NOTES ON SET THEORY - euclid.colorado.edueuclid.colorado.edu/~monkd/full.pdf · NOTES ON SET THEORY J. Donald Monk March 6, 2018 TABLE OF CONTENTS LOGIC 1. Sentential logic.....1

NOTES ON SET THEORYJ. Donald MonkMarch 6, 2018

TABLE OF CONTENTS

LOGIC

1. Sentential logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12. First-order logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .163. Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294. The completeness theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

ELEMENTARY SET THEORY

5. The axioms of set theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 776. Elementary set theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .807. Ordinals, I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 898. Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 949. Ordinals, II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10310. The axiom of choice. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12711. Cardinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

MODELS OF SET THEORY

12. The set-theoretical hierarchy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16113. Absoluteness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .18114. Checking the axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19115. Reflection theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19716. Consistency of no inaccessibles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20417. Constructible sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .205

INFINITE COMBINATORICS

18. Real numbers in set theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22119. The Cichon diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27220. Continuum cardinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29121. Linear orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31622. Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35423. Clubs and stationary sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39024. Infinite combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42325. Martins axiom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43426. Large cardinals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476

FORCING

27. Boolean algebras and forcing orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50528. Generic extensions and forcing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53829. Forcing and cardinal arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56230. General theory of forcing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .577

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31. Iterated forcing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60632. p = t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 690

PCF

33. Cofinality of posets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72534. Basic properties of PCF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .75935. Main cofinality theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 781

ADDITIONAL SECTIONS

36. Various forcing orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .80537. More examples of iterated forcing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82838. Consistency results concerning P()/fin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 838

REAL NUMBERS

39. The integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84740. The rationals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85441. The reals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 860Index of symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .870Index of words . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .875

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1. Sentential logic

We go into the mathematical theory of the simplest logical notions: the meaning of and,or, implies, if and only if and related notions. The basic idea here is to describe aformal language for these notions, and say precisely what it means for statements in thislanguage to be true. The first step is to describe the language, without saying anythingmathematical about meanings. We need very little background to carry out this develop-ment. is the set of all natural numbers 0, 1, 2, . . .. Let + be the set of all positiveintegers. For each positive integer m let m = {0, . . . , m 1}. A finite sequence is afunction whose domain is m for some positive integer m; the values of the function canbe arbitary.

To keep the treatment strictly mathematical, we will define the basic symbols ofthe language to just be certain positive integers, as follows:

Negation symbol: the integer 1.Implication symbol: the integer 2.Sentential variables: all integers 3.

Let Expr be the collection of all finite sequences of positive integers; we think of thesesequences as expressions. Thus an expression is a function mapping m into +, for somepositive integer m. Such sequences are frequently indicated by 0, . . . , m1. The casem = 1 is important; here the notation is .

The one-place function mapping Expr into Expr is defined by = 1 for anyexpression . Here in general is the sequence followed by the sequence .

The two-place functionmapping ExprExpr into Expr is defined by = 2for any expressions , . (For any sets A,B, A B is the set of all ordered pairs (a, b)with a A and b B. So Expr Expr is the set of all ordered pairs (, ) with , expressions.)

For any natural number n, let Sn = n+ 3.Now we define the notion of a sentential formulaan expression which, suitably inter-

preted, makes sense. We do this definition by defining a sentential formula construction,which by definition is a sequence 0, . . . , m1 with the following property: for eachi < m, one of the following holds:

i = Sj for some natural number j.

There is a k < i such that i = k.

There exist k, l < i such that i = (k l).

Then a sentential formula is an expression which appears in some sentential formula con-struction.

The following proposition formulates the principle of induction on sentential formulas.

Proposition 1.1. Suppose that M is a collection of sentential formulas, satisfying thefollowing conditions.

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(i) Si is in M , for every natural number i.(ii) If is in M , then so is .(iii) If and are in M , then so is .

Then M consists of all sentential formulas.

Proof. Suppose that is a sentential formula; we want to show that M . Let0, . . . , m be a sentential formula construction with t = , where 0 t m. We proveby complete induction on i that for every i m, i M . Hence by applying this to i = twe get M .

So assume that for every j < i, the sentential formula j is in M .Case 1. i is Ss for some s. By (i), i M .Case 2. i is j for some j < i. By the inductive hypothesis, j M , so i M by

(ii).Case 3. i is j k for some j, k < i. By the inductive hypothesis, j M and

k M , so i M by (iii).

Proposition 1.2. (i) Any sentential formula is a nonempty sequence.(ii) For any sentential formula , exactly one of the following conditions holds:

(a) is Si for some i .(b) begins with 1, and there is a sentential formula such that = .(c) begins with 2, and there are sentential formulas , such that = .

(iii) No proper initial segment of a sentential formula is a sentential formula.(iv) If and are sentential formulas and = , then = .(v) If , , , are sentential formulas and = , then = and

= .

Proof. (i): Clearly every entry in a sentential formula construction is nonempty, so(i) holds.

(ii): First we prove by induction that one of (a)(c) holds. This is true of sententialvariablesin this case, (a) holds. If it is true of a sentential formula , it is obviously trueof ; so (b) holds. Similarly for , where (c) holds.

Second, the first entry of a formula differs in cases (a),(b),(c), so exactly one of themholds.

(iii): We prove this by complete induction on the length of the formula. So, supposethat is a sentential formula and we know for any formula shorter than that no properinitial segment of is a formula. We consider cases according to (ii).

Case 1. is Si for some i. Only the empty sequence is a proper initial segment of in this case, and the empty sequence is not a sentential formula, by (i).

Case 2. is for some formula . If is a proper initial segment of and it is aformula, then begins with 1 and so by (ii), has the form for some formula . Butthen is a proper initial segment of and is shorter than , so the inductive hypothesisis contradicted.

Case 3. is for some formulas and . That is, is 2. If is a properinitial segment of which is a formula, then by (ii), has the form 2 for someformulas , . Now = , so is an initial segment of or is an initial segment

2

of . Since and are both shorter than , it follows from the inductive hypothesis that = . Hence = , and = , contradiction.

(iv) is rather obvious; if = , then and are both the sequence obtained bydeleting the first entry.

(v): Assume the hypothesis. Then is the sequence 2, and isthe sequence 2

. Since these are equal, and start at the same place in thesequence. By (iii) it follows that = . Deleting the initial segment 2 from thesequence, we then get = .

Parts (iv) and (v) of this proposition enable us to define values of sentential formulas,which supplies a mathematical meaning for the truth of formulas. A sentential assignmentis a function mapping the set {0, 1, . . .} of natural numbers into the set {0, 1}. Intuitivelywe think of 0 as false and 1 as true. The definition of values of sentential formulas isa special case of definition by recursion:

Proposition 1.3. For any sentential assignment f there is a function F mapping the setof sentential formulas into {0, 1} such that the following conditions hold:

(i) F (Sn) = f(n) for every natural number n.(ii) F () = 1 F () for any sentential formula .(iii) F ( ) = 0 iff F () = 1 and F () = 0.

Proof. An f -sequence is a finite sequence (0, 0), . . . , (m1, m1) such that eachi is 0 or 1, and such that for each i < m one of the following holds:

(1) i is Sn for some n , and i = f(n).

(2) There is a k < i such that i = k and i = 1 k.

(3) There are k, l < i such that i = k l, and i = 0 iff k = 1 and l = 0.

Now we claim:

(4) For any sentential formula and any f -sequences (0.0), . . . , (m1, m1) and(0.

0), . . . , (

n1,

n1) such that m1 =

n1 = we have m1 =

n1.

We prove (4) by induction on , thus using Proposition 1.1. If = Sn, then m1 = f(n) =n1. Assume that the condition holds for , and consider . There is a k < m 1such that = m1 = k. By Proposition 1.2(iv) we have k = . Similarly, thereis an l < n 1 such that = n1 =

l and so

l = . Applying the inductive

hypothesis to and the sequences 0, . . . , k and 0, . . . , l we get k =

l. Hence

m1 = 1 k = 1 l = n1.

Now suppose that the condition holds for and , and consider . There arek, l < m 1 such that ( ) = (k l). By Proposition 1.2(v) we have k = and l = . Similarly there are s, t < n 1 such that ( ) = (s

t). By

Proposition 1.2(v) we have s = and t = . Applying the inductive hypotheis to

and the sequences 0, . . . , k and 0, . . . ,

s we get k =

s. Similarly, we get l =

t.

Hence

m1 = 0 iff k = 1 and l = 0

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iff s = 1 and t = 0

iff n1 = 0.

This finishes the proof of (4).

(5) If 0, . . . , m1 is a sentential formula construction, then there is an f -sequence ofthe form (0, 0), . . . , (m1, m1).

We prove this by induction on m. First suppose that m = 1. Then 0 must equal Sn forsome n, and (0, f(n)) is as desired. Now suppose that m > 1 and the statement is true

for m 1. So let def= (0, 0), . . . .(m2, m2) be an f -sequence.

Case 1. m1 = Sp. Then (m1, f(p)) is as desired.

Case 2. There is a k < m such that m1 = k. Then (m1, 1 k) is asdesired.

Case 3. There are k, l < m such that m1 = k l. Then (m1, ) is asdesired, where = 0 iff k = 1 and l = 0.

Thus (5) holds. Now we can define the function F required in the Proposition. Let be any sentential formula. Let 0, . . . , m1 be a sentential formula construction suchthat m1 = . By (5), let (0, 0), . . . , (m1, m1) be an f -sequence. We defineF () = m1. This is unambiguous by (4).

Case 1. = Sn for some n. Then by the definition of f -sequence we have F () =f(n).

Case 2. There is a k < m such that = m1 = k. Then (0, 0), . . . , (k, k)is an f -sequence, so F (k) = k. So

F () = F (m1) = m1 = 1 k = 1 F (k).

Case 3. There are k, l < m such that = m1 = k l. Then (0, 0), . . . ,(k, k) is an f -sequence, so F (k) = k; and (0, 0), . . . , (l, k) is an f -sequence, soF (l) = l. So

F () = 0 iff F (m1) = 0 iff em1 = 0 iff

k = 1 and l = 0 iff F (k) = 1 and F (l) = 0.

