Equivalence (RE): If a formula ψ1 occurs as a part of a for-mula ϕ1, and if ψ1 ≡ ψ2 (i.e. ψ1 and ψ2 are logically equivalent), then
ϕ1
ψ1 ≡ ψ2
ϕ2
where ϕ2 is created by substituting ψ2 on the place of ψ1 in ϕ1.
Rule of Identity (I): If ϕ(t1) is a formula, and St1t2 = ϕ(t2), where t1 are terms,
then
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ϕ(t1)t1 = t2ϕ(t2)
Example 3.10. Find a deduction from the premises to the conclusion in the fol-lowing inference:
No ducks are willing to waltz. No officers are unwilling to waltz. All mypoultry are ducks. Therefore, none of my poultry are officers.
For formalizing the sentences, D(x) stands for "x is a duck", W (x) standsfor "x is willing to waltz", O(x) stands for "x is an officer", and P (x) stands for"x’s are my poultry". We have the following deduction:
As in the case of propositional logic, one can use logical equivalences to manipu-late logical expressions. Specifically, ifA is a logical expression, one can replaceany subexpression B of A with a subexpression C, as long as B and C are logi-cally equivalent.
We give a collection of useful logical equivalents in the following list.
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1. ∀xA ≡ A if x not free in A1d. ∃xA ≡ A if x not free in A2. ∀xA ≡ ∀ySx
yA if y not free in A2d. ∃xA ≡ ∃ySx
yA if y not free in A3. ∀xA ≡ Sx
t A ∧ ∀xA for any term t3d. ∃xA ≡ Sx
t A ∨ ∃xA for any term t4. ∀x(A ∨B) ≡ A ∨ ∀xB if x not free in A4d. ∃x(A ∧B) ≡ A ∧ ∃xB if x not free in A5. ∀x(A ∧B) ≡ ∀xA ∧ ∀xB5d. ∃x(A ∨B) ≡ ∃xA ∨ ∃xB6. ∀x∀yA ≡ ∀y∀xA6d. ∃x∃yA ≡ ∃y∃xA7. ¬∃xA ≡ ∀x¬A7d. ¬∀xA ≡ ∃x¬A
Example 3.11. In the expression ∀xP (x)∨∀xQ(x), the bound variable x appearsunder two different scopes. By using law 2, one can change the x of the seconduniversal quantifier to y, which yields
∀xP (x) ∨ ∀xQ(x) ≡ ∀xP (x) ∨ ∀yQ(y).
Example 3.12. In the expression P (x) ∧ ∃xQ(x), the variable x appears firstfree and then bound. By using law 2d, one can write this as P (x) ∧ ∃yQ(y).
Examples 3.11 and 3.12 show how to standardize the variables apart.
Definition 3.4. Renaming the variables in an expression such that distinct vari-ables have distinct names is called standardizing the variables apart.
Example 3.13. Standardize all variables apart in the following expression:
Some operations cannot be done for expressions containing negated quan-tifiers. To remove negated quantifiers, laws 7 and 7d are useful, as the nextexample demonstrates.
Example 3.14. Apply laws 7 and 7d to remote all negations in front of the quan-tifiers of the following expression:
¬∀z(∃xP (x, z) ∧ ¬∀xQ(x, z))
We have
¬∀z(∃xP (x, z) ∧ ¬∀xQ(x, z))
≡ ∃z¬(∃xP (x, z) ∧ ¬∀xQ(x, z)) Law7d
≡ ∃z(¬∃xP (x, z) ∨ ∀xQ(x, z)) DeMorgan
≡ ∃z(∀x¬P (x, z) ∨ ∀xQ(x, z)) Law7
3.6.2 Other Important Equivalences
Universal quantifiers cannot be dropped unless they are at the beginning of theexpression. In this respect, the following equivalence is of interest.
∀xP (x) ∨ ∀yQ(y) ≡ ∀x∀y(P (x) ∨Q(y)).
To prove this law, rewrite law 4 as
(∀xB) ∨ A ≡ ∀x(B ∨ A). (3.1)
Now we have
∀xP (x) ∨ ∀yQ(y) ≡ ∀x(P (x) ∨ ∀yQ(y)) (3.1) with A := ∀y(Qy)≡ ∀x∀y(P (x) ∨Q(y)) Law 4 with A := P (x)
Note that the condition that A must not contain the bound variable of thequantifier in question applies. ∀yQ(y) does not contain the bound variable x.Moreover, when A := P (x), the bound variable is y, and P (x) does not containy.
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Consider the statement "If somebody talks, it will be in the news tomorrow".If C(x) stands for "x talks" and ifQ stands for "it will be in the news tomorrow",one can translate this sentence as
∃xC(x) → Q. (3.2)
Another translation, which is not obvious, is
∀x(C(x) → Q). (3.3)
Both expressions are logically equivalent, as we will prove next. The first versionis the version normally used in natural language. However, for logical deriva-tions, the second version is preferable. Unfortunately, a verbal translation of thesecond version is difficult. One would have to say something like this: for eachx, it is true that if x talks then it will be in the news tomorrow.
The following law shows that the expressions given by (3.2) and (3.3) areindeed logically equivalent:
∀x(B → A) ≡ ∃xB → A. (3.4)
The proof of this equivalency is as follows:
∀x(¬B ∨ A) ≡ (∀x¬B) ∨ A ≡ ¬∃xB ∨ A ≡ ∃xB → A
Note again that A must not contain x.
Example 3.15. Clearly, if x < y and y < z, then x < z. If G(x, y) means thatx < y, then this translates into G(x, y) ∧ G(y, z) → G(x, z). This obviouslyholds for all x, y, and z; that is
∀x∀y∀z(G(x, y) ∧G(y, z) → G(x, z)). (3.5)
On the other hand, one can say that x < z if there is a y such that x < y andy < z. This can be expressed as follows:
∀x∀z(∃y(G(x, y) ∧G(y, z)) → G(x, z)). (3.6)
Are these two expressions logically equivalent?
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We solve this question as follows. Interchange the second and third quantifierin (3.5) to obtain
∀x∀z(∀y(G(x, y) ∧G(y, z) → G(x, z)))
The innermost quantifier is an instance of the left of (3.4), which implies that
∀x∀z(∃y(G(x, y) ∧G(y, z)) → G(x, z)).
