Part A: Propositional Logic

Truth AssignmentsEnderton Pg. 20-24

First, read the handout on wffs, induction and recursion.

Definition: A truth assignment for a set of sentence symbols is a function

assigning either F or T to each symbol in .

Now let be the set of wffs that can be built up from by the five formula-buolding

operations. We want an extension of , ie. which assigns the correct truth

value to each wff in . See pg. 20-21 Enderton for the rules. You should already know

them.

1.

Truth TablesEnderton Pg. 24-26

See Problem Set problems for truth table questions.

2.

TautologiesEnderton Pg. 24-26

Let be the empty set. It is vacuously true that any truth assignment satisfies every

member of the empty set. Hence we are left with iff every truth assignment for the

sentence symbols in will satisfy . We thus call a tautology.

3.

Tautological ImplicationEnderton Pg. 24-26

Definition: tautologically implies (written ) iff every truth assignment for the

sentence symbols in and that satisfies every member of also satisfies .

If and , then are tautologically equivalent.

4.

Part B: First Order Logic

First Order LanguagesEnderton Pg. 67-79

Assume that we are given the following objects of a language: 1) "Logical Symbols"

(parentheses, sentential connective symbols, infinitely many variables , the

equality symbol ) and 2) "Parameters" (the quantifier symbol , predicate symbols,

constant symbols and function symbols. From these we may construct "first order

languages".

Definition: An expression is any finite sequence of symbols. From these, there are two

types of expressions: terms and wffs. Terms are expressions that can be built up from

the constant symbols and the variables by applying (zero or more times) the n-place

building operations on expressions ( ).

Example: terms in the language of number theory are: .

An atomic formula is an expression of the form where is an n-place predicate

symbol and terms. Example: atomic since predicate.

Definition: The well-formed formulas are expressions that can be built up from the

atomic formulas by use (zero or more times) of the connective symbols and the

quantifier symbols.

1.

Models and SatisfactionEnderton Pg. 80-89

Models/Structures are the first-order-logic analogues of the sentential-logic idea of

"truth assignments". Basically, since we are given the logical symbols and parameters, a

model/structure will give an "interpretation" for these symbols.

Let be a wff of our language and a structure for our first-order language. Let

map all variables in our first-order language into the universe of objects

in the structure. We then have holds iff the translation of determined by is true. (where the variable is translated as wherever it occurs free (ie,

unbounded by a quantifier).Read: pg 83-85 Enderton for more details.

Theorem 22A: Assume that and are functions from into , which agree at all

variables (if any) that occur free in the wff. Then iff .

Corollary 22B: For a sentence , either satisfies with every function from into

, or does not satisfy with any such function.

Definition: Logical Implication: Let be a set of wffs, a wff. Then logically implies , ie. iff for every structure for the language and every function such

that satisfies every member of with , also satisfies with .

2.

Exam Notes for Mathematical Logic

Page 1

that satisfies every member of with , also satisfies with .

Corollary 22C: For a set of sentences, iff every model of is also a model of

. A sentence is valid iff it is true in every structure.

Non-Standard ModelsEnderton Pg. 147-158

Enderton Pg. 187-190

Enderton Pg. 202-210

See Problem Set solutions.

3.

DeductionsEnderton Pg. 109-128

See Problem Set solutions.

4.

Soundness & CompletenessEnderton Pg. 131-145

Read your notes.

5.

CompactnessEnderton Pg. 59-60

Compactness for sentential logic:

Theorem: Let be an infinite set of wffs such that for any finite subset of , there is a

truth assignment that satisfies every member of . Then there is a truth assignment

that satisfies every member of .

6.

Effective ProceduresEnderton Pg. 61-65

See Problem set & Midterm solutions.

7.

Decidability & Effective EnumerabilityEnderton Pg. 61-65

8.

Definability in a StructureEnderton Pg. 90-92

See Midterm Question 2.

9.

Equivalence RelationsSee Problem Sets

10.

Gödel's Incompleteness TheoremSee Handout

11.

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Practice Problems & Solutions

Assignment #01

#1 - §1.1, #2 - Show that there are no wffs of length 2, 3, or 6, but that any other positive

length is possible.

We denote the length of an expression by . Note that appying the first formula-

building operation to gives a wff of length , and the other four operations

give one of length .

Consider the following examples, where are any sentence symbols. By indiction,

we have:

wff length example

1 4 5 7 8 9 Apply any of the last four operations to a wff of length Hence we proved that there are wffs of any length 2, 3, 6.