With f a sentential assignment, and with F as in this proposition, for any sententialformula we let [f ] = F (). Thus:

Si[f ] = f(i);

()[f ] = 1 [f ];

( )[f ] =

{0 if [f ] = 1 and [f ] = 0,

1 otherwise.

The definition can be recalled by using truth tables:

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1 0

0 1

1 1 1

1 0 0

0 1 1

0 0 1

Other logical notions can be defined in terms of and . We define

= ( ). = . = ( ) ( ).

Working out the truth tables for these new notions shows that they mean approximatelywhat you would expect:

1 1 0 0 1 0 1 1 1 1

1 0 1 1 0 0 1 0 1 0

0 1 0 1 0 1 1 1 0 0

0 0 1 1 0 1 0 1 1 1

(Note that corresponds to non-exclusive or: or , or both.)

The following simple proposition is frequently useful.

Proposition 1.4. If f and g map {0, 1, . . .} into {0, 1} and f(m) = g(m) for every msuch that Sm occurs in , then [f ] = [g].

Proof. Induction on . If is Si for some i, then the hypothesis says that f(i) = g(i);hence Si[f ] = f(i) = g(i) = Si[g]. Assume that it is true for . Now Sm occurs in iff it occurs in . Hence if we assume that f(m) = g(m) for every m such thatSm occurs in , then also f(m) = g(m) for every m such that Sm occurs in , so()[f ] = 1 [f ] = 1 [g] = ()[g]. Assume that it is true for both and , andf(m) = g(m) for every m such that Sm occurs in . Now if Sm occurs in , then italso occurs in , and hence f(m) = g(m). Similarly for . It follows that

( )[f ] = 0 iff ([f ] = 1 and [f ] = 0) iff ([g] = 1 and [g] = 0) iff ( )[g] = 0.

This proposition justifies writing [f ] for a finite sequence f , provided that f is long enoughso that m is in its domain for every m for which Sm occurs in .

A sentential formula is a tautology iff it is true under every assignment, i.e., [f ] = 1for every assignment f .

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Here is a list of common tautologies:

(T1) .(T2) .(T3) ( ) .(T4) ( ) ( ).(T5) ( ).(T6) ( ) [( ) ( )].(T7) [ ( )] [( ) ( )].(T8) ( ) ( ).(T9) ( ) .(T10) ( ) .(T11) [ ( )].(T12) ( ).(T13) ( ).(T14) ( ) [( ) (( ) )].(T15) ( ) ( ).(T16) ( ) ( ).(T17) [ ( )] [( ) ].(T18) [ ( )] [( ) ].(T19) [ ( )] [( ) ( )].(T20) [ ( )] [( ) ( )].(T21) ( ) ( ).(T22) ( ).(T23) ( ).

Now we describe a proof system for sentential logic. Formulas of the following form aresentential axioms; , , are arbitrary sentential formulas.

(1) ( ).

(2) [ ( )] [( ) ( )].

(3) ( ) ( ).

Proposition 1.5. Every sentential axiom is a tautology.

Proof. For (1):

( )

1 1 1 1

1 0 1 1

0 1 0 1

0 0 1 1

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For (2): Let denote this formula:

( ) ( ) ( )

1 1 1 1 1 1 1 1 1

1 1 0 0 0 1 0 0 1

1 0 1 1 1 0 1 1 1

1 0 0 1 1 0 0 1 1

0 1 1 1 1 1 1 1 1

0 1 0 0 1 1 1 1 1

0 0 1 1 1 1 1 1 1

0 0 0 1 1 1 1 1 1

For (3):

(3)

1 1 0 0 1 1 1

1 0 0 1 1 1 1

0 1 1 0 0 0 1

0 0 1 1 1 1 1

If is a collection of sentential formulas, then a -proof is a finite sequence 0, . . . , msuch that for each i m one of the following conditions holds:

(a) i is a sentential axiom.

(b) i .

(c) There exist j, k < i such that k is j i. (Rule of modus ponens, abbreviated MP).

We write if there is a -proof with last entry . We also write in place of .

Proposition 1.6. (i) If , f is a sentential assignment, and [f ] = 1 for all ,then [f ] = 1.

(ii) If , then is a tautology.

Proof. For (i), let 0, . . . , m be a -proof. Suppose that f is a sentential assign-ment and [f ] = 1 for all . We show by complete induction that i[f ] = 1 for alli m. Suppose that this is true for all j < i.

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Case 1. i is a sentential axiom. Then i[f ] = 1 by Proposition 1.5.Case 2. i . Then i[f ] = 1 by assumption.Case 3. There exist j, k < i such that k is j i. By the inductive assumption,

k[f ] = j[f ] = 1. Hence i[f ] = 1.(ii) clearly follows from (i),

Now we are going to show that, conversely, if is a tautology then . This is a kindof completeness theorem, and the proof is a highly simplified version of the proof of thecompleteness theorem for first-order logic which will be given later.

Lemma 1.7. .

Proof.(a) [ [( ) ]] [[ ( )] ( )] (2)(b) [( ) ] (1)(c) [ ( )] ( ) (a), (b), MP(d) ( ) (1)(e) (c), (d), MP

Theorem 1.8. (The deduction theorem) If {} , then .

Proof. Let 0, . . . , m be a ( {})-proof with last entry . We replace each iby several formulas so that the result is a -proof with last entry .

If i is a logical axiom or a member of , we replace it by the two formulas i ( i), i.

If i is , we replace it by the five formulas in the proof of Lemma 1.7; the last entryis .

If i is obtained from j and k by modus ponens, so that k is j i, we replacei by the formulas

[ (j i)] [( j) ( i)]

( j) ( i)

i

Clearly this is as desired.

Lemma 1.9. ( ).

Proof. By axiom (1) we have {,} . Hence axiom (3) gives {,} , and hence {,} . Now two applications of Theorem 1.8 give the desiredresult.

Lemma 1.10. ( ) [( ) ( )].

Proof. Clearly { , , } , so three applications of Theorem 1.8 give thedesired result.

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Lemma 1.11. ( ) .

Proof. Clearly { ,} and also { ,} , so by Lemma 1.9,{( ,} ( ). Then Theorem 1.8 gives { } ( ), andso using axiom (3), { } ( ) . Hence by Lemma 1.7, { } , andso Theorem 1.8 gives the desired result.

Lemma 1.12. ( ) [( ) ].

Proof.

{ , ,} using axiom (1)

{ , ,} using axiom (3)

{ , ,} using Lemma 1.10

{ , ,} by Lemma 1.11

{ , ,}

{ , } by Theorem 1.8

{ , } by Lemma 1.11

Now two applications of Theorem 1.8 give the desired result.

Theorem 1.13. There is a sequence 0, 1, . . . containing all sentential formulas.

Proof. One can obtain such a sequence by the following procedure.

(1) Start with S0.

(2) List all sentential formulas of length at most two which involve only S0 or S1; they areS0, S1, S0, and S1.

(3) List all sentential formulas of length at most three which involve only S0, S1, or S2;they are S0, S1, S2, S0, S1, S2, S0, S1, S2, S0 S0, S0 S1, S0 S2,S1 S0, S1 S1, S1 S2, S2 S0, S2 S1, S2 S2.

(4) Etc.

Theorem 1.14. If not( ), then there is a sentential assignment f such that [f ] = 1for all , while [f ] = 0.

Proof. Let 0, 1, . . . list all the sentential formulas. We now define 0,1, . . . byrecursion. Let 0 = . Suppose that i has been defined. If not(i {i} ) then weset i+1 = i {i}. Otherwise we set i+1 = i.

Here is a detailed proof that exists. Let M = { : is a function with domain m

for some positive integer m, 1 = , and for every positive integer i with i + 1 m wehave

i+1 =

{i {i} if not(i {i} ),i otherwise.}

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(1) If , M with domains m, n respectively, with m n, then i m[i = i].

This is easily proved by induction on i.

(2) For every positive integer m there is a M with domain m.

Again this is easily proved by induction on m.Now we define i = i, where M and i < dmn(). This is justified by (1) and

(2).Now it is easily verified that the defining conditions for hold.Let =

i i. By induction we have not(i ) for each i . In fact, wehave 0 = , so not(0 ) by assumption. If not(i ), then not(i+1 ) byconstruction.

Hence also not( ), since means that there is a -proof with last entry ,and any -proof involves only finitely many formulas i, and they all appear in some j ,giving j , contradiction.

() For any formula i, either i or i .

In fact, suppose that i / and i / . Say i = j . Then by construction,i {i} and j {i} . So {i} and {i} . Henceby Theorem 1.8, i and i . So by Lemma 1.12 we get ,contradiction.

() If , then .

In fact, clearly not( {} ) by Theorem 1.8, so () follows.Now let f be the sentential assignment such that f(i) = 1 iff Si . Now we claim

( ) For every sentential formula , [f ] = 1 iff .

We prove this by induction on . It is true for = Si by definition. Now suppose that itholds for . Suppose that ()[f ] = 1. Thus [f ] = 0, so by the inductive assumption, / , and hence by (), . Conversely, suppose that . If ()[f ] = 0,then [f ] = 1, hence by the inductive hypothesis. Hence by Lemma 1.9, ,contradiction. So ()[f ] = 1.

Next suppose that () holds for and ; we show that it holds for . Supposethat ( )[f ] = 1. If [f ] = 1, then by the inductive hypothesis. By axiom(1), . Hence by (), ( ) . Suppose that [f ] = 0. Then [f ] = 0also, since ( )[f ] = 1. By the inductive hypothesis and () we have . Hence by axiom (1), so by axiom (3). So ( ) by ().

Conversely, suppose that ( ) . Working for a contradiction, suppose that( )[f ] = 0. Thus [f ] = 1 and [f ] = 0. So and by the inductivehypothesis and (). Since ( ) and , we get . Since also , weget by Lemma 1.9, contradiction.

This finishes the proof of ( ).Since , ( ) implies that [f ] = 1 for all . Also [f ] = 0 since

/ .

Corollary 1.15. If [f ] = 1 whenever [f ] = 1 for all , then .

10

Proof. This is the contrapositive of Theorem 1.14.