This shows that (3.5) and (3.6) are logically equivalent:
Equality is essential for doing algebraic operations. In fact, algebraic opera-tions are used to convert a given expression into some other expression that iseither simpler or otherwise more suitable for the purpose at hand. Equality isalso important in logic, in particular for indicating that there is only one ele-ment satisfying a certain property. For doing algebraic operations, one needs analgebra. Generally, an algebra is given by domain, such as the real numbers,together with some operations, such as + and ×. As it turns out, operationsare really functions, and to understand operations, one must know about func-tions. Generally, functions have values or images, and these images depend ontheir arguments. All arguments must be individuals, that is, they must be partof the universe, and so are their images. A function establishes in this sense arelation between individuals. What makes this relation special is that the imageof a function is always unique. The fact that the image of a function is alwaysuniquely identifiable individual makes the image a term. In other words, one canuse a function wherever one can use a constant or a variable. Moreover, withoutrequiring uniqueness, algebraic manipulations would be rather restricted, as willbe shown. Although the standard algebra, that is, the algebra in the domain ofthe reals using operations + and × (or ·), is by far the most important algebra,there are other algebras.
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3.7.2 Equality
Two elements t and r are equal if they refer to the same individual, and to expressthis, one writes t = r. Equality is really a predicate, and Eq(t, r) could be usedinstead to express that t and r are equal. However, the normal equal sign is moreconvenient, and it will therefore be used here. Hence, t = r is an atomic formula,which can be combined with other atomic formulas in the usual fashion, as in
(x = y) ∧ (y = z), or ¬(x = y).
Especially, instead of using the form ¬(x = y), the abbreviation x 6= y is gener-ally used.
Consider the ways to define equality between two objects. We have followingpossibilities:
1. Explicit assignment: objects that are equal must be enumerated. This methodis obviously restricted to finite domains.
2. Providing rules to determine whether or not two objects are equal.
3. To postulate a number of axioms that the equality predicate must satisfy.
We use the last alternative.
Obviously, every term t is equal to itself, which leads to the following axiom:
∀x(x = x). (3.8)
This axiom is called the reflexivity axiom. A predicate G(x, y) is said to bereflexive if G(x, x) is true for all x. Another important property of the equalityis the substitution property. If t and r, t = r, are two terms, the substitutionproperty allows one to substitute t in any expression by r. For example, if t = r,and P (t) is true, then P (r) must also be true. This gives rise to a rule of infer-ence. To formulate this rule, we first introduce a symbol for doing replacements.Specifically, if A is any expression, then R(n)t
tA is the expression one obtainsfrom A by replacing the nth instance of term t by r. If t occurs fewer that ntimes in A, then R(n)t
rA = A.
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Example 3.16. Determine R(1)xy(x = x), R(2)x
y(x = x), and R(3)xy(x = x).
Since the first instance of x is at the left of the equal sign, R(1)xy(x = x) is
y = x. Similarly, R(2)xy(x = x) is x = y. Finally, R(3)x
y(x = x) is x = x.
We now have the following rule:
Substitution Rule: IfA and t = r are two expressions that have been derived,one is allowed to conclude that R(n)t
rA for any n > 0. In this case, we say thatwe substitute t from t = r into A.
Example 3.17. Substitute x+ 1 from x+ 1 = 2y into x < x+ 1.According to the substitution rule, x + 1 is replaced by 2y, which yields
x < 2y. This is the same as R(1)x+12y (x < x+ 1).
Example 3.18. show that x = y substituted into x = x yields y = x.Replace the first x in x = x by y, which yields R(1)x
y(x = x) = (y = x).
The equality predicate is symmetric; that is, x = y implies that y = x. Thisis easily proved. We assume x = y, and we prove that y = x. This is done asfollows:
1. ∀x(x = x) reflexivity axiom2. x = y assumption3. x = x instantiate line 1 with x := x4. y = x substitute line 2 into line 35. x = y → y = x deduction theorem, 2, 4
The proof starts with line 3, which instantiates the reflexivity axiom given online 1. Specifically, the bound variable x of line 1 is replaced by free variable x,and this is indicated by writing x := x. Line 3 is now obtained by substituting thefirst x by y. According to the deduction theorem, one concludes that x = y →y = x. This completes the proof. The resulting expression can be universallyquantified, which yields
∀x∀y((x = y) → (y = x)).
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Consequently, given t = u, not only can one replace t by u, but one canalso replace u by t. (To be completely rigorous, one would have to add thecorresponding steps 2, 3, and 4 of the preceding derivation. For simplicity, weomit these steps.)
Next we show that x = y and y = z imply that x = z. This property is calledthe transitivity property of equality. To show that this is the case, x = y andy = z are used as premises. This allows one to derive x = z as follows:
1. x = y premise2. y = z premise3. x = z substitute line 2 into line 1.
Hence,(x = y) ∧ (y = z) → (x = z).
Since this is valid, one can generalize to obtain
∀x∀y∀z((x = y) ∧ (y = z) → (x = z)).
Substitutions of the type used in the proofs for reflexivity and transitivity arefrequent. We therefore add the following example.
Example 3.19. After each iteration in a certain loop, the condition s > 3i is true.Moreover, once the loop terminates, i = 10. Prove that when the loop terminatess > 30.
When the loop terminates, both s > 3i and i = 10 hold. By using theseconditions as premises, one easily finds s > 30 by replacing i in s > 3i by 10,which completes the proof. For the sake of completeness, we give the formalproof:
1. s > 3i premise2. i = 10 premise3. s > 3 · 10 substitute line 2 into line 1.
The substitution rule can be applied to subexpressions embedded in otherexpressions. No other rule of inference discussed so far can do this. This means
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that the substitution rule is very efficient. Whereas the application of all otherrules requires that the higher layers of a formula must first be peeled away toallow access to the target subexpressions, the substitution rule can be applieddirectly. This saves many steps in the derivation. The substitution rule can beapplied directly. This saves many steps in the derivation. The substitution rule istherefore used extensively in mathematical derivations.