#2 - §1.1, #3 - Let be a wff; let be the number of places at which binary connective

symbols ( ) occur in . Let be the number of places at which sentence symbols

occur in . Show by using the induction principle that .

Induct on the length of .

Base Case: , ie. , where is a sentence symbol. , so Inductive step: Suppose for all wff with , where is the

number of places at which sentence symbols occur in , etc. We look at the ancestral

tree of : It must be constructed either from one wff (where ) by applying

one of the other four formula-building operations.

In the first case, , .

In the second case: .

But , by induction hypothesis. Hence in both cases,

#3 - §1.2, #3 - (a) Determine whether or not is a tautology.

(b) Determine whether or not tautologically implies

.

(a) T T T T

T F F T F T T

F T T T

F T F T

observe that any truth value assignment over the sentence symbols in satisfies . Hence it is indeed a tautology.

For (b): See Assignment #01 printed.

Yes. Given a formula (with parentheses replaced by |), we prove by induction on the

#4 - §1.3, #7 - Suppose that left and right parentheses are indistinguishable. Thus, instead of

, we have . Do formulas still have unique decomposition?

Page 3

Yes. Given a formula (with parentheses replaced by |), we prove by induction on the

length of .

Base case: . is a molecular sentence, which clearly has a unique

decomposition (Note that the | next to here denotes the length, and are not those |

replacing the parenthesis.)

Assume now that every formula (with parentheses replaced by |) of length smaller than

has a unique decomposition. Considering the formula-building operations, we see

that either , or , where is a binary connective symbol.

Respectively, we could decompose into and , or into only, which in turn

themselves have unique decomposition by the inductive hypothesis. Therefore the

decomposition of is unique.

PART B: Read and understand the proof of the Recursion Theorem on page 42. Briefly

explain where freeness is used.

See Assignment #01 printed. TA solution: 'By freeness, an element is precisely

one of the following: , for a unique pair where , or

for unique . So can be built up step-by-step without conflict.

Assignment #02

(a) By Theorem 17E: "A set of expressions is effectively enumerable iff there exists

an effective procedure that, given any expression , produces the answer "yes" iff "

hence a set is effectively enumerable iff it is semidecidable iff there exists an effective

semidecidable procedure. Let be effectively enumerable, with corresponding

effective semidecidable procedures respectively. Let be a procedure that

executes steps of and those of alternatively. (ie. Do step 1 of followed by step

1 of followed by step 2 of , etc) and give "yes" if either or gives "yes". is an

effective procedure since both and are, (check definition of effective procedure)

and says "yes" for an expression whenever . Hence is semidecidable,

and hence is effectively enumerable.

(b) Let be the same as above, except that says "yes" only if both and say "yes".

#5 - §1.7, #11 - (a) Explain why the union of two effectively enumerable sets is again

effectively enumerable (b) Explain why the intersection of two effectively enumerable sets is

again effectively enumerable.

#6 - §1.7, #12 - For each of the following conditions, give an example of an unsatisfiable set of formulas that meets the condition.

(a) Each member of is -by itself- satisfiable.

(b) For any two members and of , the set { , } is satisfiable.

(c) For any three members , and of , the set { , , } is satisfiable.

TA answer: Let . This works for (a)-(c).

#7 - §2.1, #1 - Assume that we have a language with the following parameters: , intended to

mean "for all things"; , intended to mean "is a number"; , intended to mean "is interesting";

<, intended to mean "is less than"; and 0, a constant symbol intended to denote zero.

Translate into this language the English sentences listed below. If the English sentence is

ambiguous, you will need more than one translation.

(a) Zero is less than any number.

(b) If any number is interesting, then zero is interesting.

(c) No number is less than zero.

(d) Any uninteresting number with the property that all smaller numbers are interesting

certainly is interesting.

(e) There is no number such that all numbers are less than it.

(f) There is no number such that no number is less than it.

Page 4

(f) There is no number such that no number is less than it.

(a) (b) " " or " "

(c) (d) (e) (f)

TA Answer:

#8 - §2.1, #3 - Translate the following sentence into the first-order language specified. Make

full use of the notational conventions and abbreviations to make the end result as readable as

possible.

"Neither nor is a member of every set".

( , for all sets, , is a member of; , ; , .)

(a) (b) (c)

#9 - §2.1, #9 - Assume that the language has equality and a two-place predicate symbol . For

each of the following conditions, find a sentence such that the structure is a model of iff the condition is met.

(a) | | has exactly two members.

(b) is a function from | | into | |

(c ) is a permutation of | | (ie. is a one-to-one function with domain and range equal

to | |

Midterm

#10 - 1 - State the Compactness Theorem for sentential logic.

