Math 1090 Part two: Predicate (First-Order) Logicsaeedgh/Math1090/PL.pdf · 2018. 10. 10. · 3 =...

Math 1090Part two: Predicate (First-Order) Logic

Saeed Ghasemi

York University

5th July 2018

Saeed Ghasemi (York University) Math 1090 5th July 2018 1 / 95

Question. Is propositional logic rich enough to do mathematics andcomputer science?The answer is “Absolutely not”!Mathematics and computer science deals with structures like sets, strings,numbers, matrices, trees, graphs, programs, Turing machines and manyothers. The propositional logic is not rich enough to deal with thisstructures.For example, try expressing the statements “there is a rational numberstrictly between two given rational numbers” or “every natural number hasa unique prime factorization”. We cannot even express these statements inthe Boolean Logic, let alone proving them!


Number Theory-Peano’s Arithmetic

In order to express formulas in number theory:

We need to be able to refer to numbers, N = {1, 2, 3, . . . } asvariables. we need to have variable symbols that would refer tonumbers, like n,m, n′, . . . . Let’s call them “object variables”.

We need to be able to say when two numbers are equal, that’s whywe need a symbol “=” for the equality between the variable objects.

We need to be able to say expressions like “for all natural numbers...”. So let’s have a symbol ∀, which says “for all”.

We need a special constant denoting 0.

We also need to be able to add and multiply numbers. So lets havefunction + : N× N→ N and × : N× N→ N for addition andmultiplication, respectively (we say these functions have arity 2because they take two inputs, so the function f (x , y , z) = ... has arity3).


Number Theory-Peano’s Arithmetic

We need a unary (of arity one) function S : N→ N defined byS(n) = n + 1. This is called the “successor function”.

Lastly we need a predicate symbol “<” expressing formulas like“2 < 3”.

We can abbreviate the structure obtained above by (N, 0,+,×, S , <). Wecould also include = there, but since equality is going to be always part ofour languages, we just presume it is there. No need to write it!But why did we need the constant 0 or the function S?Suppose we want to express 2 < 3. We need specific numbers, 2 and 3!Well, we can obtain them using 0 and S , e.g., 2 = S(S(0)) and3 = S(S(S(0))). So we can express any specific number.Now add to the Boolean Logic our new symbols as part of the alphabet.Then we can formulate many things in our new language. For instance ifwe want to express n ≤ m, then we can write (n = m) ∨ (n < m).


The Alphabet of the General First-Order Language

A language for first order logic consists of the following symbols:

(I) Logical symbols

1 All of V; the Boolean Logic Alphabet.2 Object variables: x , y , u, v , v0, v1, v2, v3, ..., x0, x1, x2, x3, ...3 The equality symbol: =4 The universal quantifier symbol: ∀

(II) Non-logical symbols1 Object constants; denoted by a, b, c , . . .2 Function symbols; denoted by f , g , h, . . .3 Predicate symbols; denoted by φ, ψ, φ0, . . .


Note:

The logical symbols belong to every first order language. However,the choice of non-logical symbols depends on the “theory” that wewant to study.

The equality is only for object variables (e.g., x = y) and not forBoolean variables (“≡ “does the similar job for Boolean variables).

In applications, each function and predicate has a fixed arity.

Each of the sets of non-logical symbols (object constants, functionsymbols and predicate symbols) can be empty.

Examples.(1) For example the first order language for Number Theory consists of{0,+,×,S , <} as non-logical symbols; one constant symbol 0, two binary(arity 2) functions +,×, one unary function S and one binary predicate <.(2) The first order language for Set Theory only has one non-logicalsymbol, namely {∈}.


Definition (Term)

For a first-order language (alphabet) L a term or an L-term is definedrecursively as follows:

1 Every object variable and object constant is a term.

2 If t1, t2, . . . tn are terms and f is an n-ary (of arity n) function symbol(in L) then f (t1, t2, . . . , tn) is also a term.

The set of all terms is denoted by Term. We denote terms byt, s, t ′, t1, . . . .Examples.In the language of Number Theory these are some terms:

n

0

n + m

S(S(n))

(n ×m) + S(m′)


Definition (Term-Calculation/ Term-Parse)

A term-calculation (or term-parse) is any finite sequence of strings that wemay write respecting the following two requirements:

1 At each step we may write any object variable and object constant.

2 If t1, t2, . . . tn are already written and f is an n-ary function symbolthen we may write f (t1, t2, . . . , tn).

Proposition

An string of symbols from our alphabet (language) L is a term iff it canbe written in a step of a term-calculation.


Definition (Atomic formulas)

For a first-order language (alphabet) L an atomic formula or an atomicL-formula is defined recursively as follows:

1 Any Boolean variable and Boolean constant is an atomic formula.

2 If t, s are terms then t = s is an atomic formula.

3 For any predicate symbol φ (in L) of arity n, and any n termst1, t2, . . . , tn the string φ(t1, t2, . . . , tn) is an atomic formula.

We denote the set of all atomic formulas by AF .For example m + s(0) < n × 3 and 1 + 0 = 1 are atomic formulas in thelanguage of Number theory.


First-Order Formulas

Definition (First-Order Formulas)

For a first-order language (alphabet) L a formula or an L-formula isdefined recursively as follows:

1 Any atomic formula is a formula.

2 If A is a formula, then (¬A) is a formula.

3 If A and B are formulas, then so are (A ∧ B), (A ∨ B) , (A→ B) and(A ≡ B).

4 If A is a formula, then for any choice of (object) variable x the string((∀x)A) is a formula.

Note: The sub-formula A in the formula ((∀x)A) is called the scope of(∀x).Note: We are only allowed to quantify over first-order (object) variables.We can not quantify over functions or predicates. For example in a firstorder language we are not allowed to say “for all functions f ...”.Saeed Ghasemi (York University) Math 1090 5th July 2018 10 / 95

Definition (Formula-Calculation/ Formula-Parse)

A formula-calculation (or formula-parse) is any finite sequence of stringsthat we may write respecting the following four requirements:

1 At each step we may write any atomic formula.

2 At any step we may write (¬A), given that A is already written.

3 At any step we may write (A ∧ B), (A ∨ B), (A→ B) and (A ≡ B) ,given that A and B are already written.

4 At any step, and for any variable x , we may write the string ((∀x)A),given that we have already written A.

Proposition

An string of symbols from our alphabet (language) L is a formula iff it canbe written in a step of a formula calculation.