3.7.3 Equality and Uniqueness
Consider the statement "The lion is a mammal". Examine, whether this statementcan be expressed as
lion = mammal.
The answer to this question is no. To see why, add the statement
bear = mammal.
By substituting the first statement into the second, one obtains
bear = lion,
and this is obviously false. This shows that the word "is" cannot always betranslated to ’=’. Generally, the equal sign cannot be used if the left side of theexpression can refer to different objects. If x1 = y and x2 = y, then one canalways conclude that x1 = x2; that is, x1 and x2 must also be the same. Therecan be only one x for each y such that x = y. Equality necessitates uniqueness.
Conversely, to express uniqueness, one uses equality. The statement that indi-vidual a is the only element with the property P can be reworded as "if x is nota, then P (x) cannot be true". This translates into
∀x(¬(x = a) → (x = a)).
Example 3.20. Translate "only a could have forgotten the meeting" into logic.Let P (x) be "x forgot". Then one has
∀x(P (x) → x = a).
In words, "If somebody forgot, then it must have been a".
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The fact that there is one, but only one individual with the property P cannow be expressed as
∃x(P (x) ∧ ∀y(P (y) → (y = x))).
This expresses the fact that there is an x that makes P (x) true, and P (y) is trueonly if y = x. To avoid having to write such a lengthy expression, one defines∃1xP (x) to indicate that only one element satisfies P . In other words, one has
∃1xP (x) ≡ ∃x(P (x) ∧ ∀y(P (y) → y = x)). (3.9)
Example 3.21. Translate the statement "The company has exactly one CEO"into logic, both with and without using the quantifier ∃1.
Let CEO(x) express the fact that x is CEO. Then one has
∃1xCEO(x) ≡ ∃x(CEO(x) ∧ ∀y(CEO(y) → y = x)).
Example 3.22. Let C(x), S(x), and Q(x) indicate that x is a city, x is a capital,and x is a country, respectively. Assume that the universe of discourse is the setof all cities and the set of all countries. Express the statement "All countries haveexactly one capital" in terms of logic.
By using the quantifier ∃1, this statement translates into
∀x(Q(x) → ∃1y(S(y) ∧ C(y))).
To write this expression without the symbol ∃1, one uses (3.9), except that allvariables must be renamed properly. Moreover, care must be taken that the vari-ables do not clash. In the case considered here, one finds that
∀x(Q(x) ∧ ∃y(S(y) ∧ ∀z(S(z) ∧ C(z) → z = y))).
There is a second way to express uniqueness. Clearly, if P (x) and P (y)always imply that x = y, then there can be at most one x such that P (x) is true.If, in addition to this, there is an element with property P , then this element isunique. Consequently,
∃1xP (x) ≡ ∃xP (x) ∧ ∀x∀y(P (x) ∧ P (y) → x = y). (3.10)
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This method of expressing uniqueness is logically equivalent to the one given by(3.9). In fact, one can derive (3.9) from (3.10), and one can derive (3.10) from(3.9).
Example 3.23. Use (3.10) to express "There is exactly one carpenter in the vil-lage".
If C(x) stands for x is carpenter, and if there is at most one carpenter, thenC(x)∧C(y) implies x = y for all possible x and y. In logic, this can be expressedas ∀x∀y(C(x)∧C(y) → x = y). If there is exactly one carpenter, one thereforehas
∃xC(x) ∧ ∀x∀y(C(x) ∧ C(y) → x = y).
3.7.4 Functions and Equational Logic
Functions are extremely important for doing equational logic. To define a func-tion, it must first be given a name, such as f . One now has the following:
Definition 3.5. A function f with one argument associates with each individualx a unique individual y, which is referred to as f(x). The value y = f(x) iscalled the image of x.
The fact that each function has a unique image is essential. The reason is thatfunctions are used as terms in logic, and fallacies arise if a term can refer to twoor more different individuals.
Definition 3.6. A function with n arguments is said to have an arity of n. Thearity of a function is fixed. A function f with an arity of n associates with eachlist of n individuals x1, x2, . . . , xn a unique individual y = f(x1, x2, . . . , xn).The individual y is said to be the image of x1, x2, . . . , xn.
The fact that in a function f each x has a unique image y is crucial. Withoutthis condition, one cannot even write y = f(x) without risking fallacies. Todemonstrate this, consider the following example, which pretends to show that1 = −1.
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Example 3.24. Let y = f(x) if x = y2 or, alternatively, f(x) = ±√y. By usingf(x) as a function, one can consider a "proof" that 1 = −1. Find the fallacy inthe argument:
1. 1 = 1 reflexivity of equality2. 1 = f(1) 1 = ±
√1
3. −1 = f(1) −1 = ±√
14. 1 = −1 substitute f(1) in line 2 by line 3.
Clearly, f(x) is not a function, because for each x there are two values forf(x), that is, +
√x and−
√x. Consequently, one is not allowed to write 1 = f(1)
or −1 = f(1). When ignoring this fact, fallacies like the preceding can beobtained.
The definition of a function implies that there is an image for every argumentor, if the function has n arguments, for every possible n-tuple of arguments. Thisis essential in logic because, otherwise, predicates that use functions as termsmay be undefined. Outside logic, it is sometimes convenient to remove thisrestriction. For example, technically, 1/x is not a function of x in the domain ofthe real numbers, yet 1/x shares many properties with functions. One thereforedefines the following:
Definition 3.7. A partial function f of arity 1 associates with each individual xat most one individual y, which, if it exists, is referred to as f(x). Similarly, apartial function g of arity n associates with each n-tuple of individuals x1, x2,. . . , xn at most one individual g(x1, x2, . . . , xn).
Note that every function is a partial function, but not every partial functionis a function. For example, on the domain of integers, f(x) = x/2 is a partialfunction. However, since one cannot associate any individual with f(x) if x isodd, f(x) = x/2 is not a function, according to Definition 3.5.
Definition 3.8. A partial function that is not a function is called a strict partialfunction.
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The definitions of the terms function and partial function are not uniform.Some authors include partial functions among the functions. In this case, theyuse total function for the term function as defined in Definition 3.5. For clarity,we will frequently use the phrase "total function", even though according to ourdefinition, the word "function" alone would suffice.