Theorem 1.16. iff is a tautology.

Proof. is given by Proposition 1.6(ii). follows from Corollary 1.15 by taking = .

Proposition 1.17.S0 S1 = 2, 3, 1, 4

and(S0 S1) (S1 S0) = 2, 2, 3, 4, 2, 1, 4, 1, 3.

Proof.

S0 S1 = 2S0 S1

= 231S1

= 2, 3, 1, 4;

(S0 S1) (S1 S0) = 2(S0 S1)

(S1 S0)

= 22S0 S1 2

S1 S0

= 2, 2, 3, 4, 21S1 1S0

= 2, 2, 3, 4, 2, 1, 4, 1, 3.

Proposition 1.18. There is a sentential formula of each positive integer length.

Proof. If m is a positive integer, then

m1 times

1, 1, . . . , 1, S0

is a formula of length m, it ism1 times S0.

Proposition 1.19. m is the length of a sentential formula not involving iff m is odd.

Proof. : We prove by induction on that if is a sentential formula not involving, then the length of is odd. This is true of sentential variables, which have length 1.Suppose that it is true of and , which have length 2m+1 and 2n+1 respectively. Then , which is 1, has length 1 + 2m+ 1 + 2n+ 1 = 2(m+ n+ 1) + 1, which isagain odd. This finishes the inductive proof.. We construct formulas without with length any odd integer by induction. S0

is a formula of length 1. If has been constructed of length 2m+ 1, then S0 , whichis 1, S0, has length 2m+ 3. This finishes the inductive construction.

11

Proposition 1.20. The truth table for a sentential formula involving n basic formulashas 2n rows.

Proof. We prove this by induction on n. For n = 1, there are two rows. Assume thatfor n basic formulas there are 2n rows. Given n+ 1 basic formulas, let be one of them.For the others, by the inductive hypothesis there are 2n rows. For each such row thereare two possibilities, 0 or 1, for . So for the n + 1 basic formulas there are 2n 2 = 2n+1

rows.

Proposition 1.21. The formula

( ) ( )

is a tautology.

Proof.

( ) ( )

1 1 1 0 1 1

1 0 0 0 0 1

0 1 1 1 1 1

0 0 1 1 1 1


[ ( )] [( ) ( )]

is a tautology.

Proof. Let be the indicated formula.

( ) ( ) ( )

1 1 1 1 1 1 1 1 1

1 1 0 1 1 1 0 1 1

1 0 1 1 1 1 0 1 1

1 0 0 1 1 1 0 1 1

0 1 1 1 1 1 1 1 1

0 1 0 1 0 0 0 0 1

0 0 1 0 1 0 0 0 1

0 0 0 0 0 0 0 0 1

12


( ) ( )

is not a tautology.

Proof.

( ) ( )

1 1 1 0 0 0

Proposition 1.24. The following is a tautology:

S0 (S1 (S2 (S3 S1))).

Proof. Suppose that f is an assignment making the indicated formula false; we worktowards a contradiction. Thus

(1) S0[f ] = 1 and

(2) (S1 (S2 (S3 S1)))[f ] = 0.

From (2) we get

(3) S1[f ] = 1 and

(4) (S2 (S3 S1))[f ] = 0.

From (4) we get

(5) S2[f ] = 1 and

(6) (S3 S1)[f ] = 0.

From (6) we get S1[f ] = 0, contradicting (3).

Proposition 1.25. The following is a tautology.

({[( ) ( )] } ) [( ) ( )].

Proof. Suppose that f is an assignment which makes the given formula false; wewant to get a contradiction. Thus we have

(1) ({[( ) ( )] } )[f ] = 1 and

(2) [( ) ( )][f ] = 0.

By (2) we have

(3) ( )[f ] = 1 and

13

(4) ( )[f ] = 0.

By (4) we have

(5) [f ] = 1 and

(6) [f ] = 0.

By (3) and (6) we get

(7) [f ] = 0.

By (1) and (7) we get

(8) {[( ) ( )] }[f ] = 0.

It follows that

(9) [( ) ( )][f ] = 1 and

(10) [f ] = 0.

Now by (6) we have

(11) ( )[f ] = 1,

and hence by (9),

(12) ( )[f ] = 1.

By (5) we have

(13) ()[f ] = 0,

and hence by (12),

(14) ()[f ] = 0.

This contradicts (10).

Proposition 1.26. The following statements are logically consistent: If the contract isvalid, then Horatio is liable. If Horation is liable, he will go bankrupt. Either Horatio willgo bankrupt or the bank will lend him money. However, the bank will definitely not lendhim money.

Proof. Let S0 correspond to the contract is valid, S1 to Horatio is liable, S2 toHoratio will go bankrupt, and S3 to the bank will lend him money. Then we want tosee if there is an assignment of values which makes the following sentence true:

(S0 S1) (S1 S2) (S2 S3) S3.

We can let f(0) = f(1) = f(2) = 1 and f(3) = 0, and this gives the sentence the value1.

Proposition 1.27. {} .

14

Proof. Following the proof of Lemma 1.9, the following is a {,}-proof:

(a) (b) ( ) (1)(c) (a), (b), MP(d) ( ) ( ) (3)(e) (c), (d), MP(f) (g) (e), (f), MP

Now applying the proof of the deduction theorem, the following is a {}-proof:

(a) [ [( ) ]] [[ ( )] ( )] (2)

(b) [( ) ] (1)(c) [ ( )] ( ) (a), (b), MP(d) ( ) (1)(e) (c), (d), MP(f) [ ( )] [ [ ( )]] (1)(g) ( ) (1)(h) [ ( )]] (f), (g), MP(i) [( ) ( )] [ [( ) ( )]] (1)(j) ( ) ( ) (3)(k) [( ) ( )] (i), (j), MP(l) [ [( ) ( )]] [[ ( )]

[ ( )]] (2)(m) [ ( )] [ ( )] (k), (l), MP(n) ( ) (g), (m), MP(o) ( ) (1)(p) (q) (o), (p), MP(r) [ ( )] [( ) ( )] (2)(s) ( ) ( ) (n), (r), MP(t) (q), (s), MP

15

2. First-order logic

Although set theory can be considered within a single first-order language, with only non-logical constant , it is convenient to have more complicated languages, corresponding tothe many definitions introduced in mathematics.

All first-order languages have the following symbols in common. Again, as for senten-tial logic, we take these to be certain natural numbers.

1 (negation)2 (implication)3 (the equality symbol)4 (the universal quantifier)5m for each positive integer m (variables ranging over elements, but not subsets, of a givenstructure) We denote 5m by vm1. Thus v0 is 5, v1 is 10, and in general vi is 5(i+ 1).

Special first-order languages have additional symbols for the functions and relations andspecial elements involved. These will always be taken to be some positive integers notamong the above; thus they are positive integers greater than 4 but not divisible by 5. Sowe have in addition to the above logical symbols some non-logical symbols:

Relation symbols, each of a certain positive rank.Function symbols, also each having a specified positive rank.Individual constants.

Formally, a first-order language is a quadruple (Rel, Fcn, Cn, rnk) such that Rel, Fcn, Cnare pairwise disjoint subsets of M (the sets of relation symbols, function symbols, andindividual constants), and rnk is a function mapping RelFcn into the positive integers;rnk(S) gives the rank of the symbol S.

Now we will define the notions of terms and formulas, which give a precise formu-lation of meaningful expressions. Terms are certain finite sequences of symbols. A termconstruction sequence is a sequence 0, . . . , m1, m > 0, with the following properties:for each i < m one of the following holds:

i is vj for some natural number j.

i is c for some individual constant c.

i is F0 1

n1 for some n-place function symbol F, with each j equal to kfor some k < j, depending upon j.

A term is a sequence appearing in some term construction sequence. Note the similarityof this definition with that of sentential formula given in Chapter 1.

Frequently we will slightly simplify the notation for terms. Thus we might writesimply vj , or c, or F0 . . . n1 for the above.

The following two propositons are very similar, in statement and proof, to Propositions1.1 and 1.2. The first one is the principle of induction on terms.

Proposition 2.1. Let T be a collection of terms satisfying the following conditions:

16

(i) Each variable is in T .(ii) Each individual constant is in T .(iii) If F is a function symbol of rank m and 0, . . . , m1 T , then also F0 . . . m1

T .

Then T consists of all terms.

Proof. Let be a term. Say that 0, . . . , m1 is a term construction sequenceand i = . We prove by complete induction on j that j T for all j < m; hence T . Suppose that j < m and k T for all k < j. If j = vs for some s, thenj T . If j = c for some individual constant c, then sj T . Finally, suppose thatj is Fk0 . . . kn1 with each kt < j. Then kt T for each t < n by the inductivehypothesis, and it follows that j T . This completes the inductive proof.

Proposition 2.2. (i) Every term is a nonempty sequence.(ii) If is a term, then exactly one of the following conditions holds:

(a) is an individual constant.(b) is a variable.(c) There exist a function symbol F, say of rank m, and terms 0, . . . , m1 such

that is F0 . . . m1.(iii) No proper initial segment of a term is a term.(iv) If F and G are function symbols, say of ranks m and n respectively, and if

0, . . . , m1, 0, . . . , n1 are terms, and if F0 . . . m1 is equal to G0, . . . n1, thenF = G, m = n, and i = i for all i < m.

Proof. (i): This is clear since any entry in a term construction sequence is nonempty.(ii): Also clear.(iii): We prove this by complete induction on the length of a term. So suppose that

is a term, and for any term shorter than , no proper initial segment of is a term.We consider cases according to (ii).

Case 1. is an individual constant. Then has length 1, and any proper initialsegment of is empty; by (i) the empty sequence is not a term.

Case 2. is a variable. Similarly.Case 3. There exist an m-ary function symbol F and terms 0, . . . , m1 such that

is F0 . . . m1. Suppose that is a term which is a proper initial segment of . By(i), is nonempty, and the first entry of is F. By (ii), has the form F0 . . . m1 forcertain terms 0, . . . , m1. Since both 0 and 0 are shorter terms than , and one ofthem is an initial segment of the other, the induction hypothesis gives 0 = 0. Let i < mbe maximum such that i = i. Since is a proper initial segment of , we must havei < m 1. But i+1 and i+1 are shorter terms than and one is a segment of the other,so by the inductive hypthesis i+1 = i+1, contradicting the choice of i.