Examples of formula-calculation

These are some formula-calculations:

p, q,⊥, (p ∨ q), ((p ∨ q)→ ⊥)

>, (¬>), p,⊥, ((¬>) ∧ p), q, (q ∨ ⊥), ((¬>) ∧ p) ≡ (q ∨ ⊥)

p, x = c , ((∀y)p), u = v , (¬x = c), (((∀y)p)∨u = v), (¬x = c)→ (((∀y)p)∨u = v)

Therefore any string appearing in each of these sequences is a formula.We sometimes use the same notations WFF, to denote the set of all first-ordered (well-formed) formulas. We simply say WFF is the set of allformulas and denote its members by wff.


Definition (The Complexity of Formulas)

The complexity of a wff is the total number of occurrences of∀,¬,∧,∨,→,≡, in the formula.

For example, p and x = y have complexity 0.((∀x)((∀y)(¬x = z))) has complexity 3.

Definition

We introduce a new symbol in the metatheory (an abbreviation, not aformal symbol) ∃, called the existential quantifier.For any formula A, the string (∃x)A) abbreviates the formula(¬(∀x)(¬A)).


Priorities and Bracket Reductions

1 Outermost brackets are redundant.

2 The priorities, decreasing from left to right, is agreed to be: ((∀x)and ¬ have equal priorites.){

(∀x)¬

},∧,∨,→,≡

3 If ◦ ∈ {∧,∨,→,≡} and a formula is like · · · ◦ A ◦ . . . then the right ◦acts first. That means · · · ◦ (A ◦ . . . .

¬¬A is short for ¬(¬A).

¬(∀x)A is short for ¬((∀x)A).

(∀x)¬A is short for (∀x)(¬A).

(∀x)(∀x)A is short for (∀x)((∀y)A).

(∃x)(∀x)A is short for (∃x)((∀y)A).


More Examples of Formulas

Instead of (u = v → (((∀x)x = a) ∧ p)) we write

u = v → (∀x)x = a ∧ p

Instead of ((∀z)(¬x = y)) we write

(∀z)¬x = y


Definition (Subformulas)

The concept of “B is a subformula of A” is defined inductively:

1 A is atomic: then A and B are identical strings.

2 A is (¬C ): then either B is the same string as A or B is a subformulaof C .

3 A is C ◦ D where ◦ ∈ {∧,∨,→,≡}: then either B is the same stringas A or B is a subformula of C or of D or both.

4 A is (∀x)C : then either B is the same string as A or B is asubformula of C .


From now on whenever we say “variable” we mean “object variable”, andfor “Boolean variables” we always say and write the word “Boolean”.

Definition (Bound and Free Occurrences of variables)

An occurrence of a variable x in a formula A is called bound iff either thatoccurrence is in the substring (∀x) of it is in the scope of some (∀x) in A.Otherwise we say the occurrence of x is free.

Example: In the formula below the bound occurrences of x are boxed andthe rest are free.

x = y → x = y ∨ (∀ x ) x = z)


Abstraction of First-Order Formulas

The abstraction of a first-order formula A is when we identify the“shortest” subformulas that contain each non-Boolean symbols (the objectvariables and object constants, functions, predicates, =, and ∀) in theformula A and replace (re-name) them by “new” (fresh) Boolean variables.Examples:

The abstraction the formula

p → x = y ∨ (∀x)φx ∧ q

is p → p′′ ∨ p′′′ ∧ q.The abstraction the formula

x = y → x = y ∨ z = v

is p → p ∨ q.The abstraction the formula

(∀x)(x = y → (∀z)z = a ∨ q)

is p.Saeed Ghasemi (York University) Math 1090 5th July 2018 18 / 95

The abstraction the formula

p → x = y ∨ (∀x)(φx ∧ q)

is p → p′ ∨ p′′.

Note

Exactly the atomic formulas and formulas of the form ((∀x)A) getabstracted.

Proposition

The abstraction of any first-order formula is a Boolean formula

Proof.

By induction on the complexity of formulas.


The “abstraction” of formulas enables us to use the semantics (truthtable) syntactic (proof) techniques from Boolean logic.

In first-order logic Boolean variables are used to denote thestatements about objects that we either do not know what it says orwe don’t care about what it says.

Definition (Tautologies and Tautological implications)

We say a first-order formula A is a tautology, and write |=taut A, iff itsabstraction is a tautology. In first-order logic we write Γ |=taut A iff theabstraction of the formulas in Γ tautologically imply the abstraction of A.


Definition (Substitution of Terms into Variables)

If s, t are terms and x is a variable we define s[x := t] inductively asfollows:

s[x := t] is

s if s is a constant or a variable

(not x)

t if s is x

f (s1[x := t], . . . , sn[x := t]) if s is f (s1, . . . , sn)


Definition (Substitution of Terms into Variables)

If A is a first-order formula and x is a variable we define A[x := t]inductively as follows:

A[x := t] is

φ(s1[x := t], . . . , sn[x := t]) if A is φ(s1, . . . , sn)

s1[x := t] = s2[x := t] if A is s1 = s2

¬C [x := t] if A is ¬C

C [x := t] ◦ D[x := t] if A is C ◦ D

A if A is one of the p,>,⊥, (∀x)B

(∀y)B[x := t] if A is (∀y)B, where -

y (not x) does not occur in -

t or x is not free in B.

undefined if A is (∀y)B, where -

y (not x) does occur in t and -

x is free in B.Saeed Ghasemi (York University) Math 1090 5th July 2018 22 / 95

The intuition behind the definition of A[x := t] is that no freevariable has to be “captured” (to fall into the scope of a quantifier)as a result of this substitution.

The operations [x := t] takes place in the metatheory and has thehighest priority against all the other formal and informal operationssuch as ∀,∃,¬,∧,∨,→,≡.

Examples.(1) t = s[x := s ′] is actually t = (s[x := s ′])(2) (∀x)A[x := t] is (∀x)(A[x := t])(3) (x = y)[y := x ] is x [y := x ] = y [y := x ] which is x = x(4) (∀x)x = y [y := x ] is (∀x)x = x(5) ((∀x)x = y)[y := x ] is “undefined”, because otherwise the x that wasfree (the third one from left) would get captured.(6) (∀x)(x = y)[y := x ] is (∀x)x = x


(7)((∀x)(∀y)φ(x , y)

)[y := x ] is (∀x)

((∀y)φ(x , y)

)[y := x ] which is

(∀x)(∀y)φ(x , y)(8) (z = a ∨ (∀x)x = y)[y := x ] is (z = a)[y := x ] ∨ ((∀x)x = y)[y := x ],but ((∀x)x = y)[y := x ] is undefined, so the whole thing is undefined.Now we want to substitute Boolean variables with formulas. There are twoalmost identical definitions, “conditional substitution” and “unconditionalsubstitution”.