All floating-point operations on computers that may lead to exponent under-flow and overflow are strict partial functions. They are partial functions becausefor each set of operands, the result, if it exists, will ne unique. However, in caseof overflow or underflow, there is no result, and the operation is therefore not atotal function. Division on the integers is not a total function either. To be total,all divisions have to be defined. However, a division by zero is undefined.
3.7.5 Function Compositions
If f is a function with one variable, then z = f(y) is the image of y under thefunction f . Since each individual y has an image, one can set y = g(x), whereg is some function with one variable. This construct associates with each x aunique value z, where z = f(g(x)).
Definition 3.9. Let f and g be two functions with one argument. The functionthat associates with each x the value f(g(x)) is called the composition of f andg, and it is written f ◦ g. it follows that (f ◦ g)(x) = f(g(x)).
We will occasionally use the word composition for the composition of partialfunctions. Hence, if f and g are two partial functions, f ◦g is the partial functionthat associates the individual f(g(x)) with x, provided such an individual exists.
Example 3.25. Let m be the function that associates with each x his or hermother, and let f be the function that associates with each x his or her father.Then f(m(x)) is the father of the mother, f(f(x)) is the father of the father,m(f(x)) is the mother of the father, and m(m(x)) is the mother of the motherof x. Consequently, f ◦m is the function that associates with each individual itsmaternal grandfather, f ◦ f is the function that associates with each individualits paternal grandfather, and so on. Note, that f ◦m and m◦ f are distinct. f ◦mis the maternal grandfather function, whereas m ◦ f is the paternal grandmother
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function.
One can also form compositions involving functions with several variables,as shown by the next example.
Example 3.26. Let s(x, y) be the function that associates with each pair (x, y)the sum x+ y, and let p(x, y) be similarly the function that associates with eachpair (x, y) the product x·y. Then p(s(x, y), z), s(z, p(x, y)), and s(s(z, x), y) areall function compositions: they all associate with each triple (x, y, z) a uniqueresult, as indicated by the following equations:
p(s(x, y), z) = (x+ y) · zs(z, p(x, y)) = (z + (x · y))s(s(z, x), y) = (z + x) + y
Operations are functions. However, expressions tend to be much clearer whenthey are written by using operations than when they are written as functions. Forexample, (x + y) · z is much clearer than p(s(x, y), z). Hence, we will now useoperation symbols rather than functions. Strictly speaking, only total functionsare operations. However, we will occasionally use operations to express par-tial functions. A domain, together with one or more operations constitutes analgebra. Since all operations are functions or at least partial functions, algebraicexpressions are really compositions, or even compositions of compositions. Inthis sense, one can say that algebras deal with function compositions.
Algebraic expressions must be unambiguous, because fallacies arise other-wise. For example, 3 · 4 + 5 may be interpreted as either (3 · 4) + 5 or 3 · (4 + 5),and if the wrong interpretation is used, the resulting derivation is erroneous. Thisis demonstrated by the following derivation, which pretends to show that 12 = 6.
1. 3 = 2 + 1 premise2. 12 = 3 · 4 premise3. 12 = 2 + 1 · 4 substitute 2 + 1 for 3 on line 2.
Hence, 12 = 2 + 1 · 4 = 2 + 4 = 6, which is obviously false. To avoidsuch errors, one can use fully parenthesized expressions. If line 1 had been
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written as 3 = (2 + 1), and line 2 as 12 = (3 · 4), the result would have been12 = ((2+1) · 4), which does not admit the erroneous conclusion. We note, thatgrouping terms in the wrong way is one of the most frequent errors when doingalgebraic manipulations.
3.7.6 Properties of Operations
This section deals with operations and their properties, a topic of great impor-tance in algebra. The properties of the operations in standard algebra are ofcourse well known. For example, everyone knows that x + y = y + x, and thatx · y = y · x. This is the commutative property of addition and multiplication,respectively.
Definition 3.10. A function f of arity 2 is commutative if f(x, y) = f(y, x).Similarly, if ◦ is an operation, the ◦ is said to be commutative if x ◦ y = y ◦ x.
When speaking about an operation, one always implies a certain domain.Note, however, that the same operation symbol may be used in different domains.For example, the operation + may be used for natural numbers, integers, and realnumbers. In this case, it is assumed that there are really three different oper-ations, one corresponding to each domain. Often, the domain is given by thecontext.
Example 3.27. Let max be the binary operation that selects the maximum valueof two integers. For example, 3 max 4 = 4. The operation max is commutative,since for all x and y, xmax y = ymaxx, one concludes that the operation maxis commutative.
The operation ’−’ is not commutative, because x − y = y − x does notgenerally hold.
All logical connectives can be thought of as operations. Obviously, ∨ and ∧are commutative operations, while→ is not commutative. If one deals with files,one can define several operations. A merge operation can be used to indicate thattwo files are merged. The merge operation is typically commutative.
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If ◦ is an operation that is defined for a finite universe, one can express theresults of the operation by means of a table. Such a table is often called an oper-ation table. For example, the truth tables of logical connectives are operationtables.
Example 3.28. Examine, whether the operation ◦ defined in the table is commu-tative.
◦ a b c d
a a b d cb a b c dc d c b ad a b a b
The operation ◦ is not commutative. To be commutative, the operation musthave the property that x ◦ y = y ◦ x for all x and y. This is not the case here. Forexample, d ◦ a = a, yet a ◦ d = c. Generally, an operation is only commutativeif its operation table is symmetric, and this is not the case in this example.
Definition 3.11. If ◦ is an operation, then ◦ is said to be associative, if, for all x,y, and z,
x ◦ (y ◦ z) = (x ◦ y) ◦ z.
Clearly, the operation + is associative, and the operation − is not. If anoperation ◦ is associative and if an expression contains ◦ as its only operation,one can drop all parentheses. This makes expressions more readable.
To find if an operation ◦ given by its operation table is associative, one mustverify that x ◦ (y ◦ z) = (x ◦ y) ◦ z holds for each possible combination of x,y, and z. Often the only way to do this is to enumerate all these combinations,and this is a lengthy process. In the case of the table in the previous example,there are 43 = 64 such combinations, and to show that ◦ is associative, one mustenumerate all these 64 combinations.