(iv): F is the first entry of F0 . . . m1 and G is the first entry of G0, . . . n1, soF = G. Then by (ii) we get m = n. By induction using (iii), each i = i.

We now give the general notion of a structure. This will be modified and extended for settheory later. For a given first-order language L = (Rel, Fcn, Cn, rnk), an L -structure isa quadruple A = (A,Rel, F cn, Cn) such that A is a nonempty set (the universe of the

17

structure), Rel is a function assigning to each relation symbol R a rnk(R)-ary relationon A, i.e., a collection of rnk(R)-tuples of elements of A, Fcn is a function assigning toeach function symbol F a rnk(F)-ary opeation on A, i.e., a function assigning a value in Ato each rnk(F)-tuple of elements of A, and Cn is a function assigning to each individual

constant c an element of A. Usually instead of Rel(R), Fcn(F) and Cn(c) we write RA,

FA, and cA.Now we define the meaning of terms. This is a recursive definition, similar to the

definition of the values of sentential formulas under assignments:

Proposition 2.3. Let A be a structure, and a a function mapping into A. (A is theuniverse of A.) Then there is a function F mapping the set of terms into A with thefollowing properties:

(i) F (vi) = ai for each i .

(ii) F (c) = cA for each individual constant c.

(iii) F (F0 . . . m1) = FA(F (0), . . . , F (m1)) for every m-ary function symbol F

and all terms 0, . . . , m1

Proof. An (A, a)-term sequence is a sequence (0, b0), . . . , (m1, bm1) such thateach bi A and for each i < m one of the following conditions holds:

(1) i is vj and bi = aj .

(2) i is c for some individual constant c, and bi = cA.

(3) i = Fk(0) k(n1) and bi = F

A(bk(0), . . . , bk(n1)) for some n-ary function

symbol F and some k(0), . . . , k(n 1) < i.

Now we claim

(4) For any term and any (A, a)-term sequences

(0, b0), . . . , (m1, bm1) and (0, b

0), . . . , (

n1, b

n1)

such that m1 = n1 = we have bm1 = b

n1.

We prove (4) by induction on , thus using Proposition 2.1. If = vi, then bm1 =

ai = bn1. If is an individual constant c, then bm1 = cA = bn1. Finally, if =

F0 p1, then we have:

m1 = Fk(0)

k(p1) and bm1 = FA(bk(0), . . . , bk(p1));

n1 = F l(0)

l(p1) and bm1 = F

A(bl(0), . . . , bl(p1))

with each k(s) and l(t) less than i. By Proposition 2.2(iv) we have k(s) = l(s) for every

s < p. Now for every s < p we can apply the inductive hypothesis to k(s) and thesequences

(0, b0), . . . , (k(s), bk(s)) and (0, b

0), . . . , (

l(s), b

l(s))

18

to obtain bk(s) = bl(s). Hence

bm1 = FA(bk(0), . . . , bk(p1)) = F

A(bl(0), . . . , bl(p1)) = b

n1,

completing the inductive proof of (4).

(5) If 0, . . . , m1 is a term construction sequence, then there is an (A, a)-term sequenceof the form (0, b0), . . . , (m1, bm1.

We prove this by induction on m. For m = 1 we have two possibilities.Case 1. 0 is vj for some j . Then (0, bj) is as desired.

Case 2. 0 is c, an individual constant. Then (0, cA) is as desired.

Now assume the statement for m 1 1. By the induction hypothesis there is an

(A, a)-term sequence of the form def= (0, b0), . . . , (m2, bm2). Then we have three

possibilities:Case 1. m1 is vj for some j . Then (m1, bj) is as desired.

Case 2. m1 is c, an individual constant. Then (m1, cA) is as desired.

Case 3. m1 is Fk(0) k(p1) for some p-ary function symbol F with each

k(s) < i. Then (m1,FA(bk(0), . . . , bk(p1)) is as desired.

So (5) holds.Now we can define F as needed in the Proposition. Let be a term. Let 0, . . . , m1

be a term construction sequence with m1 = . By (5), let (0, b0), . . . , (m1bm1) bean (A, a)-term sequence. Then we define F () = bm1. This definition is unambigu-ous by (4). Now we check the conditions of the Proposition. Let be a term, and let(0, b0), . . . , (m1, bm1) be an (A, a)-term sequence with m1 = .

Case 1. = vj for some j . Then F () = bm1 = aj .

Case 2. = c for some individual constant c. Then F () = bm1 = cA.

Case 3. = F0 p1 with F a p-ary function symbol and each s a term.

Then there exist c(0), . . . , c(p 1) < m 1 such that s = c(s) for every s < p. Then

F (c(s)) = bc(s) = Ac(s) for each s < p, and hence

F () = bm1 = FA(bs(0), . . . , bs(p1)) = F

A(As(0), . . . , As(p1)) = F

A(A0 , . . . , Ap1).

With F as in Proposition 2.3, we denote F () by A(a). Thus

vAi (a) = ai;

cA(a) = cA;

(F0 . . . m1)A(a) = FA(A0 (a), . . . ,

Am1(a)).

Here vi is any variable, c any individual constant, and F any function symbol (of somerank, say m).

19

What A(a) means intuitively is: replace the individual constants and function sym-bols by the actual members of A and functions on A given by the structure A, and replacethe variables vi by coresponding elements ai of A; calculate the result, giving an elementof A.

Proposition 2.4. Suppose that is a term, A is a structure, a, b assignments, and

a(i) = b(i) for all i such that vi occurs in . Then A(a) = A(b).

Proof. By induction on :

cA(a) = cA = cA(b);

vAi (a) = a(i) = b(i) = vAi (b);

(F0 . . . m1)A(a) = FA(A0 (a), . . . ,

Am1(a))

= FA(A0 (b), . . . , Am1(b))

= (F0 . . . m1)A(b).

The last step here is the induction step (many of them, one for each function symbol andassociated terms). The inductive assumption is that a(i) = b(i) for all i for which vi occursin F0 . . . m1; hence also for each j < m, a(i) = b(i) for all i for which vi occurs in j ,so that the inductive hypothesis can be applied.

This proposition enables us to simplify our notation a little bit. If n is such that each

variable occurring in has index less than n, then in the notation A(a) we can just usethe first n entries of a rather than the entire infinite sequence.

We turn to the definition of formulas. For any terms , we define = to be the sequence3 . Such a sequence is called an atomic equality formula. An atomic non-equalityformula is a sequence of the form R0

m1 where R is an m-ary relation symboland 0, . . . m1 are terms. An atomic formula is either an atomic equality formula or anatomic non-equality formula.

We define , a function assigning to each sequence of symbols of a first-order

language the sequence def= 1. is the function assigning to each pair (, ) of

sequences of symbols the sequence def= 2. is the function assigning to

each pair (i, ) with i and a sequence of symbols the sequence videf= 4, 5i+5.

A formula construction sequence is a sequence 0, . . . , m1 such that for each i < mone of the following holds:

(1) i is an atomic formula.

(2) There is a j < i such that i is j

(3) There are j, k < i such that i is j k.

(4) There exist j < i and k such that i is vkj .

A formula is an expression which appears as an entry in some formula construction se-quence.

20

The following is the principle of induction on formulas.

Proposition 2.5. Suppose that is a set of formulas satisfying the following conditions:(i) Every atomic formula is in .(ii) If , then .(iii) If , , then ( ) .(iv) If and i , then vi .

Then is the set of all formulas.

Proof. It suffices to take any formula construction sequence 0, . . . , m1 and showby complete induction on i that i for all i . So, suppose that i < m and j for all j < i. By the definition of formula construction sequence, we have the followingcases.

Case 1. i is an atomic formula. Then i by (i).Case 2. There is a j < i such that i is j . By the inductive hypothesis, j .

Hence by (ii), i .Case 2. There are j, k < i such that i is j k. By the inductive hypothesis,

j and k . Hence by (iii), i .Case 4. There exist j < i and k such that i is vkj . By the inductive

hypothesis, j . Hence by (iv), i .This completes the inductive proof.

Proposition 2.6. (i) Every formula is a nonempty sequence.(ii) If is a formula, then exactly one of the following conditions holds:

(a) is an atomic equality formula, and there are terms , such that is = .(b) is an atomic non-equality formula, and there exist a positive integer m, a

relation symbol R of rank m, and terms 0, . . . , m1, such that is R0 . . . m1.(c) There is a formula such that is .(d) There are formulas , such that is .(e) There exist a formula and a natural number i such that is vi.

(iii) No proper initial segment of a formula is a formula.(iv) (a) If is an atomic equality formula, then there are unique terms , such that

is = .(b) If is an atomic non-equality formula, then there exist a unique positive integer

m, a unique relation symbol R of rank m, and unique terms 0, . . . , m1, such that isR0 . . . m1.

(c) If is a formula and the first symbol of is 1, then there is a unique formula such that is .

(d) If is a formula and the first symbol of is 2, then there are unique formulas, such that is .

(e) If is a formula and the first symbol of is 4, then there exist a unique naturalnumber i and a unique formula such that is vi.

Proof. (i): First note that this is true of atomic formulas, since an atomic formulamust have at least a first symbol 3 or some relation symbol. Knowing this about atomic

21

formulas, any entry in a formula construction sequence is nonempty, since the entry iseither an atomic formula or else begins with 1,2, or 4.

(ii): This is true on looking at any entry in a formula construction sequence: eitherthe entry begins with 3 or a relation symbol and hence (a) or (b) holds, or it begins with1, 2, or 4, giving (c), (d) or (e). Only one of (a)(e) holds because of the first symbol inthe entry.

(iii): We prove this by complete induction on the length of the formula. Thus supposethat is a formula of length m, and for any formula of length less than m, no properinitial segment of is a formula. Suppose that is a proper initial segment of and isa formula; we want to get a contradiction. By (ii) we have several cases.