Definition (Unconditional Substitution)

The unconditional substitution of a formula B into all occurrences of aBoolean variable p in a formula A is denoted by A[p B] and is definedinductively as follows:

A[p\B] is

B if A is p

A if A is Atomic Formula but is not p

¬C [p\B] if A is ¬C

C [p\B] ◦ D[p\B] if A is C ◦ D

(∀x)C [p\B] if A is (∀x)C


Definition (Conditional Substitution)

The Conditional substitution of a formula B into all occurrences of aBoolean variable p in a formula A is denoted by A[p := B] and is definedinductively as follows:

A[p := B] is

B if A is p

A if A is Atomic Formula but is not p

¬C [p := B] if A is ¬C

C [p := B] ◦ D[p := B] if A is C ◦ D

(∀x)C [p := B] if A is (∀x)C and x is not free in B

undefined else

Examples.• ((∀x)p)[p\x = y ] is (∀x)x = y• ((∀x)p)[p := x = y ] is undefined.


Induction on the complexity of First-order formulas

Suppose P(A) is an statement about the (first-order) formula A andsuppose we want to prove by induction on the formula complexity thatP(A) holds for every formula A. Then

(Basis): Show that P(A) holds for every atomic formula A.

Assume P(B) and P(C ) hold for formulas B and C (InductionHypothesis). Show that P(A) holds of A is one of the ¬B, B ◦ C ,where ◦ ∈ {∧,∨,→,≡} and for (∀x)B.

If you can follow these steps then the Induction Principle implies thatP(A) holds for every formula.


Proposition

If z is a fresh variable (does not occur) in A, then A[x := z ][z := x ] is A.

Proof carries out by induction on the complexity of the formula A, butbefore going on with the proof, lets note that this is not true if z is notfresh in A. For example let A be the formula x = z , then(x = z)[x := z ][z := x ] is (z = z)[z := x ], which is x = x . That is s not A.

Proof.

Suppose A is an atomic formula, that is A is t = s or φ(t1, . . . tn). Weneed to compute t[x := z ][z := x ] first, when z does not occur in t, sincez is fresh in A. We claim that for every such t we have that

(?) t[x := z ][z := x ] is t

To show this we also use Induction (this time on the complexity of terms),Basis has two casesCase 1) t is x then t[x := z ][z := x ] is t.


proof continues:

Case 2) t is y (not x) or t is a constant. Then since t is also not z wehave t[x := z ][z := x ] is x .If t is f (s1, . . . , sn). Then t[x := z ][z := x ] isf (s1[x := z ][z := x ], . . . , sn[x := z ][z := x ]). By I. H. for any si we havethat si [x := z ][z := x ] is si . Therefore t[x := z ][z := x ] is f (s1, . . . , sn),which is t again. This finishes the proof of the claim.Basis. It is clear that if A is a Boolean variable or a Boolean constantthen A[x := z ][z := x ] is A. Suppose A is an atomic formula t = s, thenA[x := z ][z := x ] is t[x := z ][z := x ] = s[x := z ][z := x ], which by (?) isA. If A is φ(s1, . . . , sn) then A[x := z ][z := x ] isφ(s1[x := z ][z := x ], . . . , sn[x := z ][z := x ]), which by (?) is nothing butA.Inductive step. If A is one of ¬B or B ◦ C , it is clear from the definitionof A[x := z ][z := x ] and I.H. that A[x := z ][z := x ] is A.


proof continues:

To deal with the stage of formation with (∀), assume the result is true forB (I.H.) and consider two cases where A is either (∀x)B and (∀w)B,where w is not x . Remember that since our assumption was that z doesnot occur in A, so it does not occur in B either. The first case: A is(∀x)B. Then ((∀x)B)[x := z ][z := x ] is ((∀x)B)[z := x ], which is(∀x)B[z := x ] ( z is not free in B, nor it is bound!), which is A, since zdoes not occur in B.The second case: A is (∀w)B. Then ((∀w)B)[x := z ][z := x ] is((∀w)(B[x := z ])

)[z := x ] (by definition and the fact that z is not w !),

which is (∀w)B[z := x ][z := x ] (by definition and the fact that x is notw !). By I. H. we know that B[z := x ][z := x ] is B, therefore(∀w)B[z := x ][z := x ] is (∀w)B, which is A. Done!


Axioms of First-Order Logic

Ax1. All tautologies: meaning all the first-order formulas whoseabstraction is a Boolean tautology.Ax2. (specialization or substitution axiom) : All the formulas of the form(∀x)A→ A[x := t].Ax3. All the formulas of the form (∀x)(A→ B)→ (∀x)A→ (∀x)B.Ax4. All the formulas of the form A→ (∀x)A, where x does not occurfree in A.Ax5. (Identity axiom): All the formulas of the form x = x .Ax6. (Leibniz axiom for identity): All the formulas of the formt = s → (A[x := t] ≡ A[x := s]).Ax7.(Partial Generalization) If A is an axiom then (∀x)A is also an axiom,for any choice of variable x .

Every “instance” of these axioms is an axiom, so there are in fact infinitelyaxioms.

• We show the set of all first-order axioms by Λ1.Saeed Ghasemi (York University) Math 1090 5th July 2018 31 / 95

Primary Rules of Inference

We have the same of rules of inference as the Boolean logic.1. Boolean Leibniz rule

A ≡ B

C [p := A] ≡ C [p := B]

Provided that p is not in the scope of any quantifiers in C .2. Equanimity rule

A,A ≡ B

B

We will abbreviate Boolean Leibniz by BL.

We apply these rules to abstractions of formulas.

In BL it is important that p does not belong to a scope of anyquantifier in C , otherwise p will disappear in the abstraction ofC [p := A] and C [p := B].


The concept of “theorem calculation” and theorems remains unchanged.

Definition (Theorem Calculation or Proof)

Let Γ be a given set of (first-order) formulas. A “theorem-calculation ” or“proof” from Γ is a finite ordered sequence of formulas such that: In anystage we may write down

1 Any member of Λ1 or Γ

2 As long as all the formulas of the numerator of an “instance” ofLeibniz rule or equanimity are already written down, we may write thedenominator for the same instance.

Any formula in Γ-proof is called “assumption” or “hypothesis” or“nonlogical axiom”.

Any Γ-proof has to finish in finitely many steps. Therefore it only usesfinitely many formulas in Γ and axioms.


Definition

Any formula A that appears in in a Γ-proof is called a Γ-theorem. If A is aΓ-theorem we write Γ ` A.

If A is a Γ-theorem and Γ = ∅, we write ` A and we call A an“absolute theorem” or a “logical theorem”.

Definition (Inductive definition of theorems)

A formula E is a Γ-theorem iff one of the following holds.

Th1. E belongs to Γ ∪ Λ1

Th2. If (an instance of) A ≡ B is a Γ-theorem, then E is equal toC [p := A] ≡ C [p := B], provided that p does not appear in scope ofany formula C , for any formula C and variable p.

Th3. If both A and A ≡ E are Γ-theorems.