Example 3.29. Consider a universe of discourse that contains only the two indi-viduals a and b. In this universe, the operation ◦ is defined by the followingtable:
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◦ a b
a a bb a a
The values x ◦ (y ◦ z) and (x ◦ y) ◦ z are calculated in the following table:
xyz y ◦ z x ◦ (y ◦ z) x ◦ y (x ◦ y) ◦ zaaa a a a aaab b b a baba a a b aabb a a b abaa a a a abab b a a bbba a a a abbb a a a b
Since the values for these two expressions do not always agree, the operation◦ is not associative.
Because of rounding errors, floating-point addition is not associative. If (x+y)+z is evaluated, then the operation x+y is done first, and the result is rounded.To this intermediate result, z is added, and rounding takes place once more. Inthe expression x+ (y + z), on the other hand, y + z is calculated first, rounded,and then x is added. This final result is then rounded. Typically, the roundingerrors in these two evaluations are different, which means that (x+ y) + z is notequal to x + (y + z). To minimize rounding errors in floating-point arithmetic,one should make sure that all intermediate results are as small as possible. Thereason is that rounding errors tend to be proportional to the numbers that arerounded, and smaller intermediate results therefore typically give rise to smallerabsolute errors. Specifically, if x, y, and z are all positive reals and if x and y areboth smaller than z, it is best to add x and y first, because the intermediate resultx+ y is smaller than y + zand the same tends to be true for the rounding error.
If several operations are defined within the same domain, then the relationsbetween the operations become important. In arithmetic, one has addition andmultiplication, and in statement algebra, one has ∧ and ∨. It is therefore impor-tant to consider systems that have two operations, say ◦ and �.
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Definition 3.12. Let ◦ and � ne two operations. The operation ◦ is said to be leftdistributive over � if, for all x, y, and z, one has
x ◦ (y�z) = (x ◦ y)�(x ◦ z).
The operation ◦ is said to be left distributive over � if, for all x, y, and z,
(y�z) ◦ x = (y ◦ x)�(z ◦ x).
An operation which is both right and left distributive is said to be distributive.
If ◦ is left distributive over � and if ◦ is commutative, then ◦ is also rightdistributive over �. To see this, interchange the order of all terms containing xin the applicable definition.
Example 3.30. In arithmetic, multiplication is distributive over addition, butaddition is not distributive over multiplication. This means that
Example 3.31. In logic, ∧ is distributive over ∨, and ∨ is distributive over ∧.
Example 3.32. Let us denote xmax y = max{x, y} to be the maximum of xand y, and let min{x, y} be similarly the minimum of x and y. We show thatmax is distributive over min.
Assume first that y ≥ z. In this case, min{y, z} = z, and equation (3.11)becomes
max{x, z} = min{max{x, y},maxx, z.
Since y ≥ z, max{x, y} ≥ max{x, z}, which proves that both sides of theequation are equal. This settles the case y ≥ z. The case y < z can be dealt within the same way. Hence, by the law of cases, (3.11) is always true.
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3.7.7 Identity and Zero Elements
In this section we discuss two elements, the identity element and the zero ele-ment, that, if they exist, turn out to be very useful for doing algebraic manipula-tions. In particular, they allow one to simplify algebraic expressions.
Definition 3.13. Let ◦ be an operation, defined for some universe of discourse.If there is an individual er with the property that, for all x, x ◦ er = x, then er iscalled a right identity. Similarly, if there is an individual el such that, for all x,el ◦ x = x, then el is called a left identity. An individual e that is both a right anda left identity is called an identity.
If an operation possesses both a left and a right identity, then the two identityelements must be equal. In fact, any identity that is both a right and a left identityis unique. The important point now is that, if we talk about the identity, then weimplicitly assume that this identity is both a right and a left identity and that it istherefore unique.
Example 3.33. Consider a domain with the three elements a, b, and c, togetherwith the operation ◦, as defined by the following operation table:
◦ a b c
a a b cb c b ac a b c
We find all left and right identities. Since a ◦ x = x, no matter whether xis a, b, or c, a is a left identity. Similarly, c is a left identity. There is no rightidentity, however. In fact, the existence of two left identities precludes that thesebe a right identity.
If an operation is commutative, the right identity must be equal to the leftidentity. This is shown as follows: If el is a left identity of ◦, and if ◦ is commu-tative, then el◦x = x◦el, which makes el a right identity. Hence, for commutativeoperations there is only one identity, which is both a right an a left identity.
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Example 3.34. Find the identities for addition and multiplication. Since additionand multiplication are both commutative, there is only one identity. Since ∀x(x+0 = x) is true, 0 is the identity for +. Similarly, 1 is the identity for multiplication,because ∀x(x · 1 = x).
Suppose that the operation ◦ has both a right and a left identity e. Any propersubexpression of the form x ◦ y satisfying x ◦ y = e can obviously be deleted.To facilitate such deletions, one defines inverses as follows:
Definition 3.14. If ◦ is an operation and if x is an individual, then y is called aleft inverse of x if y ◦ x = e. Similarly, if x ◦ y = e, then y is called a rightinverse of x. An individual that is both a left and a right inverse of x is called aninverse of x.
Example 3.35. Consider the operation ◦ given in the table. Find the right inversesand the left inverses of all elements.
◦ a b c d
a a b c db b a a cc c b d cd d a b c
According to the table, the identity element is a. To find the inverses, one iden-tifiers all combinations x ◦ y that yield a. One has
a ◦ a = a, b ◦ b = a, b ◦ c = a, d ◦ b = a.
Hence, a is its own inverse, and so is b. In addition, b is the left inverse of c,and c is the right inverse of b. Similarly, d is the left inverse of b, and b is theright inverse of d. There is no right inverse of c; that is, there is no x such thatc ◦ x = a.
Example 3.36. Let x be a real number. Find the additive inverse, that is, theinverse of + for all numbers x. Also, find the multiplicative inverse of x if itexists.