Case 1. is an atomic equality formula = for certain terms , . Thus is3 . Since is a formula which begins with 3 (since is an initial segment of andis nonempty by (i)), (ii) yields that is 3 for some terms , . Hence = .Thus is an initial segment of or is an initial segment of . By Proposition 2.2(iii) itfollows that = . Then also = , so = , contradiction.

Case 2. is an atomic non-equality formula R0 . . . m1 for some m-ary relationsymbol R and some terms 0, . . . , m1. Then is a formula which begins with R, and sothere exist terms 0, . . . , m1 such that is R0 . . . m1. By induction using Proposition2.2(iii), i = i for all i < m, so = , contradiction.

Case 3. is for some formula . Then 1 is the first entry of , so by (ii) hasthe form for some formula . Thus is a proper initial segment of , contradicting theinductive hypothesis, since is shorter than .

Case 4. is for some formulas , , i.e., it is 2. Then starts with 2,so by (ii) has the form 2 for some formulas , . Now both and are shorterthan , and one is an initial segment of the other. So = by the inductive assumption.Then is a proper initial segment of , contradicting the inductive assumption.

Case 5. is 4, 5(i+ 1) for some i and some formula . Then by (ii), is4, 5(i+ 1) for some formula . So is a proper initial segment of , contradiction.

(iv): These conditions follow from Proposition 2.2(iii) and (iii).

Now we come to a fundamental definition connecting language with structures. Again thisis a definition by recursion; it is given in the following proposition. First a bit of notation.If a : A, i , and s A, then by ais we mean the sequence which is just like aexcept that ais(i) = s.

Proposition 2.7. Suppose that A is an L -structure. Then there is a function G assigningto each formula and each sequence a : A a value G(, a) {0, 1}, such that

(i) For any terms , , G( = , a) = 1 iff A(a) = A(a).

(ii) For each m-ary relation symbol R and terms 0, . . . , m1, G(R0 . . . m1, a) =

1 iff A0 (a), . . . , Am1(a) R

A.

(iii) For every formula , G(, a) = 1G(, a).

(iv) For all formulas , , G( , a) = 0 iff G(, a) = 1 and G(, a) = 0.

(v) For all formulas and any i , G(vi, a) = 1 iff for every s A, G(, ais) = 1.

22

Proof. An (A, a)-formula sequence is a sequence (0, b0), . . . , (m1, bm1) such

that each bs is a function mapping Mdef= {a : a : A} into {0, 1} and for each i < m

one of the following holds:

(1) i is an atomic equality formula = , and a M [bi(a) = 1 iff A(a) = A(a)].

(2) i is an atomic nonequality formula R0 . . . m1, and

a M [bi(a) = 1 iff A0 (a), . . . ,

Am1(a) R

A].

(3) There is a j < i such that i = j , and a M [bi(a) = 1 bj(a)].

(4) There are j, k < i such that i = j k, and a M [bi(a) = 0 iff bj(a) = 1 andbk(a) = 0].

(5) There are j < i and k such that i = vkj , and a M [bi(a) = 1 iff u A[bj(a

ku) = 1]].

Now we claim

(6) For any formula and any (A, a)-formula sequences

(0, b0), . . . , (m1, bm1) and (0, b

0), . . . , (

n1, b

n1)

such that m1 = n1 = we have bm1 = b

n1.

We prove (6) by induction on , thus using Proposition 2.5. First suppose that is anatomic equality formula = . Then the desired conclusion is clear. Similarly for atomicnonequality formulas. Now suppose that is . Then by Proposition 2.6(c) there arej < m and k < n such that = j =

k. By the inductive hypothesis we have bj = b

k,

and hence a M [bm1(a) = 1 bj(a) = 1 bk(a) = bn1(a)], so that bm1 = bn1. Next

suppose that is . Then by Proposition 2.6(d) there are j, k < m 1 such that = j and = k, and there are s, t < n 1 such that = s and =

t. Then bj = b

s

and bk = bt by the inductive hypothesis. Hence for any a M ,

bm1(a) = 0 iff bj(a) = 1 and bk = 0 iff bs = 1 and b

t = 0 iff b

n1(a) = 0.

Thus bm1 = bn1. Finally, suppose that is vk. Then by Proposition 2.6(e) there are

j, s < i such that j = and s = . So by the inductive hypothesis bj = b

s. Hence for

any a M we have

bm1(a) = 1 iff for every u A[bj(aku) = 1]

iff for every u A[bs(aku) = 1]

iff bn1(a) = 1.

Thus bm1 = bn1, finishing the proof of (6).

(7) If 0, . . . , m1 is a formula construction sequence, then there is an (A, a)-formulasequence of the form (0, b0), . . . , (m1, bm1).

23

We prove (7) by induction on m. For m = 1 we have two possibilities.

Case 1. 0 is an atomic equality formula = . Let b0(a) = 1 iff A(a) = A(a).

Case 2. 0 is an atomic nonequality formula R0 . . . , xm1. Let b0(a) = 1 iff

A0 (a), . . . , Am1 R

A.

Now assume the statement in (7) for m 1 1. By the inductive hypothesis there is an

(A, a)-formula sequence of the form def= (0, b0), . . . , (m2, bm2). Then we have these

possibilities for m1.

Case 1. m1 is = for some terms , . Define bm1(a) = 1 iff A(a) = A(a).

Then (m1, bm1) is as desired.Case 2. m1 is R0 . . . p1 for some terms 0, . . . , p1. Define bm1(a) = 1 iff

A0 (a), . . . , Ap1(a) R

A. Then (m1, bm1) is as desired.Case 3. m1 is i with i < m 1. Define bm1(a) = 1 bi(a) for any a. Then

(m1, bm1) is as desired.Case 4. m1 is i j with i, j < m 1. Define bm1(a) = 0 iff bi(a) = 1 and

bj(a) = 0. Then (m1, bm1) is as desired.

Case 5. m1 is vki with i < m 1. Define bm1(a) iff for all u A, bi(aku) = 1.Then (m1, bm1) is as desired.

This completes the proof of (7).Now we can define the function G needed in the Proposition. Let be a formula

and a : A. Let 0, . . . , m1 be a formula construction sequence with m1 = .By (7) let (0, b0), . . . , (m1, bm1) be an (A, a)-formula sequence. Then we defineG(, a) = bm1(a). The conditions in the Proposition are clear.

With G as in Proposition 2.7, we write A |= [a] iff G(, a) = 1. A |= [a] is read: A isa model of under a or A models under a or is satisfied by a in A or holdsin A under the assignment a. In summary:

A |= ( = )[a] iff A(a) = A(b). Here and are terms.

A |= (R0 . . . m1)[a] iff the m-tuple A0 , . . . , Am1 is in the relation R

A. Here R is anm-ary relation symbol, and 0, . . . , m1 are terms.

A |= ()[a] iff it is not the case that A |= [a].

A |= ( )[a] iff either it is not true that A |= [a], or it is true that A |= [a].(Equivalently, iff (A |= [a] implies that A |= [a]).

A |= (vi)[a] iff A |= [ais] for every s A.

We define some additional logical notions:

is the formula ; is called the disjunction of and .

is the formula ( ); is called the conjunction of and .

is the formula ( ) ( ); is called the equivalence between and.

24

Now we want to apply the material of Chapter 1 concerning sentential logic. By definition,a tautology in a first-order language is a formula such that there exist formulas 0, 1, . . .and a sentential tautology such that is obtained from by replacing each symbol Sioccurring in by i, for each i < .

Theorem 2.9. If is a tautology in a first-order language, then holds in every structurefor that language.

Proof. Let A be any structure, and b : A any assignment. We want to showthat A |= [b]. Let formulas 0, 1, . . . , be given as in the above definition. For eachsentential formula , let be the first-order formula obtained from by replacing eachsentential variable Si by i. Thus

is . We define a sentential assignment f by setting,for each i ,

f(i) =

{

1 if A |= i[b],0 otherwise.

Then we claim:

(*) For any sentential formula , A |= [b] iff [f ] = 1.

We prove this by induction on :

If is Si, then is i, and our condition holds by definition. If inductively is , then

is , and

A |= [b] iff not(A |= [b])

iff not( [f ] = 1)

iff [f ] = 0

iff [f ] = 1.

Finally if inductively is , then is , and

A |= [b] iff (A |= [b] implies that A |= [b]

iff [f ] = 1 implies that [f ] = 1

iff [f ] = 1.

This finishes the proof of (*).Applying (*) to , we get A |= [b], i.e., A |= [b].

A language for the structure (,

Proposition 2.11. In any first-order language, the sequence v0, v0 is not a term.

Proof. Suppose that v0, v0 is a term. This contradicts Proposition 2.2(ii).

Proposition 2.12. In the language for (, S, 0,+, ), the sequence +, v0, v1, v2 is not aterm. Here S(i) = i+ 1 for any i .

Proof. Suppose it is a term. By Proposition 2.2(ii)(c), there are terms , such that+, v0, v1, v2 is + . Thus v0, v1, v2 = . So the term v0 is an initial segmentof the term . By Proposition 2.2(iii) it follows that v0 = . Hence v1, v2 = . Thiscontradicts Proposition 2.2(ii).

The structure (, S, 0,+, ) can be put in the general framework of structures as follows.It can be considered to be the structure (,Rel, F cn, Cn) where Rel = , Cn is thefunction with domain {8} such that Cn(8) = 0, and Fcn is the function with domain{6, 7, 9} such that Fcn(6) = S, Fcn(7) = +, and Fcn(9) = .

Proposition 2.13. In the language for the structure (,+), a term has length m iff m isodd.

Proof. First we show by induction on terms that every term has odd length. Thisis true for variables. Suppose that it is true for terms and . Then also + has oddlength. Hence every term has odd length.

Second we prove by induction on m that for all m, there is a term of length 2m+ 1.A variable has length 1, so our assertion holds for m = 0. Assume that there is a term of length 2m+ 1. Then + v0 has length 2m+ 3. This finishes the inductive proof.

Proposition 2.14. In the language for (Q,+, ) the formula def= v1[v0 v1 = v1] is such

that for any a : Q, (Q,+, ) |= [a] iff a0 = 1.

Proposition 2.15. The following formula holds in a structure, under any assignment,iff the structure has at least 3 elements.

v0v1v2((v0 = v1) (v0 = v2) (v1 = v2)).