Remarks:

All the (Γ-) theorems from Boolean logic carry over unchanged, suchas Redundant true, Modus Ponens, de Morgan and etc.

The deduction theorem is also true for the first order logic (with atiny bit of change in the proof!).

Most importantly, the Post’s theorem carries over from Boolean Logicto predicate logic: If Γ |=taut A then Γ ` A.The special case where Γ = ∅ follows from Ax1.: If |=taut A, then Abelongs to Ax1, therefore ` A.

However, soundness needs some adjustments! For example, we knowthat ` x = x hols for any variable x (Ax5), but the abstraction ofx = x is just p, for a Boolean variable p, well we know that 6|=taut p,so 6|=taut x = x .This does not mean our first-order logic is not sound. In fact, it issound. However in predicate logic the concept of “truth” is narrowerthan tautologies and tautological implications.


Two Equivalent Logics

Definition

Two logics over the same first-order language are called equivalent iff theyhave the same absolute theorems.

• Two equivalent logic also have the same relative theorems. Meaningthat if in one logic Γ proves A, so does the other one. This is an easyconsequence of the deduction theorem. If both logics extend Booleanlogic, then both do enjoy the deduction theorem.here are the two logics that we want to compare:

(1) The first-order logic that we introduced (lets call it Logic (1)).

(2) The same logic, except that this logic has Modeus Ponens as the onlyprimary rule of inference (lets call it Logic (2)).


Lemma

Post’s theorem holds for logic (2) for finite Γ.

Proof.

Suppose A1,A2, . . . ,An |=taut B. We need to show that in logic (2) wehave

(?) A1,A2, . . . ,An ` B

By induction on natural numbers, it is easy to show that|=taut A1 → A2 → · · · → An → B. Lets give a Hilbert style proof for(?).


(1) A1 (Hyp.)

(2) A2 (Hyp.)

......

(n) An (Hyp.)

(n + 1) A1 → A2 → · · · → An → B (Axiom)

(n + 2) A2 → · · · → An → B ((1), (n + 1) + MP)

(n + 3) A3 → · · · → An → B ((2), (n + 2) + MP)

......

(n + 1) An → B ((n − 1), (n + n − 1) + MP)

(n + n + 1) B ((n), (n + n) + MP)


Lemma

Logic (2) has BL and Eqn as derived rules of inference.

Proof.

Suppose A, B and C are first-order formulas. By the soundness theoremfor Boolean logic we have that A,A ≡ B |=taut B andA ≡ B |=taut C [p := A] ≡ C [p := B] (note that the soundness is appliedfor the Boolean abstraction of the corresponding theorems, just as therules of inference for first-order logic do). By the previous lemma we canreplace |=taut with ` since there are only finitely many hypotheses, givingus the Eqn and BL as derived rules of inference.


Theorem

Logic (1) and (2) are equivalent.

Proof.

Well, they both have same axioms. The axioms and rule of inference Eqnand BL of the logic (1) can imply MP and conversely the axioms and MPof logic (2) implies Eqn and BL.

It is sometimes easier to work with Logic (2) than logic (1) one to provetheorems, especially when proving by induction on the length of the proof.By the theorem above it does not matter which logic we choose.


Theorem (Weak Generalization)

Suppose Γ ` A and x is a variable which does not occur free in anyformula in Γ, then Γ ` (∀x)A.

Proof.

Prove by induction on the length of the Γ-proof.Basis. A has a proof of length 1. The A is either an axiom or it belongs toΓ.A is an axiom: Then A→ (∀x)A is also an axiom (Ax7.), thereforeΓ ` (∀x)A.A ∈ Γ: Since A ∈ Γ, we know that x is not free in A, therefore A→ (∀x)Ais an axiom (Ax4). Since also Γ ` A, by MP we have we know Γ ` (∀x)A.Now suppose the theorem is true for any formula that appears in a Γ-proofof length ≤ n (I.H.). Suppose Γ ` A and A has a proof of length n + 1.This means that at some steps before n + 1 of the proof for A, there wereformulas B and B → A such that A is derived from them via MP. By I.H.we have Γ ` (∀x)B and Γ ` (∀x)(B → A). By Ax3 we have


Γ ` (∀x)(B → A)→ (∀x)A→ (∀x)A. By twice application of MP wehave Γ ` (∀x)A.

Corollary

If Γ ` A and all the formulas that are used in a Γ-proof of A do not have afree occurrence of x , then Γ ` (∀x)A.

Corollary

If ` A, then ` (∀x)A.

Note. The last corollary does NOT say for any A we have A ` (∀x)A.One can prove the latter if x is not free in A. The weak generalizationtheorem says that if A is an “absolute theorem” and (∀x)A is an absolutetheorem. In other words, the weak generalization theorem says that if youhave written down A in a proof, then you may write down (∀x), but onlyif A is an absolute theorem.


Theorem (Specialization Rule)

(∀x)A ` A[x := t] (of course only if A[x := t] is defined)

Proof.

(1) (∀x)A (Hyp.)

(2) (∀x)A→ A[x := t] (Ax2)

(3) A[x := t] ((1), (2) + MP)

Corollary

(∀x)A ` A

In a proof we will abbreviate Specialization Rule and its Corollary by“Spec”.Saeed Ghasemi (York University) Math 1090 5th July 2018 43 / 95

Theorem (Distributivity of ∀ over ∧)

` (∀x)(A ∧ B) ≡ (∀x)A ∧ (∀x)B

Proof.

By a theorem (Ping-Pong Theorem) it is enough to prove` (∀x)(A ∧ B)→ (∀x)A ∧ (∀x)B and ` A ∧ (∀x)B → (∀x)(A ∧ B).First direction (→):

(1) (∀x)(A ∧ B) (Hyp.)

(2) A ∧ B ((1) + Spec.)

(3) A ((2) + Ax1)

(4) B ((2) + Ax1)

(5) (∀x)A ((3) + Gen. : Hypothesis has no free x)

(6) (∀x)B ((4) + Gen. : Hypothesis has no free x)

(7) (∀x)A ∧ (∀x)B ((5, 6) + Ax1)


proof continues.

For the other direction (←):

(1) (∀x)A ∧ (∀x)B (Hyp.)

(2) (∀x)A ((1) + Ax1.)

(3) (∀x)B ((1) + Ax1)

(4) A ((2) + Spec.)

(5) B ((3) + Spec.)

(6) A ∧ B ((4, 5) + Ax1.)

(7) (∀x)(A ∧ B) ((6) + Gen. : Line (1) has no free x)


Theorem

` (∀x)(∀y)A ≡ (∀y)(∀x)A

Proof.

Another Ping-Pong argument.(→)

(1) (∀x)(∀y)A (Hyp.)

(2) (∀y)A ((1) + Spec.)