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The identity for + is 0. The additive left inverse must therefore satisfy y+x =0. Clearly, y = −x, which makes (−x) the left inverse of x. One easily verifiesthat −x is also the right inverse of x. Hence, there is only one inverse, which issimultaneously the right and the left inverse. The inverse of x in multiplicationis the reciprocal of x, which we denote by x−1. The right and the left inversesare again equal, and the inverse, if it exists, is unique. However, the number 0has no inverse.
The following theorem deals with the uniqueness of an inverse.
Metatheorem 3.1. If the operation ◦ has the (right and left) identity e, if ◦ isassociative, and if the right and left inverses of x are equal, then x has a uniqueinverse.
Example 3.37. Use the table in Example 3.35 to demonstrate that it is not suffi-cient to show that the right and left inverses are equal in order to prove that theinverse is unique.
In the table of Example 3.35, b is both a right and a left inverse of itself, yet bhas other inverses besides b.
In expressions containing inverses, one can rearrange the terms in such a waythat the inverses can be combined to yield the identity element, which can thenbe dropped. For example, the expression (x + y) + (−x) can be written as(y+x) + (−x) by applying the commutative law, and because of the associativelaw, this yields in turn y + (x + (−x)). since x + (−x) = e, this yields y + e,or y. Of course, in an expression like x + (−x) = e, nothing is left after e hasbeen dropped, and in this case e must be retained. Except for this case, one cancancel all inverses, provided that the expression contains only one operation andthis operation is commutative and associative.
Example 3.38. Simplify aba−1bcb−1 in the domain of real numbers, where a−1
and b−1 denote the inverse of a and b, respectively.Since multiplication is associative and commutative, one can cancel a against
a−1 and b against b−1, and the result is bc.
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Another important element is the zero element. To avoid the need to distin-guish between a left and a right zero, we assume that the operation in question iscommutative.
Definition 3.15. An element d is called a zero element of a commutative opera-tion ◦ if, for all x, x ◦ d = d.
Example 3.39. Find the zero element of multiplication in the domain of integers.Since ∀x(x · 0 = 0), one concludes that the zero for multiplication is d = 0.
Not all operations have zeros. In particular, the operation + for integers hasno zero. There is no number d such that x+ d = d for all possible values of x.
Suppose that ◦ is a commutative and associative operation with the zero ele-ment d. Then x1 ◦ x2 ◦ . . . ◦ xn = d if a single term xi for some i = 1, 2, . . . , n,is equal to the zero element d.
Example 3.40. In the case of multiplication of integers, a · b · c = 0 if either a,b, or c is zero.
In expressions with two operations, further possibilities for creating identityelements and zero elements arise. In the domain of the real numbers, the identityof the addition becomes the zero of the multiplications, which allows one tosimplify expressions such as (x+ (−x)) · y, which simply become 0.
3.7.8 Derivations in Equational Logic
Equational logic will now be used to give formal derivations for results from theprevious section. Moreover, some new results will be derived in a formal way.In equational logic, as in other formal derivations, unification plays a major role.
It was stated that if an operation has a right and a left identity then these twoidentities must be equal, and there is no other identity. Hence, if u and u′ are twoleft identities and if v and v′ are two right identities, then u = u′ = v = v′. Thisresult is now formally derived. We need the following premises, which are directconsequences of the definitions (steps 1 - 4). In the steps 5 - 7 we unify the leftsides of line 1 and 2 and equate the results. In the steps 8 - 10 we unify lines 3
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and 4, one similarly finds u′ = v′. Finally, in the steps 11 - 13 we unify lines 1and 4 to show that u = v′.
1. u ◦ x = x u is a left identity, x is a true variable.2. x ◦ v = x v is a right identity, x is a true variable.3. u′ ◦ x = x u′ is a left identity, x is a true variable.4. x ◦ v′ = x v′ is a right identity, x is a true variable.5. u ◦ v = v Instantiate line 1 with x := v.6. u ◦ v = u Instantiate line 2 with x := u.7. u = v Substitute u ◦ v on line 5 by line 6.8. u′ ◦ v′ = v′ Instantiate line 3 with x := v′.9. u′ ◦ v′ = u′ Instantiate line 4 with x := u′.
10. u′ = v′ Substitute u′ ◦ v′ on line 8 by line 9.11. u ◦ v′ = v′ Instantiate line 1 with x := v′.12. u ◦ v′ = u Instantiate line 4 with x := u.13. u = v′ Substitute u ◦ v′ on line 11 by line 12.
The equalities of lines 7, 10, and 13 imply that u, v, v′, and u′ are all equal.To be rigorous, one would have to prove equality for each combination of thesefour variables; that is, one would have to prove that u = v, u = v′, u = u′,v = v′, and so on, but since these proofs are easy, they are omitted.
Let d be the inverse of c. Then one has
∀x((x ◦ c) ◦ d = x). (3.12)
For a formal proof, all premises must be stated. They are as follows:
1. x ◦ e = x e is an identity, x is a true variable.2. c ◦ d = e d is the inverse of c, where c, d are fixed.3. (x ◦ y) ◦ z = x ◦ (y ◦ z) Operation is associative and x, y, z are true variables.
Line 1 indicates that to have an inverse one must have an identity. This iden-tity is denoted by e. Line 2 defines d to be the inverse of c, and line 3 indicatesthat ◦ is associative. To derive (x ◦ c) ◦ d = x, first unify this (x ◦ c) ◦ d with theleft of line 3. This yields line 4. The remaining lines of the derivation are easyto trace.
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4. (x ◦ c) ◦ d = x ◦ (c ◦ d) Instantiate line 3 with y := c and z := d.5. x ◦ c) ◦ d = x ◦ e Replace c ◦ d on line 4 by line 2.6. x ◦ c) ◦ d = x Substitute line 1 into line 5.
Since x is a true variable, one can generalize the last line to obtain (3.12) asrequired.
In some cases, there is no premise that can be applied directly. In such cases,one uses the reflexivity axiom to generate an equality that can then be modifiedby substitutions.
Example 3.41. Show that a◦ (b◦e) = a◦b. Here, e is the identity. The premisesare the reflexivity axiom and the definition of the identity. The reflexivity axiomis used first, with x instantiated to the left of the desired conclusion. The fol-lowing derivation results.