A set of wffs is satisfiable iff it's finitely satisfiable. Ie. Every finite subset of is

satisfiable.

#11 - 2 - Prove that each of the sets below is definable in the given structure by providing a

formula in first-order logic that defines the set:

is prime : is even in in

#12 - 3 - (a) Explain under which conditions a set of expressions is effectively enumerable

(b) Briefly explain why the union of two effectively enumerable sets of expressions is again

effectively enumerable

(a) is effectively enumerable iff it is semidecidable, iff there exists an effective

semidecision procedure

(b) Let be effectively enumerable, with corresponding effective semidecidable

procedures respectively. Let be a procedure that executes steps of and

those of alternatively. (ie. Do step 1 of followed by step 1 of followed by step

2 of , etc) and give "yes" if either or gives "yes". is an effective procedure

since both and are, (check definition of effective procedure) and says "yes" for

an expression whenever . Hence is semidecidable, and hence is effectively enumerable.

Page 5

#13 - 4 - Let and be sentence symbols and let be wffs in propositional logic. (a)

Prove that but and (b) Prove that if or , then .

(a) Consider the truth table:

T T T T F

T F F F

F T F T

F F T T T

There are two evaluations, namely , such that . And note that

, but and . Hence the

result holds.

(b) Consider an evaluation function such that . It is sufficient to prove

. If then . Otherwise, . In either case,

.

#14 - 5 - The length of a wff of sentential logic is the number of symbols in the formula

including parentheses, sentence symbols and logical connectives. (a) Show that if , are

wffs of length and respectively with , then the length of is of

the form

for some . (b) Supose is a wff not containing the negation symbol. Show by

induction on formulas that the length of is of the form , where

(a) Length of where (b) Prove by induction on complexity of .

Base case: where is a sentential symbol. Length of Induction: where length of , length of , then we

have a length equal to the case in (a). Length of =

length of where

, .

#15 - 6 - Let be a set of wffs of sentential logic such that 1) Every finite subset of is

satisfiable, and 2) For every wff , either or . Define the truth assignment

if , if } for each sentence symbol . Show by induction that for

every wff , iff . For this problem, you may assume that we only use the

connectives .

Prove by induction on the complexity of .

Case 1: iff , result follows by definition of and that for

sentential symbols Case 2: where the statement holds for and .

ie.: iff , iff .

Note that iff iff and (by inductive

hypothesis) iff . We justify this last step below:

Proof by contradiction:

If then by (ii), so is satisfiable

by (i), but this is clearly not true.

If but then by (ii), so is

satisfiable by (i), but this is clearly not true. Hence . By a similar argument, we

also have .

Case 3: and the statement holds for :

iff iff (by inductive hypothesis) iff by (ii).

Hence all cases hold, and the result follows.

Page 6

Hence all cases hold, and the result follows.

Assignment #03

(a) : definable by the formula (b) : definable by the formula

(c ) is the successor of in : definable by the formula

for and representing and respectively and representing "1" in .

TA answer: , meaning

(d) in : definable by the formula

for and representing and respectively and any element of .

TA answer: ,

meaning " for some

#16 - §2.2, #11 - For each of the following relations, give a formula which defines it in

. (The language is assumed to have equality and the parameters , and ).

(a) Give a formula that defines in the interval .

One such formula is . Indeed, consider the graph of ,

which has domain . TA: "Squares are non-negative"

(b) Give a formula that defines in the set .First, we need to define {1}.

A formula which defines {1} is Hence a formula which defines {2} is

TA answer: meaning " is the

nonzero root of .

(c) Show that any finite union of intervals, the endpoints of which are algebraic, is

definable in .

TA answer: A heuristic argument. We may define the number 1 as in above, and get

any natural number from this definition by successive addition of 1. Also, we can get

all the negative integers. Ie. can be defined by since we

already defined .

An 'algebraic number' is definable: suppose is the nth root (ordered by ) of the

polynomial . Wlog, the coefficients of are integers. Thus, as we can define

every integer, we are able to define and hence .

We can define intervals with algebraic endpoints: For simplicity, we directly write instead of using the formulas defining them. Write instead of using the

definition, instead of using the definition, etc. Hence is defined by

. Similarly for .

We can define unions of two definable sets: Suppose are sets that are defined by

the formulas respectively. Then is defined by .

For a finite union of definable sets, we repeat the construction of the union of two

definable sets finitely many times.