(3) A ((2) + Spec.)

(4) (∀x)A ((3) + Gen. : Line (1) has no free x)

(5) (∀x)(∀y)A ((4) + Gen. : Line (1) has no free y)

(←) is very similar to the above.


Theorem (∀-Monoticity)

If Γ ` A→ B, then Γ ` (∀x)A→ (∀x)B, provided that x does not occurfree in any formula in Γ

Proof.

(1) A→ B (Γ− theorem)

(2) (∀x)(A→ B) ((1) + Gen : x does not occur free in Γ)

(3) (∀x)(A→ B)

→ (∀x)A→ (∀x)B (Ax3)

(4) (∀x)A→ (∀x)B ((2, 3) + MP)


Corollary

If ` A→ B, then ` (∀x)A→ (∀x)B.


Theorem

If Γ ` A ≡ B, then Γ ` (∀x)A ≡ (∀x)B, provided that x does not occurfree in any formula in Γ

Proof.

(1) A ≡ B (Γ− theorem)

(2) A→ B ((1) |=taut (2),Ax1)

(3) B → A ((1) |=taut (3),Ax1)

(4) (∀x)A→ (∀x)B ((2) + ∀ −mon)

(5) (∀x)B → (∀x)A ((3) + ∀ −mon)

(6) (∀x)A ≡ (∀x)B ((4.5) |=taut (6),Ax1)


Corollary

If ` A ≡ B, then ` (∀x)A ≡ (∀x)B.

An example of substitution. Suppose that we have a formula x = y andwant to use substitution to obtain y = x . Note that

(x = y)[x := y ][y := x ] is x = x

and(x = y)[y := x ][x := y ] is y = y

One way to obtain y = x via a substitution from x = y is to introduce twonew variables z and w and do:

(x = y)[x := z ][y := w ][z := y ][w := x ] is y = x


Definition (Simultaneous Substitution)

The expressionA[x1, x2, . . . , xr := t1, t2, . . . , tr ]

denotes the simultaneous substitution of the terms t1, . . . , tr into thevariables x1, . . . , xr and it is defined as

A[x1 := z1] . . . [xr := zr ][z1 := t1] . . . [zr := tr ]

where z1, . . . , zr are distinct new variables that do not occur in any ofA, t1, . . . , tr at all.

• In simultaneous substitution, because z1, . . . , zr do not occur in any ofA, t1, . . . , tr , in the sequence of the substitutions the previous substitutionsdo not have any effect on the next one. Therefore simultaneoussubstitution is order independent.


Theorem (Substitution Theorem)

If ` A and t1, . . . , tr are any terms, then ` A[x1, x2, . . . , xr := t1, t2, . . . , tr ].

Proof.

If ` A then ` (∀x)A (a metatheorem). Then by the specialization rule wehave ` A[x := z1]. Apply this argument 2r times to get

` A[x1 := z1] . . . [xr := zr ][z1 := t1] . . . [zr := tr ]

which is` A[x1, x2, . . . , xr := t1, t2, . . . , tr ]


Theorem

If Γ ` A and there is a Γ-proof for A which none of the formulas appearingin that proof have free occurrences of x1, . . . , xr , thenΓ ` A[x1, . . . , xr := t1, . . . , tr ].

Proof.

Suppose C1, . . . ,Cn is a subset of Γ which x1, . . . , xr do not occur free inthem and C1, . . . ,Cn ` A. Then by applying the deduction theorem ntimes, we get ` C1 → . . .Cn → A. By the substitution theorem we get

` (C1 → · · · → Cn → A)[x1 := z1] . . . [xr := zr ][z1 := t1] . . . [zr := tr ]

but since none of xi or zi appear free in C1, . . . ,Cn we have

` C1 → · · · → Cn → A[x1 := z1] . . . [xr := zr ][z1 := t1] . . . [zr := tr ]


proof continues.

Apply MP n times to get

C1, . . . ,Cn ` A[x1 := z1] . . . [xr := zr ][z1 := t1] . . . [zr := tr ]

By strengthening hypotheses we have

Γ ` A[x1 := z1] . . . [xr := zr ][z1 := t1] . . . [zr := tr ]


Theorem (Weak Leibniz–”WL”)

If ` A ≡ B then ` C [p\A] ≡ C [p\B].

Proof.

The proof is by induction on the complexity of C .Basis. Complexity zero C . There are two cases to consider.(1) C is p. Then we have to show that if ` A ≡ B then ` A ≡ B, which istrivial.(2) C is not p. Then C is one of the q (not p), t = s, φ(t1, . . . , tn), >, ⊥.Then we must show if ` A ≡ B then ` C ≡ C . But the latter is a theoremso it holds by (Ax1).Inductive step. (i) C is ¬D. By I.H. we have ` D[p\A] ≡ D[p\B]. Thenwe have ` ¬D[p\A] ≡ ¬D[p\B] by tautological implication (verify thatA ≡ B is tautologically equivalent to ¬A ≡ ¬B, for any formulas A,B).Thus by the definition ` (¬D)[p\A] ≡ (¬D)[p\B].


proof continued.

(ii) C is D ◦ E , where ◦ ∈ {∧,∨,→,≡}: By I. H. ` D[p\A] ≡ D[p\B] and` E [p\A] ≡ E [p\B]. Hence by tautological implication` D[p\A] ◦ E [p\A] ≡ D[p\B] ◦ E [p\B] (check the tautologicalimplication!). Thus by definition A ≡ B ` (D ◦ E )[p\A] ≡ (D ◦ E )[p\B].(iii) C is (∀x)D. By I.H. we have ` D[p\A] ≡ D[p\B]. Therefore by atheorem that we proved, we have ` (∀x)D[p\A] ≡ (∀x)D[p\B]. Again bythe definition ` ((∀x)D)[p\A] ≡ ((∀x)D)[p\B].

Note.

In order to apply WL, A ≡ B has to be an absolute theorem. We do NOThave A ≡ B ` D[p\A] ≡ D[p\B] in general.


Here is a more generous WL theorem which allows A ≡ B to be a relativetheorem.

Theorem (Weak Leibniz–”WL”)

If Γ ` A ≡ B and none of the bound variables of C occur free in any of theformulas in Γ, then Γ ` C [p\A] ≡ C [p\B].

Proof.

The proof is again by induction on C . The basis and inductive steps areexactly the same as the previous WL theorem, replacing ` with Γ ` in theproof. Only the case where C is (∀x)D is different. By I.H. we haveΓ ` D[p\A] ≡ D[p\B]. Since x is bound in C , by our assumption it doesnot occur free in Γ. Hence by a theorem Γ ` (∀x)D[p\A] ≡ (∀x)D[p\B].Again by the definition Γ ` ((∀x)D)[p\A] ≡ ((∀x)D)[p\B].