1. ∀x(x = x) Reflexivity axiom.2. ∀x(x ◦ e = x) Definition of the identity e.3. a ◦ (b ◦ e) = a ◦ (b ◦ e) Instantiate line 1 with x := a ◦ (b ◦ e).4. b ◦ e = b Instantiate line 2 with x := b.5. a ◦ (b ◦ e) = a ◦ b Replace second b ◦ e in line 3 by line 4.
It is well known that one can add a constant to both sides of an equation andthat one can multiply both sides of an equation by a constant. More generally,from a = b, one can conclude that a ◦ c = b ◦ c for any operation ◦. One has
1. ∀x(x = x) Reflexivity axiom.2. a = b Premise.3. a ◦ c = a ◦ c Instantiate line 1 with c := a ◦ c.4. a ◦ c = b ◦ c Substitute from line 2 into line 3.
The step from a = b into a ◦ c = b ◦ c is called postmultiplication. In spiteof its name, postmultiplication is not restricted to multiplication; it can involveany operation, including addition, Boolean operations, and others. One can alsopremultiply equations. For example, if a = b, then premultiplication with c leadsto c ◦ a = c ◦ b. Note that a, b, and c are true variables, which means that one can
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generalize as follows:
∀x∀y∀z((x = y) → (z ◦ x = z ◦ y)) (3.13)∀x∀y∀z((x = y) → (x ◦ z = y ◦ z)) (3.14)
3.7.9 Equational Logic in Practice
In practice, algebraic manipulations are often abbreviated. The commutativeand associative laws are often used like rules of inference, and to get from oneexpression to the next, one often applies several rules simultaneously. Moreover,to express the two equations x = y and y = z, one frequently writes x = y = z.Because of the transitive property of equality, x = y and y = z imply thatx = z, and for this reason, x = y = z also means that x = z. This observationobviously generalizes. For example, to prove that x1 = x4, one can show thatx1 = x2 = x3, and x3 = x4. One abbreviates this as x1 = x2 = x3 = x4. Thefollowing example shows how this idea is applied.
Example 3.42. Simplify a ◦ (b ◦ a−1), where ◦ is commutative and associative,and a−1 is the inverse of a.
One has
a ◦ (b ◦ a−1) = a ◦ (a−1 ◦ b) Commutativity= (a ◦ a−1) ◦ b Associativity= e ◦ b Inverse= b Identity
Note the strategy used: in each step, the distance between a and a−1 is reducedin some sense, until a◦a−1 is obtained. Generally, one should always try to keepthe objective of the derivation in mind and try to narrow the gap between whatone has and when one wants to accomplish. It also helps to identify intermediategoals, especially when dealing with derivations.
The process of writing a sequence of expressions, each equal to the pre-vious one, with the objective to reach some type of a goal is used extensivelyin functional programming languages, where it is known as rewriting. Although
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rewriting is extremely important, it sometimes leads to inefficiencies because itmay require that some subexpression be evaluated repeatedly. To show this, con-sider the calculation of the Fibonacci numbers Fn, which are defined as follows:
F0 = 1,
F1 = 1,
Fn = Fn−1 + Fn−2 , n > 1.
(3.15)
For example, F2 = F1 +F0, F3 = F2 +F1, and so on. Suppose now that F4 is tobe calculated. This can be done in two ways. First, one can write
F2 = F1 + F0 = 1 + 1 = 2 since F1 = 1, F0 = 1
F3 = F2 + F1 = 2 + 1 = 3 since F2 = 2, F1 = 1
F4 = F3 + F2 = 3 + 2 = 5 since F3 = 3, F2 = 2
Alternatively, one can rewrite F4 as follows:
F4 = F3 + F2
= (F2 + F1) + (F1 + F0) Expand F3 and F2.
= (F2 + 1) + (1 + 1) F1 = F0 = 1.
= ((F1 + F0) + 1) + 2 Expand F2. 1 + 1 = 2.
= ((1 + 1) + 1) + 2 F1 = F0 = 1.
= (2 + 1) + 2 = 5.
Note that, in the derivation, F2 = F1+F0 = 1+1 = 2 has been calculated twice.This can happen unless special precautions are taken. It will be shown that thiscan slow down execution times dramatically. Generally, it pays to watch forcommon subexpressions appearing more that once in order to avoid evaluatingthem repeatedly.
An important result is the cancellation rule, which allows one to infer undercertain conditions that a = b from a ◦ c = b ◦ c. To see why restrictions apply,consider the following applications of the cancellation rule. Clearly, if a + c =b + c, then a = b. In the case a · c = b · c, one can conclude that a = b only ifc 6= 0. If c = 0, the cancellation law does not apply. This suggests that for thecancellation law to hold the term to be cancelled must have an inverse, and this
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is indeed the case. Hence, the derivation must reflect that c has an inverse, sayc−1. The premise is
a ◦ c = b ◦ c.Postmultiplying by c−1 yields
(a ◦ c) ◦ c−1 = (b ◦ c) ◦ c−1.
We can now simplify both sides separately. The left side yields
(a ◦ c) ◦ c−1 = a ◦ (c ◦ c−1) Associativity= a ◦ e Inverse= a Identity (3.16)
A similar argument shows that the right side yields b, and we conclude that a = b.If c has an inverse and if ◦ is associative, then a ◦ c = b ◦ c implies that a = b.
3.7.10 Boolean Algebra
For statement algebra, the notion of commutativity and associativity have alreadybeen introduced earlier. Indeed, ∧ and ∨ are both commutative and associative.Moreover, P ∧ T = P (where T is the truth value ’true’) for all P , which makesT the identity of ∧. The identity of ∨ is simirarly F (where F is the truth value’false’). Also, by the law of domination, P ∧ F = F , which makes F the zeroof ∧. The zero of ∨ is similarly T . Finally, the distributivity laws indicate that ∧is distributive over ∨ and that ∨ is distributive over ∧.
However, there is a difference in the philosophy of statement algebra andother algebras. As the name implies, statement algebra deals with statements,and it distinguishes between atomic and compound statements. Two statementsof a different form are not considered equal, even if they are equivalent. Forexample, P ∨Q is not considered equal to Q ∨ P . For this reason, we formallyintroduce a number of new operations and new symbols. In fact, we introduce aBoolean algebra.