Thus, any finite union of intervals with algebraic endpoints is definable in

#17 - §2.2, #12 - Let be the structure . (The language is assumed to have equality

and the parameters , and . is the structure whose universe is the set of real numbers

and such that and are the usual addition and multiplication operations.).

#18 - §2.4, #2 - To which axiom groups, if any, do each of the following formulas belong?

(See Axiom groups, pg. 112 Enderton)

Page 7

(a) (Group 1)

(a) Suppose ⊢ .

Then by the Generalization Theorem:

⊢ Recall from Axiom Group 3, we have: " "This, together with the following, gives a deduction from to :

1 - 2 - ; by MP

This, together with the deduction of , gives a deduction of from . So ⊢ .

(b) It is not true in general that ⊨ :

Consider with .

Then since

On the other hand: but it is not true that

(b) (Group 2, where )

(c) Not an axiom. Note that is not substitutable for in

#19 - §2.4, #6 - (a) Show that if ⊢ then ⊢ (b) Show that it is not in general true that ⊨

To show ⊢ :First, we prove ⊢ , and then prove the other direction.

By 'Deduction', it is sufficient to prove And by 'Contraposition', it is sufficient to prove Starting the proof:

; (Group 1)

; (Group 3)

; MP

; Hypothesis

; MP

; (Group 1)

; (Group 3)

; MP

; MP

Similarly, we can get

(where is just ) ; (Group 2)

; (Group 1)

; MP

Hence we proved and thus the result holds.

To show , note that the following are equivalent:

So the result holds immediately.

#20 - §2.4, #8 - (Q2b) Assume that does not occur free in . Show that

. Also show that, under the same assumption,

we have Q3a: .

Page 8

Assignment #04

Note that (if is consistent, then is satisfiable) is equivalent to (If is unsatisfiable,

then is inconsistent), which can be written as: (If then ).

We prove (If then ) (If then Clear, by setting .

Note the following:

If then ;

If then by deduction theorem.

However, , so by MP. Ie. We have (if then )

Thus, if implies , then implies .

Hence it is sufficient to have which we have by assumption.

For an elaborate answer, see Assignment #04 printout.

#21 - §2.5, #2 - Prove the equivalence of parts (a) and (b) of the completeness theorem.

#22 - §2.5, #8 - Assume the language (with equality) has just the parameters and , where

is a two-place predicate symbol. Let be the structure with | | , the set of integers

(positive, negative, and zero) and with iff . Thus looks like an

infinite graph:

Show that there is an elementarily equivalent structure that is not

connected.

Add constants to the language. Let:

"Any path of length from does not go to "

"Any path of length from does not go to "

Consider .Note that is finitely satisfiable: A given finite subset of is

satisfied by the structure , (where is exactly the same as except in that we added

the constants to the language), with

being integers that are sufficiently far

apart. Hence, by the Compactness Theorem, is itself satisfiable. So there exists a

model for . Now, reduce to the original language (without adding the constants

). We can check then that for any sentence , iff . Hence is

elementarily equivalent to .

Note: Being connected means that for every two members of , there is a path between

them. A path - of length - from to is a sequence with ,

and for each .)

#23 - §2.6, #7 Consider a language with a two-place predicate symbol , and let

Add constants to the language. Define sentences, for :

and let .As in the previous question, we let be the same model as except that we add

countably many constants to the language. Note that is finitely

satisfiable: Suppose is finite. Then but for .

Let be the same as except in that we also have

for all . Hence

is satisfied by , and by the Compactness Theorem, is satisfiable.

Let be a model of , and be the reduction of to the original language

. We now prove that .

Since , we have that for every sentence , if , then .

Suppose that ⊭ , then , so . Hence , therefore

be the structure consisting of the natural numbers with their usual ordering.

Show that there is some elementarily equivalent to such that has a descending chain.

(That is, there must be in | | such that for all .)

Page 9

Suppose that ⊭ , then , so . Hence , therefore

⊭ .

Suppose for every finite number , there is a model for such that is not true in

and that has at least elements. We aim for a contradiction.

For , let "There are at least elements"

Then is satisfiable. We prove this:

Suppose finite with , but for .

Then has a model , so is satisfiable. Thus by the Compactness Theorem, is

satisfiable.

Hence there is a model of . , so must be an infinite model.

Thus contradicts the fact that is true in every infinite model of .

#24 - §2.6, #8 Assume that is true in all infinite models of a theory . Show that there is a

finite such that is true in all models of for which | | has or more elements.

Let be the relation given by iff for all

1)

Let be the relation given by iff for all

2)

Let be the relation given by iff for all

3)

1) is the set of all such that , notice that

only satisfies this relation for real numbers,

so is geometrically the point at the origin.

is the set of all such that . This

equivalence class is hence the set of points in that form a circle of radius 5 about

the origin.