The next theorem allows us to drop the side condition on BL rule (the onesaying that p can not be in the scope of any quantifiers of C ).

Theorem (Strong Leibniz–”SL”)

A ≡ B ` C [p := A] ≡ C [p := B] (of course only when both C [p := A] andC [p := B] are defined)

Proof.

The proof is by induction on the complexity of C .Basis. Complexity zero C . There are two cases to consider.(1) C is p. Then we have to show that A ≡ B ` A ≡ B, which holds,since it is a theorem from Boolean logic (Ax1).(2) C is not p. Then C is one of the q (not p), t = s, φ(t1, . . . , tn), >, ⊥.Then we must show A ≡ B ` C ≡ C . Since ` C ≡ C , we know thatA ≡ B ` C ≡ C follows from (Ax1).Inductive step. (i) C is ¬D. By I.H. we haveA ≡ B ` D[p := A] ≡ D[p := B]. Then we haveA ≡ B ` ¬D[p := A] ≡ ¬D[p := B] by tautological implication.


proof continued.

Thus by the definition A ≡ B ` (¬D)[p := A] ≡ (¬D)[p := B].(ii) C is D ◦ E , where ◦ ∈ {∧,∨,→,≡}: By I. H.A ≡ B ` D[p := A] ≡ D[p := B] and A ≡ B ` E [p := A] ≡ E [p := B].Hence by tautological implicationA ≡ B ` D[p := A] ◦ E [p := A] ≡ D[p := B] ◦ E [p := B] (check thetautological implication!). Thus by definitionA ≡ B ` (D ◦ E )[p := A] ≡ (D ◦ E )[p := B].(iii) C is (∀x)D. By I.H. we have A ≡ B ` D[p := A] ≡ D[p := B]. SinceC [p := A] and C [p := B] are defined x is not free in either A or B.Therefore x is not free in A ≡ B. Therefore by a theorem that we proved,we have A ≡ B ` (∀x)D[p := A] ≡ (∀x)D[p := B], which again bydefinition (since x is not free in either A or B) we haveA ≡ B ` ((∀x)D)[p := A] ≡ ((∀x)D)[p := B].


Corollary

D → (A ≡ B) ` D → (C [p := A] ≡ C [p := B])

Proof.

By deduction theorem it is enough to prove

D → (A ≡ B),D ` C [p := A] ≡ C [p := B]

Then

(1) D (Hyp.)

(2) D → (A ≡ B) (Hyp.)

(3) A ≡ B ((12) + MP)

(4) C [p := A] ≡ C [p := B] ((3) + SL)


Theorem

` (∀x)(A→ B) ≡ (A→ (∀x)B), provided that x is not free in A.

Proof.

We use a Ping-Pong argument again.(→) We show that ` (∀x)(A→ B)→ (A→ (∀x)B).Applying thededuction theorem twice, this is equivalent to show that(∀x)(A→ B),A ` (∀x)B.

(1) (∀x)(A→ B) (Hyp.)

(2) A (Hyp.)

(3) A→ B ((1) + Spec)

(4) B ((2, 3) + MP)

(5) (∀x)B ((4) + Gen : there are no free x in lines (1,2))


Proof.

(←) We show that ` (A→ (∀x)B)→ (∀x)(A→ B). Applying thededuction theorem, this is equivalent to show thatA→ (∀x)B ` (∀x)(A→ B). Since Hypotheses A→ (∀x)B has no free x ,by the weak generalization theorem, it is enough to proveA→ (∀x)B ` A→ B. Applying the deduction theorem again, this isequivalent to show that A→ (∀x)B,A ` B

(1) A→ (∀x)B (Hyp.)

(2) A (Hyp.)

(3) (∀x)B ((1, 2) + MP)

(4) B ((3) + Spec)


Note.

Of course we can write equational-style proofs in predicate logic too, butin most cases Hilbert-style proofs are more convenient in predicate logic.However, equational-style proofs can prove both sides of a Ping-Pongargument ((→) and (←)) in a single blow. So they shorten this kind ofproofs.


Corollary

` (∀x)(A ∨ B) ≡ A ∨ (∀x)B, given that x is not free in A.

Proof.

(Equational-style)

(∀x)(A ∨ B)

⇐⇒ 〈WL and ` A ∨ B ≡ ¬A→ B (tautology) : C − part is (∀x)p〉(∀x)(¬A→ B)

⇐⇒ 〈theorem, x is not free in A〉¬A→ (∀x)B

⇐⇒ 〈tautology〉A ∨ (∀x)B


Corollary

` (∃x)(A ∧ B) ≡ A ∧ (∃x)B, given that x is not free in A.

Proof.

(∃x)(A ∧ B)

⇐⇒ 〈Definition of ∃〉¬(∀x)¬(A ∧ B)

⇐⇒ 〈WL and deMoegan : C − part is ¬(∀x)p〉¬(∀x)(¬A ∨ ¬B)

⇐⇒ 〈WL and theorem − x is not free in ¬A : C − part is ¬p〉¬(¬A ∨ (∀x)¬B)

⇐⇒ 〈deMorgan〉¬¬A ∧ ¬(∀x)¬B

⇐⇒ 〈WL double negation : C − part is p ∧ ¬(∀x)¬B〉Saeed Ghasemi (York University) Math 1090 5th July 2018 65 / 95

proof continued.

A ∧ ¬(∀x)¬B

⇐⇒ 〈WL definition of ∃ : C − part is A ∧ p〉A ∧ (∃x)B

Note. We might use one sided arrow (⇒) in an equational-style proof. IfA→ B is a (relative) theorem, in some parts of an equational-style proofwe may write

A

⇒〈Annotation〉B


Theorem (Empty range)

` (∀x)(⊥ → A) ≡ >.

Proof.

By redundant true it is enough to prove ` (∀x)(⊥ → A). We have` ⊥ → A (Ax1), so by the weak generalization theorem we conclude that` (∀x)(⊥ → A).


Theorem (Another Distributivity of ∀ over ∧ )

` (∀x)(A→ B) ∧ (∀x)(A→ C ) ≡ (∀x)(A→ B ∧ C ).

Proof.

(∀x)(A→ B) ∧ (∀x)(A→ C )

⇐⇒ 〈Theorem : Distributivity of ∀ over ∧ 〉(∀x)((A→ B) ∧ (A→ C ))

⇐⇒ 〈WL : tautology (Ax1) : C − part is (∀x)p〉(∀x)(A→ B ∧ C )


Theorem

` (∀x)(A ∨ B → C ) ≡ (∀x)(A→ C ) ∧ (∀x)(B → C ).

Proof.