Proposition 2.16. The following formula holds in a structure, under any assignment,iff the structure has exactly 4 elements.

v0v1v2v3((v0 = v1) (v0 = v2) (v0 = v3) (v1 = v2) (v1 = v3) (v2 = v3)

v4(v0 = v4 v1 = v4 v2 = v4 v3 = v4)).

Proposition 2.17. The following formula in the language for (,


v0 = v1 (Rv0v2 Rv1v2)

is universally valid, where R is a binary relation symbol.

Proof. Let A be a structure and a : A an assignment. Suppose that A |= (v0 =

v1)[a]. Then a0 = a1. Also suppose that A |= Rv0v2[a]. Then (a0, a2) RA. Hence

(a1, a2) RA. Hence A |= Rv1v2[a], as desired.


v0 = v1 v0(v0 = v1)

is not universally valid.

Proof. Consider the structure Adef= (,

3. Proofs

The purpose of this chapter is give the definition of a mathematical proof, and give thesimplest proofs which will be needed in proving the completeness theorem in the nextchapter. Given a set of formulas in a first-order language, and a formula in thatlanguage, we explain what it means to have a proof of from .

The following formulas are the logical axioms. Here , , are arbitrary formulasunless otherwise indicated.

(L1a) ( ).(L1b) [ ( )] [( ) ( )].(L1c) ( ) ( ).(L2) vi( ) (vi vi), for any i .(L3) vi for any i such that vi does not occur in .(L4) vi(vi = ) if is a term and vi does not occur in .(L5) = ( = = ), where , , are terms.(L6) = ( = = ), where , , are terms.(L7) = F0 . . . i1i+1 . . . m1 = F0 . . . i1i+1 . . . m1, where F is an m-aryfunction symbol, i < m, and , , 0, . . . , i1, i+1, . . . m1 are terms.(L8) = (R0 . . . i1i+1 . . . m1 R0 . . . i1i+1 . . . m1), where R is anm-ary relation symbol, i < m, and , , 0, . . . , i1, i+1, . . . m1 are terms.

Theorem 3.1. Every logical axiom is universally valid.

Proof. (L1ac): Universally valid by Theorem 2.9.(L2): Assume that

(1) A |= vi( )[a] and(2) A |= vi[a];

We want to show that A |= vi[a]. To this end, take any b A; we want to show thatA |= [aib]. Now by (1) we have A |= ( )[a

ib], hence A |= [a

ib] implies that A |= [a

ib].

Now by (2) we have A |= [aib], so A |= [aib].

(L3): We prove by induction on that if vi does not occur in , and if a, b : Aare such that a(j) = b(j) for all j 6= i, then A |= [a] iff A |= [b]. This will imply that(L3) is universally valid.

is = . Thus vi does not occur in or in . Then

A |= ( = )[a] iff A(a) = A(a)

iff A(b) = A(b) by Proposition 2.4

iff A |= ( = )[b].

is R0 . . . m1 for some m-ary relation symbol and some terms 0, . . . , m1. We areassuming that vi does not occur in R0 . . . m1; hence it does not occur in any term i.

A |= (R0 . . . m1)[a] iff A0 (a), . . . ,

Rm1(a) R

A

29

iff A0 (b), . . . , Rm1(b) R

A

(by Proposition 2.4)

iff A |= (R0 . . . m1)[b].

is (inductively).

A |= [a] iff not(A |= [a])

iff not(A |= [b]) (inductive hypothesis)

iff A |= [b].

is (inductively).

A |= [a] iff (A |= [a] implies that A |= [a])

iff (A |= [b] implies that A |= [b])

(inductive hypothesis)

iff A |= [b].

is vk (inductively). By symmetry it suffices to prove just one direction. Supposethat A |= [a]; we want to show that A |= [b]. To this end, suppose that u A; we wantto show that A |= [bku]. Since A |= [a], we have A |= [a

ku]. Now k 6= i, since vi does

not occur in . Hence (aku)(j) = (bku)(j) for all j 6= i. Hence A |= [b

ku] by the inductive

hypothesis, as desired.This finishes our proof by induction of the statement made above. Now assume that

A |= [a] and u A; we want to show that A |= [aiu]. This holds by the statement above.This finishes the proof of (L3).(L4): Suppose that is a term and vi does not occur in . To prove that A |=

(vi(vi = ))[a], we want to find u A such that A |= (vi = )[aiu]. Let u = A(a). Then

(vi)A[aiu] = u =

A(a) = A(aiu)

by Proposition 2.4 (since vi does not occur in , hence a(j) = aiu(j) for all j such that vj

occurs in ). Hence A |= (vi = )[aiu].

(L5): Assume that A |= ( = )[a] and A |= ( = )[a]. Then A(a) = A(a) and

A(a) = A(a), so A(a) = A(a), hence A |= ( = )[a].

(L6): Assume that A |= ( = )[a] and A |= ( = )[a]. Then A(a) = A(a) and

A(a) = A(a), so A(a) = A(a), hence A |= ( = )[a].

(L7): Assume that A |= ( = )[a]. Then A(a) = A(a), and so

(F0 . . . i1i+1 . . . m1)A(a) = FA(A0 (a), . . . ,

Ai1(a),

A(a), Ai+1(a), . . . , Am1(a))

= FA(A0 (a), . . . , Ai1(a),

A(a), Ai+1(a), . . . , Am1(a))

= (F0 . . . i1i+1 . . . m1)A(a);

30

it follows that A |= (F0 . . . i1i+1 . . . m1 = F0 . . . i1i+1 . . . m1)[a], hence (L7)is universally valid.

(L8): Assume that A |= ( = )[a]. Then A(a) = A(a). Assume that

A |= (R0 . . . i1i+1 . . . m1)[a]; hence

A0 (a), . . . , Ai1(a),

A(a), Ai+1(a), . . . , Am1(a) R

A; hence

A0 (a), . . . , Ai1(a),

A(a), Ai+1(a), . . . , Am1(a) R

A; hence

A |= (R0 . . . i1i+1 . . . m1)[a];

hence (L8) is universally valid.

Now let be a set of formulas. A -proof is a finite sequence 0, . . . , m1 of formulassuch that for each i < m one of the following conditions holds:

(I1) i is a logical axiom(I2) i .(I3) (modus ponens) There are j, k < i such that j is the formula k i.(I4) (generalization) There exist j < i and k such that i is the formula vkj .

Then we say that proves , in symbols , provided that is an entry in some-proof. We write in place of .

Theorem 3.2. If , then |= .

Proof. Recall the notion |= from Chapter 2: it says that for every structure Afor the implicit language we are dealing with, if A |= [a] for all and all a : A,then A |= [a] for every a : A. Now it suffices to take a -proof 0, . . . , m1 andprove by complete induction on i that |= i for each i < m.

Case 1. i is a logical axiom. Then the result follows by Theorem 3.1.Case 2. i . Obviously then |= i.Case 3. There are j, k < i such that j is k i. Suppose that A is a model of

and a : A. Then A |= k[a] by the inductive hypothesis, and also A |= (k i)[a]by the inductive hypothesis. Thus A |= k[a] implies that A |= i[a], so A |= i[a].

Case 3. There exist j < i and k such that i is vkj . Given u A, we want toshow that A |= j [aku]; but this follows from the inductive hypothesis.

One form of the completeness theorem, proved in the next chapter, is that, conversely, |= implies that .

In this chapter we will show that many definite formulas are such that . We beginwith tautologies.

Lemma 3.3. for any first-order tautology .

Proof. Let be a sentential tautology, and let 0, 1, . . . be a sequence of first-orderformulas such that is obtained from by replacing each sentential variable Si by i. Foreach sentential formula , let be obtained from by replacing each sentential variable

31

Si by i. By Theorem 1.16, (in the sentential sense). Hence there is a sententialproof 0, . . . , m with m = . We claim that 0, . . . ,

m is a first-order proof. Since

m = = , this will prove the lemma. If i m and i is a (sentential) axiom, then i is

the corresponding first-order axiom:

[ ( )] = [ ( )];

[[ ( ] [( ) ( )]] =

[[ ( )] [( ) ( )]];

[( ) ( )] = [( ) ( )].

If j, k < i and k is j i, then k is j

i.

We proceed with simple theorems concerning equality.

Proposition 3.4. = for any term .

Proof. The following is a -proof; on the left is the entry number, and on the righta justification. Let vi be a variable not occurring in .

(1) vi = (vi = = ) (L5)(2) [vi = (vi = = ) [( = ) (vi = )] (taut.)(3) ( = ) (vi = ) ((1), (2), MP)(4) vi[( = ) (vi = )] ((3), gen.)(5) vi[( = ) (vi = )] [vi( = ) vi(vi = )] (L2)(6) vi( = ) vi(vi = ) (4), (5), MP(7) ( = ) vi( = ) (L3)(8) (7) [(6) [( = ) vi(vi = )] (taut.)(9) (6) [( = ) vi(vi = )] (7), (8), MP(10) ( = ) vi(vi = ) (6), (9), MP(11) (10) [vi(vi = ) = ] (taut.)(12) vi(vi = ) = (10), (11), MP(13) vi(vi = ) (L4)(14) (13) [(12) = ] (L1)(15) (12) = ((13), (14), MP)(16) = ((12), (15), MP)

Proposition 3.5. = = for any terms , .

Proof. By (L5) we have

= ( = = );

and by Proposition 3.4 we have = . Now

= ([ = ( = = )] ( = = ))

32

is a tautology, so = = .

Proposition 3.6. = ( = = ) for any terms , , .

Proof. By (L5), = ( = = ). By Proposition 3.5, = = .Now

( = = ) ([ = ( = = )] [ = ( = = )])

is a tautology, so = ( = = ).

We now give several results expressing the principle of substitution of equals for equals.The main fact is expressed in Theorem 3.16, which says that under certain conditions theformula = ( ) is provable, where is obtained from by replacing someoccurrences of by .

Lemma 3.7. If and are terms, and are formulas, vi is a variable not occurringin or , and = ( ), then = (vi vi).

Proof.