2) is the set of all such that , notice that

only satisfies this relation for real numbers,

so is geometrically the point at the origin.

is the set of all such that . This

equivalence class is hence the set of points in that form a lopsided square about the

origin, with vertices .

3) is the set of all such that

, notice that only satisfies this relation for real numbers, so is geometrically the point at the origin.

is the set of all such that

. This equivalence class is hence the set of points in

that forms a square about the origin with vertices

.

#25 - Additional Problem: For each of the following equivalence relations on , give a

geometric description of and , the equivalence classes of and ,

respectively.

Assignment #05

#26 - §3.1,#1 - Let be the set of sentences consisting of S1, S2, and all sentences of the

form , where is a wff (in the language of )

in which no variable except occurs free. Show that . Conclude that

.

(Here is by definition . The sentence displayed above is called the induction axiom

for )

Page 10

Let S1,S2 be:

S1. , S2. and let be defined as above. Let

for

S3. ,

S4.1 ,

S4.2 …

S4. We prove that S3 and S4. ( are consequences of

.

First, we prove that : Let wff.

We present a deduction of S3:

; Tautology (This is )

; (By )

; MP

; Tautology

; Tautology

; Tautology

(note, this is )

; MP (This is , ie. S3)

We now prove that S4. : Let wff.

We present a deduction of S4. :

; By assumption S1 (note that this is )

; By S( )

; MP

; By S2 and Axiom Group 1

; Contraposition (Note, this is )

; MP

Hence we've shown that . For each , S( ) holds in ,

so .

Therefore , and hence

.

However, , so

.

#27 - §3.3, #2 - Prove Theorem 33C:

We prove by induction on "the complexity of ". That is, we consider all cases :

ie. All

possible quantifier-free sentences. In each case, assume is true in .

Case 1: Recall Lemma 33B: For any variable-free term , there is a unique natural number Such that . Hence, unique such that and

.

We know that , so we must have . Hence clearly .

Case 2: Let be as the above case. We know that , so we must have .

Now recall Lemma 33A: 1) ,

2) , Hence we must have that case 2) holds, and in particular, as , we have:

. Hence Case 3: By Axiom L3 in : , we have either or

. These cases hold given Case 2 holds above.

Case 4: By Axiom L3 as above, we must have or . These cases hold given the

cases above hold.

Case 5: Since , we have and . So by inductive hypothesis, and

"For any quantifier-free sentence true in , ."

Page 11

Since , we have and . So by inductive hypothesis, and

. Hence as desired.

Case 6: is proved similarly.

Case 7: Recall Theorem 15D: Both and are complete sets of connective symbols,

hence is a complete set of connective symbols. The case that is atomic is

considered above. We may assume that one of the following holds:

1. , Proof: Let . Ie. , so .

By inductive hypothesis, , so , ie. .

2. , Proof: Let , so . So either or is true in . By inductive hypothesis, (note that and are less complex than

), we have or . Ie. , so . A similar

proof holds for .

Hence for all possible cases.

Addition Problems For Assignment 5

#28 - 1 Let be a model of . For in define the relation:

Note that we are not necessarily in the standard model , so it's not

necessary that for some , and is not as simple as saying

or (a) We prove is reflexive, transitive and symmetric.

Reflexive: Apply to 0-many times to get .

Symmetric: The definition of is symmetric in both and .

Transitive: Suppose and .

Case 1: and in . Then in .

Case 2: and . Wlog . Then in .

(This can be proved using S2).

Case 3: and in . Wlog . Then in .

Case 4: and in . Then in .

Since all cases hold, is transitive, and hence is an equivalence relation.

(b) Only one.

iff can be applied a finite number of times to one of , to reach the other.

(a) Prove that is an equivalence relation.

(b) If is the standard model , how many equivalence classes are there?

#29 - 2 Define a relation in by iff .

(a) Check that is an equivalence relation.

(a) Reflexivity and symmetry are clear. We shall check transitivity:

Suppose and . We hence have and .

Multiplying the two gives . Since , we cancel on both sides and

get . If , we cancel on both sides and get as required.

Otherwise, so . But so . Similarly ,

thus as required.

(b) Suppose . We prove

. We have . It is sufficient

then to prove that . Note that so our result holds.

(b) be the quotient of by the equivalence relation .

That is, is the set of equivalence classes of wrt . Define by

, where denotes the equivalence class of . Show that is

well defined.

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