(∀x)(A→ C ) ∧ (∀x)(B → C )

⇐⇒ 〈Theorem〉(∀x)((A→ C ) ∧ (B → C ))

⇐⇒ 〈WL : tautology (Ax1) : C − part is (∀x)p〉(∀x)(A ∨ B → C )


Theorem (Dummy Renaming for ∀)

If z does not occur in A, then ` (∀x)A ≡ (∀z)A[x := z ].

Proof.

We use Ping-Pong. Note that since z is fresh in A, A[x := z ] is defined.(→)

(1) (∀x)A→ A[x := z ] (Ax2)

(2) (∀z)(∀x)A→ (∀z)A[x := z ] (∀ −monot)

(3) (∀x)A→ (∀z)(∀x)A (Ax4) : z is not free in (∀x)A)

(4) (∀x)A→ (∀z)A[x := z ] ((3) + Transitivity)


Proof.

(←)

(1) (∀z)A[x := z ]→ A[x := z ][z := x ] (Ax2)

(2) (∀z)A[x := z ]→ A (A[x := z ][z := x ] is A)

(3) (∀x)(∀z)A[x := z ]→ (∀x)A ((2) + ∀ −Mon)

(4) (∀z)A[x := z ]→ (∀x)A ((3) + Ax4))

Theorem (Dummy Renaming for ∃)

If z does not occur in A, then ` (∃x)A ≡ (∃z)A[x := z ].


Inserting and Removing ∃

Theorem (Dual of Ax2)

` A[x := t]→ (∃x)A

Proof.

A[x := t]→ (∃x)A

⇐⇒ 〈WL + ∃ − def : C − part is A[x := t]→ (∃x)A〉A[x := t]→ ¬(∀x)¬A

⇐⇒ 〈Tautology〉(∀x)¬A→ ¬A[x := t]


Corollary (Dual of Specialization Rule)

A[x := t] ` (∃x)A

Corollary

A ` (∃x)A

Theorem (Introduction of ∀)

If x does not occur free in Γ and A, then Γ ` A→ B iff Γ ` A→ (∀x)B.

Proof.

Suppose Γ ` A→ B, then by ∀-monotonicity theorem we haveΓ ` (∀x)A→ (∀x)B, since x is not free in Γ. And since x is not free in Awe have ` A→ (∀x)A (Ax4). Therefore Γ ` A→ (∀x)B.


proof continued.

Suppose Γ ` A→ (∀x)B. By Ax2 we have (∀x)B ` B. By transitivityΓ + A ` B. Applying the deduction theorem we conclude thatΓ ` A→ B.

Theorem (Introduction of ∃)

If x does not occur free in Γ and B, then Γ ` A→ B iff Γ ` (∃x)A→ B.

Proof.

First assume Γ ` A→ B.

A→ B

⇐⇒ 〈Tautology〉¬B → ¬A

⇐⇒ 〈Introduction of ∀〉¬B → (∀x)¬A


proof continued.

⇐⇒ 〈Tautology〉¬(∀x)¬A→ B

⇐⇒ 〈∃ − definition〉(∃x)A→ B

Now assume Γ ` (∃x)A→ B. By a theorem we have ` A→ (∃x)A.Therefore Γ + A ` (∃x)A and Γ + A ` (∃x)A→ B. By transitivityΓ + A ` B. Therefore Γ ` A→ B.


Theorem (Auxiliary Variable Metatheorem)

Assume that Γ ` (∃x)A and Γ ` A[x := z ]→ B, where z is a freshvariable with respect to Γ, A and B, then Γ ` B.

Proof.

From Introduction of ∃ theorem we have Γ ` (∃z)A[x := z ]→ B, becausez is fresh with respect to Γ and B.

(∃z)A[x := z ]→ B

⇐⇒ 〈SL : Dummy renaming for ∃ : z is fresh : C − part is p → B〉(∃x)A→ B

⇐⇒ 〈SL : (∃x)A is a Γ− theorem : C − part is p → B〉> → B

⇐⇒ 〈Tautology〉B


• Auxiliary Variable Metatheorem is very natural and used frequently inmathematics and computer science. It basically says that suppose we wantto prove B using some hypotheses Γ, and we also know that from Γ wecan deduce (∃x)A, meaning that for some x , A(x) holds, that can behelpful towards proving B. But we may not exactly know or care for whichx , A(x) holds! But we can just call this value of x by z . Then we can addA(z) to our hypotheses Γ and hope for an easier proof for B. If we proveB from Γ using this “auxiliary” assumption A(z), then we have proved itusing only Γ, since (∃x)A was provable from Γ.• For example, we use Auxiliary Variable Metatheorem in induction. Whenwe want to show that P(n) holds for arbitrary number n, using someassumptions Γ, we use Induction Hypotheses and add an extra assumptionthat P(k) is true for all k < n. Now we try to prove P(n) usingassumptions Γ + P(k) for all k < n.


Corollary

Assume that ` (∃x)A and moreover A[x := z ] ` B, where z is fresh withrespect to A and B, then ` B.

Corollary

Assume that A[x := z ] ` B, where z is fresh with respect to A and B,then (∃x)A ` B.

Auxiliary Variable Metatheorem is our ticket to remove ∃.


Example. Prove that ` (∃x)(∀y)A→ (∀y)(∃x)A.Proof 1. By the deduction theorem it is enough to prove(∃x)(∀y)A ` (∀y)(∃x)A.

(1) (∃x)(∀y)A (Hyp.)

(2) (∀y)A[x := z ] (Auxiliary hypothesis associated to (1): z is fresh)

(3) A[x := z ] ((2) + Spec.)

(4) (∃x)A ((3) + Dual of Spec .)

(5) (∀y)(∃x)A ((4) + Gen : no free y in lines (1, 2))


Proof 2.

(1) A→ (∃x)A (abs.theorem)

(2) (∀y)A→ (∀y)(∃x)A (∀ −mon.)

(3) (∃x)(∀y)A→ (∀y)(∃x)A (Introduction of ∃.)


Example. Prove that (∃x)(A→ B), (∀x)A ` (∃x)B.

(1) (∃x)(A→ B) (Hyp.)

(2) (∀x)A (Hyp.)

(3) A[x := z ]→ B[x := z ] (Aux. hypothesis associated to (1): z fresh)

(4) A[x := z ] ((2) + Ax2)

(5) B[x := z ] ((3, 4) + MP)

(6) (∃x)B ((5) + Dual of Spec .)


Properties of Equality

We explore the consequences of Ax5 and Ax6.

Lemma (symmetry of =)

` x = y → y = x

Proof.

(1) x = y → (x = x ≡ y = x) (Ax6.)

(2) x = y → x = x → y = x (Tautological Implication)

(3) x = x (Ax5)

(4) x = y → y = x ((2, 3) + tautological Implication)


Lemma (Transitivity of =)

` x = y → y = z → x = z

Proof.