(1) vi[ = ( )] (hypothesis, gen.)(2) vi( = ) vi( )] (from (1), using (L2))(3) vi( ) (vi vi) ((L2))(4) = vi( = ). ((L3))

Now putting (2)(4) together with a tautology gives the lemma.

To proceed further we need to discuss the notion of free and bound occurrences of variablesand terms. This depends on the notion of a subformula. Recall that a formula is just afinite sequence of positive integers, subject to certain conditions. Atomic equality formulashave the form = for some terms , , and = is defined to be 3 . Atomic non-equality formulas have the form R0 . . . m1 for some m, some m-ary relation symbol R,and some terms 0, . . . , m1. R is actually some positive integer k greater than 5 and notdivisible by 5, and R0 . . . m1 is the sequence k0

m1. Non-atomic formulashave the form

= 1, = 2, orvs = 4, 5(s+ 1).

Thus every formula begins with one of the integers 1,2,3,4 or some positive integer greaterthan 5 not divisible by 5 which is a relation symbol. This helps motivate the followingpropositions.

Proposition 3.8. If = 0, . . . , k1 is a term, then each i is either of the form 5mwith m a positive integer, or it is an odd integer greater than 5 which is a function symbolor individual constant.

33

Proof. We prove this by induction on , thus using Proposition 2.1. The propositionis obvious if is a variable or individual constant. Suppose that F is a function symbol ofrank m, 0, . . . , m1 are terms, and is F0 . . . m1, where we assume the truth of theproposition for 0, . . . , m1. Suppose that i < k. If i = 0, then i is F, a function symbol.If i > 0, then i is an entry in some j, and the desired conclusion follows by the inductivehypothesis.

Proposition 3.9. Let = 0, . . . k1 be a formula, suppose that i < k, and i is oneof the integers 1,2,3,4 or a positive integer greater than 5 which is a relation symbol. Thenthere is a unique segment i, i+1, . . . , j of which is a formula.

Proof. We prove this by induction on , thus using Proposition 2.5. We assume thehypothesis of the proposition. First suppose that is an atomic equality formula = with and terms. Thus = is the sequence 1 . Now by Proposition 2.2(ii), noentry of a term is among the integers 1, 2, 3, 4 or is a positive integer greater than 5 whichis a relation symbol. It follows from the assumption about i that i = 0, and hence thedesired segment of is itself. It is unique by Proposition 2.6(iii). Second suppose that is an atomic non-equality formula R0 . . . m1 with R an m-ary relation symbol and0, . . . , m1 terms. This is very similar to the first case. R0 . . . m1 is the sequenceR0

m1. By Proposition 2.2(ii) i must be 0, and hence the desired segment of is itself. It is unique by Proposition 2.6(iii).

Now assume inductively that is ; so is 1. If i = 0, then itself isthe desired segment, unique by Proposition 2.6(iii). If i > 0, then i = i1, where = 0, . . . , k1. By the inductive hypothesis there is a segment i1, i, . . . , j of which is a formula. This gives a segment i, i+1, . . . , j+1 of which is a formula; itis unique by Proposition 2.6(iii).

Assume inductively that is for some formulas , . So is 2. Ifi = 0, then itself is the required segment, unique by Proposition 2.6(iii). Now supposethat i > 0. Now we have = 1, . . . , m and = m+1, . . . , k1 for some m. If1 i m, then by the inductive assumption there is a segment i, i+1, . . . , n of which is a formula. This is also a segment of , and it is unique by Proposition 2.6(iii). Ifm+ 1 i k 1, a similar argument with gives the desired result.

Finally, assume inductively that is vs with some formula and s . If i = 0then itself is the desired segment, unique by Proposition 2.6(iii). If i > 0 then actuallyi > 1 so that i is within , and the inductive hypothesis applies.

The segment of asserted to exist in Proposition 3.9 is called the subformula of beginning

at i. For example, consider the formula def= v0[v0 = v2 v0 = v2]. The formula v0 = v2

occurs in two places in . In detail, is the sequence 4, 5, 2, 3, 5, 15, 3, 5, 15. Thus

0 = 4;1 = 5;2 = 2;3 = 3;4 = 5;5 = 15;

34

6 = 3;7 = 5;8 = 15;

On the other hand, v0 = v2 is the formula 3, 5, 15. It occurs in beginning at 3, andalso beginning at 6.

Now a variable vs is said to occur bound in at the j-th position iff with =0, . . . , m1, we have j = vs and there is a subformula of of the form vs =i, i+1, . . . , m with i + 1 j m. If a variable vs occurs at the j-th position of but does not occur bound there, then that occurrence is said to be free. We give someexamples. Let be the formula v0 = v1 v1 = v2. All the occurrences of v0, v1, v2 arefree occurrences in . Note that as a sequence is 2, 3, 5, 10, 3, 10, 15; so 0 = 2, 1 = 3,2 = 5, 3 = 10, 4 = 3, 5 = 10, and 6 = 15. The variable v0, which is the integer 5,occurs free at the 2-nd position. The variable v1, which is the integer 10, occurs free atthe 3rd and 5th positions. The variable v2, which is the integer 15, occurs free at the 6thposition.

Now let be the formula v0 = v1 v1(v1 = v2). Then the first oc-curence of v1 is free, but the other two occurrences are bound. As a sequence, is2, 3, 5, 10, 4, 10, 3, 10, 15. The variable v1 occurs free at the 3rd position, and bound atthe 5th and 7th positions.

We also need the notion of a term occurring in another term, or in a formula. The followingtwo propositions are proved much like 3.9.

Proposition 3.10. If = 0, . . . , m1 is a term and i < m, then there is a uniqueterm which is a segment of beginning at i.

Proof. We prove this by induction on . For a variable or individual constant,we have m = 1 and so i = 0, and itself is the only possibility for . Now supposethat the proposition is true for terms 0, . . . n1, F is an n-ary function symbol, and is F0 . . . n1. If i = 0, then itself begins at i, and it is the only term beginning at iby Proposition 2.2(iii). If i > 0, then i is inside some term k, and so by the inductiveassumption there is a term which is a segment of k beginning there; this term is a segmentof too, and it is unique by Proposition 2.2(iii).

Under the assumptions of Proposition 3.10, we say that occurs in beginning at i.

Proposition 3.11. If = 0, . . . , m1 is a formula, i < m, and i is a variable, anindividual constant, or a function symbol, then there is a unique segment of beginningat i which is a term.

Proof. We prove this by induction on . First suppose that is an atomic equalityformula = for some terms , . Thus is 3 . So i > 0, and hence i is inside or . If i is inside , then by Proposition 3.10, there is a term which is a segment of beginning at i; it is also a segment of , and it is unique by Proposition 2.2(iii). Similarlyfor .

Suppose inductively that is . Thus is 1. It follows that i > 0, so that iappears in ; then the inductive hypothesis applies.

35

Suppose inductively that is . Thus is 2. It follows that i > 0, sothat i appears in or ; then the inductive hypothesis applies.

Finally, suppose that is vs with a formula and s . Thus is 4, 5(s+1).Hence i > 0. If i = 1, then 5(s+1) is the desired segment, unique by Proposition 2.6(iii).Suppose that i > 1. So i is an entry in and hence by the inductive assumption, thereis a segment i, i+1, . . . m which is a term; this is also a segment of , and it is uniqueby Proposition 2.6(iii).

Under the assumptions of Proposition 3.11, we say that the indicated segment occurs in beginning at i.

We now extend the notions of free and bound occurrences to terms. Let be a term whichoccurs as a segment in a formula . Say that = 0, . . . , m1 and = i, . . . k. Wesay that this occurrence of in is bound iff there is a variable vs which occurs bound in at some place t with i t k; the occurrence of is free iff there is no such variable.

We give some examples. The term v0 + v1 is bound in its only occurrence in theformula v0(v0 + v1 = v2). The same term is bound in its first occurrence and free in itssecond occurrence in the formula v0(v0 + v1 = v2) v0 + v1 = v0.

Suppose that , , are terms, and occurs in beginning at i. By the result ofreplacing that occurrence of by we mean the following sequence . Say , , havedomains (lengths) m,n, p respectively. Then is the sequence

0, . . . , i1, 0, . . . , p1, i+n, . . . , m1.

Put another way, if is with of length i, then is .

Proposition 3.12. Suppose that , , are terms, and the sequence is obtained from by replacing one occurrence of by . Then is a term.

Proof. We prove this by induction on , thus by using Proposition 2.1. If is avariable or an individual constant, then must be itself, and is , which is a term.Now suppose that is F0 . . . m1 for some m-ary function symbol F and some terms0, . . . , m1, and the proposition holds for 0, . . . , m1. Say the occurrence of in begins at i. If i = 0, then equals , and hence equals , which is a term. If i > 0,then i is inside some j , and hence the occurrence of is actually an occurrence in j byProposition 2.2(iii). Replacing this occurrence of in j by we obtain a term by theinductive hypothesis; call this term j . It follows that is F0 . . . j1

j , j+1 . . . m1,

which is a term.

As an example, consider the term v0 (v1 + v2) in the language for (Q,+, ). Replacingthe occurrence of v1 by v0 v1 we obtain the term v0 ((v0 v1) + v2). Writing this outin detail, assuming that corresponds to 9 and + corresponds to 7, we start with thesequence 9, 5, 7, 10, 15 and end with the sequence 9, 5, 7, 9, 5, 10, 15.

Our first form of subsitution of equals for equals only involves terms:

Theorem 3.13. If , , are terms, and is a sequence obtained from by replacing anoccurrence of in by , then is a term and = = .

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Proof. is a term by Proposition 3.12. Now we proceed by induction on . If isa variable or an individual constant, then must be the same as , since has length 1and occurs in . Then is , and = = is = = , a tautology. So theproposition is true in this case.

Now assume inductively that is F0 . . . m1 with F an m-ary function symbol and0, . . . , m1 terms. There are two possibilities for the occurrence of . First, possibly isthe same as . Then is , and again we have the tautology = = , Second, theoccurrence of is within some i. Then by the inductive hypothesis, = i = i,where i is obtained from i by replacing the indicated occurrence of by . Now aninstance of (L7) is

i = i F0 . . . i1 . . . ii+1 . . . m1 = F0 . . . i1 . . .

ii+1 . . . m1.