A Boolean algebra is an algebra with two operations, which are denoted by +and ·. Both operations are total in the sense that, for all arguments x and y, x+ yand x · y are defined. Moreover, the algebra has the following properties:
1. There is an identity element for +, called 0.
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2. There is an identity element for ·, called 1.
3. The operation + is commutative.
4. The operation · is commutative.
5. The operation + is distributive over ·.
6. The operation · is distributive over +.
7. For every individual x, there is an element x′, called the complement of x,with the property that x+ x′ = 1 and x · x′ = 0.
8. The domain contains at least two elements.
The operation · is often omitted. In the simplest case, the domain of a Booleanalgebra contains two values, 0 and 1. Such Boolean algebra is called a two-valued Boolean algebra. We will show that the following assignment constitutesa two-valued Boolean algebra. In fact, it is the only such algebra.
1. The complements x′ are defined as follows:
0′ = 1, 1′ = 0.
2. The operations + and · are defined by the following operation tables:
+ 0 1
0 0 11 1 1
· 0 1
0 0 01 0 1
We now verify that all the conditions of a Boolean algebra are met. Sincethe operation tables for + and · are both symmetric, the algebra is commutative.Second, 0 is the identity element of + and 1 is the identity element of ·, as iseasily verified from the operation tables. The fact that + is distributive over · isproved by verifying that, for all x and y, x, y = 0, 1, one has
x+ (y · z) = (x+ y) · (x+ z). (3.17)
This is done in the table
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1 2 3 4 5xyz y · z x+ y · z x+ y x+ z (x+ y) · (x+ z)
In this table, x+(y · z) is given in the column labelled 2, and (x+ y) · (x+ z)on column 5. It is easy to see that the two columns are equal in all cases, whichmeans that (3.17) holds in all cases. The proof that · is distributive over + is donein a similar way. Finally, 0 + 0′ = 1 and 1 + 1′ = 1, which means that, for all x,x+ x′ = 1. The fact that x · x′ = 0 for all x is shown in a similar way.
We now show that this is the only two-valued Boolean algebra. Since 0 is theidentity for +, one must have 0 + 0 = 0 and 0 + 1 = 1 + 0 = 1, and since 1is the identity of ·, one must have 0 · 1 = 1 · 0 = 0 and 1 · 1 = 1. Moreover,since a Boolean algebra must satisfy x + x′ = 1 for all x, 0 + 0′ must be 1.This rules out that 0′ = 0 because, under this assignment, 0 + 0′ = 0 + 0 = 0.The only alternative assignment for 0′ is of course 1, and this corresponds to theBoolean algebra given earlier. A similar argument shows that 1′ = 0. The onlyvalues still open at this point are 1 + 1 and 0 · 0, the system no longer has therequired distributive properties. The reader may want to verify this. This leavesthe Boolean algebra given previously as the only possibility.
If in statement algebra all equivalences are interpreted as equalities, then thestatement algebra becomes a Boolean algebra. To see this, redefine 0, 1, +, ·,and complementation as shown in the next table. There is a dual interpretationof two-valued Boolean algebra. In this dual interpretation, one uses 0 for T, 1 forF, + for ∨, and · for ∧. This dual interpretation is the explanation for the dualrelations discussed earlier.
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Boolean Statement Naturalalgebra algebra language
0 F false1 T true+ ∨ or· ∧ andx′ ¬x not
Example 3.43. Express p ∨ ¬p↔ T and p ∧ ¬p↔ F in Boolean algebra.p ∨ ¬p↔ T becomes p+ p′ = 1, and p ∧ ¬p↔ F becomes p · p′ = 0.
In conclusion, equational logic provides a basis for Boolean algebra and indi-rectly, a basis for propositional logic. In this sense, it ties together predicate logicand propositional logic.
3.8 An Axiomatization for Fields
3.8.1 Tarski’s Axioms for Fields
We begin from the well-known field theory of real numbers.Consider a first order language P that is an extension of classical predicate
calculus. We give the axiomatization for field theory using axioms being essen-tially given by Tarski in his Introduction to Logic, New York, 1941.
Definition 3.16. Let P = {F,+,×,−, −1,0,1} be a first order calculus where Fis the set of terms of the system, + and × are binary operation symbols, − and−1 are unary operation symbols, and 0 and 1 are individual constants. A theoryhaving the following non-logical axioms is called (a first order) field theory:
A1. ∀x ∀ y (x+ y = y + x),
A2. ∀x (x+ 0 = x) (0 is the identity element of +),
A3. ∀x (x+ (−x) = 0),
A4. ∀x ∀ y ∀ z ((x+ y) + z = x+ (y + z)),
A5. ∀x ∀ y ∀ z ((x× y)× z = x× (y × z)),
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A6. ∀x (x× 1 = x) (1 is the identity element of ×),
A7. ∀x (¬(x = 0) → x× x−1 = 1),
A8. ∀x ∀ y (x× y = y × x),
A9. ∀x ∀ y ∀ z (x× (y + z) = x× y + x× z),
A10. ¬(0 = 1).
A model of a field theory is called a field.
3.8.2 The Field of Real Numbers
The set of real numbers, i.e. the first order calculus PR = {R,+,×,−, −1, 0, 1},is a field satisfying these axioms for arithmetical operations of real numbers. Thevariable symbols refer to crisp real numbers.
References
[1] Stanley N. Burris, Logic for Mathematics and Computer Science, PrenticeHall, 1998.
[2] Winfried.K. Grassmann, Jean-Paul Tremblay Logic and Discrete Mathe-matics. A Computer Science Perspective, Addison-Wesley, Pearson Educa-tion Limited, 2002.
[3] R. Johnsonbaugh, Discrete Mathematics, Prentice Hall, 4. painos, 2001.
[4] Veikko Rantala, Ari Virtanen, Logiikkaa. Teoriaa ja sovelluksia, Matemaat-tisten tieteiden laitos, Tampereen yliopisto, B 43, 1995 (In Finnish)