(1) x = y → (y = z ≡ x = z) (Ax6.)

(2) x = y → y = z → x = z (Tautological Implication)


Lemma

Suppose f is a function symbol of arity n. Then

` x = y → f (z1, . . . , zi , x , zi+2, . . . , zn) = f (z1, . . . , zi , y , zi+2, . . . , zn)

Proof.

Apply Ax6 for A being

f (z1, . . . , zi , x , zi+2, . . . , zn) = f (z1, . . . , zi , y , zi+2, . . . , zn)

then we get

` x = y →(f (z1, . . . , zi , x , zi+2, . . . , zn) = f (z1, . . . , zi , y , zi+2, . . . , zn) ≡

f (z1, . . . , zi , y , zi+2, . . . , zn) = f (z1, . . . , zi , y , zi+2, . . . , zn)

We have ` f (z1, . . . , zi , y , zi+2, . . . , zn) = f (z1, . . . , zi , y , zi+2, . . . , zn) byAx5. Then the tautological implication A→ B ≡ C ,B ` A→ C impliesthat


proof continued.

` x = y → f (z1, . . . , zi , x , zi+2, . . . , zn) = f (z1, . . . , zi , y , zi+2, . . . , zn)

Corollary

Suppose f is a function symbol of arity n. Then

` x1 = y1 → · · · → xn = yn → f (x1, . . . , xn) = f (y1, . . . , yn)

Proof.

By the deduction theorem we need to prove

x1 = y1, . . . , xn = yn `→ f (x1, . . . , xn) = f (y1, . . . , yn)


proof continued.

By the previous lemma we have

x1 = y1 `→ f (x1, x2 . . . , xn) = f (y1, x2 . . . , xn)

and

x1 = y1, x2 = y2 `→ f (x1, x2, x3 . . . , xn) = f (y1, y2, x3 . . . , xn)

and so on. After n steps we have

x1 = y1, . . . , xn = yn `→ f (x1, . . . , xn) = f (y1, . . . , yn)


Corollary

Suppose f is a function symbol of arity n and ti and si are terms. Then

` t1 = s1 → · · · → tn = sn → f (t1, . . . , tn) = f (s1, . . . , sn)

Proof.

By the last corollary we have

` x1 = y1 → · · · → xn = yn → f (x1, . . . , xn) = f (y1, . . . , yn)

Let A be the formulax1 = y1 → · · · → xn = yn → f (x1, . . . , xn) = f (y1, . . . , yn). By thesubstitution theorem we have` A[x1, . . . , xn, y1, . . . yn := t1, . . . tn, s1, . . . , sn]. This is exactly what wehad to show.


Theorem

Suppose t, t ′, s are terms. Then ` t = t ′ → s[x := t] = s[x := t ′]

Proof.

Proof is by induction on complexity of the term s.Basis Case(1) s is a constant or a variable other than x . Then we have toshow that ` t = t ′ → s = s, which is trivial.Case(2) s is x . Then we have to show that ` t = t ′ → t = t ′, which isalso trivial.Inductive step. Suppose s is f (t1, . . . , tn) and the theorem is true fort1, . . . , tn, i.e., ` t = t ′ → ti [x := t] = ti [x := t ′] for each i = 1, . . . , n.We need to show thatt = t ′ ` f (t1, . . . , tn)[x := t] = f (t1, . . . , tn)[x := t ′].


proof continued.

(1) t = t ′ (Hyp.)

(2) t1[x := t] = t1[x := t ′] ((1) + I .H.+ MP)

(3) t2[x := t] = t2[x := t ′] ((1) + I .H.+ MP)

......

...

(n + 1) tn[x := t] = tn[x := t ′] ((1) + I .H.+ MP)

(n + 2) f (t1[x := t], . . . , tn[x := t]) =

f (t1[x := t ′], . . . , tn[x := t ′]) ((1)− (n + 1) + Previous lemma)


First-Order Logic Semantics-very briefly

In order to give meanings to the so far meaningless formulas we interpretthe logical symbols as actual mathematical objects. This interpretation isnot unique and it is up to us.For a first order language, an interpretation is a pair D = (D,M), where Dis a nonempty set, called the domain or underlying set of the interpretationand M is called translator, which is a mapping that assigns an appropriatemathematical objects to each of the symbols of the language. Instead ofwriting M(...) we write ...D.


Definition (Interpreting a Language-Step 1: Translating the Alphabet)

Suppose D = (D,M) is an interpretation of a fixed first order language.

1 For each free variable x , the interpretation or translation of x ,denoted by xD (or M(x)), is some member of D.

2 For each Boolean variable p, the interpretation of p, pD is somemember of {t, f }.

3 >D = t and ⊥D = f .

4 For each object constant c of the alphabet, the translation cD issome member of D.

5 For each function symbol f of the alphabet, the translation f D is amathematical function in metatheory with the fame arity as f . f D

takes as inputs elements from D and its output value is also andelement of D.

6 For each predicate symbol φ of the alphabet, the translation φD is amathematical relation in metatheory with the fame arity as φ. phiD

takes as inputs elements from D and its output value is an element of{t, f }.

Note. Boolean connectives, = and brackets are not translated, they keeptheir meaning.


Note.

Boolean connectives, = and brackets are not translated, they keep theirmeaning.

Definition (Interpreting a Language-Step 2: Translating the Formulas)

Suppose D = (D,M) is an interpretation of a fixed first order languageand suppose that A is a formula in that language. The interpretation of Avia D is denoted by AD and constructed as follows.Logical symbols

1 We replace any occurrence of > and ⊥ with t and f , respectively.

2 We replace a Boolean variable p in A by its interpretation pD, whichis either t or f .

3 We replace any free variable x in A with the specific value xD from D.

4 We replace any (∀x) in A with (∀x ∈ D), which means for all valuesof x in D.


Non-logical symbols

1 We replace any object constant c in A with the specific value cD

from D.

2 We replace any function symbol f with the specific function f D,which has inputs from D and output from D

3 We replace any predicate symbol φ with the specific relation φD,which has inputs from D and output from {t, f }.

Definition

If AD = t for some A and interpretation D, we say A is true in theinterpretation D or D is a model of A. We denote AD by

|=D A


Definition (Universally-or Logically-valid formulas)

A first-order formula is Universally valid or just valid if |=D A holds for“every” interpretation D. Then we write

|= A

Lemma

If |=taut A then |= A

Lemma

If A is an axiom, then |= A


Theorem (Soundness in First-Order logic)

If ` A then |= A.

Theorem (Godel’s Completeness Theorem)

If |= A then ` A.


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Math 1090 Part two: Predicate (First-Order) Logicsaeedgh/Math1090/PL.pdf · 2018. 10. 10. · 3 =...

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