Putting this together with = i = i and a tautology gives = = .

Proposition 3.14. Suppose that is a formula and , are terms. Suppose that occursat the i-th place in , and if i > 0 and i1 = , then is a variable. Let the sequence be obtained from by replacing that occurrence of by . Then is a formula.

Proof. Induction on . Suppose that is = . Then by Proposition 3.13, occurs in or . Suppose that it occurs in . Let be obtained from by replacing thatoccurrence of by . Then is a term by Proposition 3.14. Since is = , is aformula. The case in which occurs in is similar. Now suppose that is R0 . . . m1with R an m-ary relation symbol and 0, . . . , m1 are terms. Then the occurrence of is within some i. Let

i be obtained from i by replacing that occurrence by . Now

is R0 . . . i1i . . . m1, so is a formula.

Now suppose that the result holds for , and is . Then occurs in , so if is obtained from by replacing the occurrence of by , then is a formula by theinductive assumption. Since is also is a formula.

Next, suppose that the result holds for and , and is . Then theoccurrence of is within or is within . If it is within , let be obtained from

by replacing that occurrence of by . Then is a formula by the inductive hypothesis.Since is , also is a formula. If the occurrence is within , let be obtainedfrom by replacing that occurrence of by . Then is a formula by the inductivehypothesis. Since is , also is a formula.

Finally, suppose that the result holds for , and is vk. If i = 1, then is vk,and by hypothesis is some variable vl. Then is vl, which is a formula. If i > 1, then occurs in , so if is obtained from by replacing the occurrence of by , then

is a formula by the inductive assumption. Since is vk also is a formula.

For the exact definition of see the description before Proposition 3.12.

Lemma 3.15. Suppose that and are terms, is a formula, and is obtained from by replacing one free occurrence of in by , such that the occurrence of that resultsis free in . Then = ( ).

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Proof. We proceed by induction on . First suppose that is an atomic equalityformula = . If the occurrence of that is replaced is in , let be the resulting term.Then by Proposition 3.13, = = . Now (L5) gives = ( = = ).Putting these two together with a tautology gives = ( = = ). Bysymmetry, = ( = = ). Hence = ( = = ).

If the occurrence of that is replaced is in , a similar argument using (L6) works.Second, suppose that is an atomic non-equality formula R0 . . . m1, with R an m-

ary relation symbol and 0, . . . , m1 terms. Say that the occurrence of that is replacedby is in i, the resulting term being

i. Then by Proposition 3.13, = i =

i.

By (L8) we have

i = i (R0 . . . m1 R0 . . . i1

ii+1 . . . m1),

so by a tautology we get from these two facts

= (R0 . . . m1 R0 . . . i1ii+1 . . . m1),

and by symmetry

= (R0 . . . i1ii+1 . . . m1 R0 . . . m1),

and then another tautology gives

= (R0 . . . m1 R0 . . . i1ii+1 . . . m1),

This finishes the atomic cases. Now suppose inductively that is . The occurrence of in that is replaced actually occurs in ; let be the result of replacing that occurrenceof by . Now the occurrence of in is free in . In fact, suppose that vi is asubformula of which has as a segment the indicated occurrence of , and vi occurs in. Then vi is also a subformula of , contradicting the assumption that the occurrenceof is free in . Similarly the occurrence of in which replaced the occurrence of is free. So by the inductive hypothesis, = ( ), and hence a tautology gives = ( ), i.e., = ( ).

Suppose inductively that is .Case 1. The occurrence of in is within . Let be obtained from by replacing

that occurrence by , such that that occurrence is free in , hence free in . By theinductive hypothesis, = ( ). Since is , a tautology gives thedesired result.

Case 2. The occurrence of in is within . Let be obtained from by replacingthat occurrence by , such that that occurrence is free in , hence free in . By theinductive hypothesis, = ( ). Since is , a tautology gives the desiredresult.

Finally, suppose that is vi. Then the occurrence of in that is replaced isin . Let be obtained from by replacing that occurrence of by . The occurrenceof in must be free since it is free in , as in the treatment of above; similarly

38

for and . Hence by the inductive hypothesis, = ( ). Now since theoccurrence of in is free, the variable vi does not occur in . Similarly, it does notoccur in . Hence by Proposition 3.7 and tautologies we get = (vi vi),i.e., = ( ).

The hypothesis that the term is still free in the result of the replacement in this propo-sition is necessary for the truth of the proposition. This hypothesis is equivalent to sayingthat the occurrence of which is replaced is not inside a subformula of of the form viwith vi a variable occurring in .

Theorem 3.16. (Substitution of equals for equals) Suppose that is a formula, is aterm, and occurs freely in starting at indices i(0) < < i(m 1). Also suppose that is a term. Let be obtained from by replacing each of these occurrences of by ,and each such occurrence of is free in . Then = ( ).

Proof. We prove this by induction on m. If m = 0, then is the same as , andthe conclusion is clear. Now assume the result for m, for any . Now assume that occurs freely in starting at indices i(0) < < i(m), and no such occurrence is inside asubformula of of the form vj with vj a variable occurring in . Let be obtained from by replacing the last occurrence of , the one beginning at i(m), by . By Proposition3.15, = ( ). Now we apply the inductive hypothesis to and the occurrencesof starting at i(0), . . . , i(m 1); this gives = ( ). Hence a tautology gives = ( ), finishing the inductive proof.

Proposition 3.17. Suppose that , , are formulas, and the sequence is obtained from by replacing an occurrence of in by . Then is a formula.

Proof. Induction on . If is atomic, then is equal to , and is equal to andhence is a formula. Suppose the result is true for and is . If = , again thedesired conclusion is clear. Otherwise the occurrence of is within the subformula .If is obtained from by replacing that occurrence by , then is a formula by theinductive hypothesis. Since is , also is a formula.

Now suppose the result is true for and , and is . If = , again thedesired conclusion is clear. Otherwise the occurrence of is within the subformula oris within the subformula . If it is within and is obtained from by replacing thatoccurrence by , then is a formula by the inductive hypothesis. Since is , also is a formula. If it is within and is obtained from by replacing that occurrenceby , then is a formula by the inductive hypothesis. Since is , also is aformula.

Finally, suppose the result is true for and is vi. If = , again the desiredconclusion is clear. Otherwise the occurrence of is within the subformula . If isobtained from by replacing that occurrence by , then is a formula by the inductivehypothesis. Since is vi, also is a formula.

For the exact meaning of see the description before Proposition 3.12.

Another form of the substitution of equals by equals principle is as follows:

39

Theorem 3.18. Let , , be formulas, and let be obtained from by replacing anoccurrence of in by . Suppose that . Then .

Proof. Induction on . If is atomic, then is the same as , and the conclusion isclear. Suppose inductively that is . If is equal to , then is equal to and theconclusion is clear. Suppose that occurs within , and let be obtained from byreplacing that occurrence by . Assume that . Then by the inductive hypothesis , so , as desired.

The case in which is is similar. Finally, suppose that is vi, and occurs within . Let be obtained from by replacing that occurrence by . Assumethat . Then by the inductive assumption. So by a tautology, ,and then by generalization vi( ). Using (L2) we then get vi vi.Similarly, vi vi. Hence using a tautology, vi vi.

Now we work to prove two important logical principles: changing bound variables, anddropping a universal quantifier in favor of a term.

For any formula , i , and term by Subfvi we mean the result of replacingeach free occurrence of vi in by . We now work towards showing that under suitableconditions, the formula vi Subf

vi is provable. The supposition expressed in the first

sentence of the following lemma will be eliminated later on.

Lemma 3.19. Suppose that vi does not occur bound in , and does not occur in the term.

Assume that no free occurrence of vi in is within a subformula of of the formvj with vj a variable occurring in . Then vi Subf

vi .

Proof.

(1) vi = ( Subfvi ) (by Proposition 3.16 and a tautology)

(2) (Subfvi (vi = )) (using a tautology)(3) vi[ (Subf

vi (vi = ))] (generalization)

(4) vi vi(Subfvi (vi = )) (using (L2))

(5) vi(Subfvi (vi = )) (viSubf

vi vi(vi = )) ((L2))

(6) vi (viSubfvi vi(vi = )) ((4), (5), a tautology)

(7) vi(vi = ) (vi viSubfvi ) ((6), a tautology)

(8) vi(vi = ) ((L4))(9) vi viSubf

vi ) ((7), (8), modus ponens)

(10) Subfvi viSubfvi ((L3))

(11) vi Subfvi ((9), (10), a tautology)

Lemma 3.20. If i 6= j, is a formula, vi does not occur bound in , and vj does notoccur in at all, then vi vjSubf

vivj.

Proof.

vi Subfvivj (by Lemma 3.19)

vjvi vjSubfvivj (using (L2) and a tautology)

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vi vjvi (by (L3)) vi vjSubf

vivj

Lemma 3.21. If i 6= j, is a formula, vi does not occur bound in , and vj does notoccur in at all, then vi vjSubf

vivj.

Proof. By Proposition 3.20 we have vi vjSubfvivj. Now vj does not occur

bound in Subfvivj and vi does not occur in Subfvivj at all. Hence by Proposition 3.20

again, vjSubfvivj viSubf

vjvi

Subfvivj. Now Subfvjvi

Subfvivj is actually just itself; so vjSubf

vivj vi. Hence the proposition follows.

For i, j and a formula, by Subbvivj we mean the result of replacing all boundoccurrences of vi in by vj . By Proposition 3.14 this gives another formula.

Proposition 3.22. If vi occurs bound in a formula , then there is a subformula vi of such that vi does not occur bound in .

Proof. Induction on . Note that the statement to be proved is an implication. If is atomic, then vi cannot occur bound in ; thus the hypothesis of the implication isfalse, and so the implication itself is true. Now suppose inductively that is , and vioccurs bound in . Then it occurs bound in , and so by the inductive hypothesis, hasa subformula vi such that vi does not occur bound in . T

NOTES ON SET THEORY - euclid.colorado.edueuclid.colorado.edu/~monkd/full.pdf · NOTES ON SET THEORY J. Donald Monk March 6, 2018 TABLE OF CONTENTS LOGIC 1. Sentential logic.....1

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