Solutions for exercises in chapter 1 = 2, 3, 1, 4 ; = 2, 2, 3, 4, 2 1 S 1 1 S 0 = 2, 2, 3, 4, 2, 1, 4,1,3. 1, 1,..., 1, S 0

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1 E1.1 Verify that and Solutions for exercises in chapter 1 S 0 S 1 = 2, 3, 1, 4 (S 0 S 1 ) ( S 1 S 0 ) = 2, 2, 3, 4, 2, 1, 4, 1,3. S 0 S 1 = 2 S 0 S 1 = S 1 = 2, 3, 1, 4 ; (S 0 S 1 ) ( S 1 S 0 ) = 2 (S 0 S 1 ) ( S 1 S 0 ) = 2 2 S 0 S 1 2 S 1 S 0 = 2, 2, 3, 4, 2 1 S 1 1 S 0 = 2, 2, 3, 4, 2, 1, 4,1,3. E1.2 Prove that there is a sentential formula of each positive integer length. If m is a positive integer, then is a formula of length m, it is m 1 times {}}{ 1, 1,..., 1, S 0 m 1 times {}}{ S 0. E1.3 Prove that m is the length of a sentential formula not involving iff m is odd. Proof. : We prove by induction on ϕ that if ϕ is a sentential formula not involving, then the length of ϕ is odd. This is true of sentential variables, which have length 1. Suppose that it is true of ϕ and ψ, which have length 2m+1 and 2n+1 respectively. Then ϕ ψ, which is 1 ϕ ψ, has length 1 + 2m n + 1 = 2(m + n + 1) + 1, which is again odd. This finishes the inductive proof.. We construct formulas without with length any odd integer by induction. S 0 is a formula of length 1. If ϕ has been constructed of length 2m + 1, then S 0 ϕ, which is 1, S 0 ϕ, has length 2m + 3. This finishes the inductive construction. E1.4 Prove that a truth table for a sentential formula involving n basic formulas has 2 n rows. We prove this by induction on n. For n = 1, there are two rows. Assume that for n basic formulas there are 2 n rows. Given n + 1 basic formulas, let ϕ be one of them. For the others, by the inductive hypothesis there are 2 n rows. For each such row there are two possibilities, 0 or 1, for ϕ. So for the n + 1 basic formulas there are 2 n 2 = 2 n+1 rows. 1

2 E1.5 Use the truth table method to show that the formula (ϕ ψ) ( ϕ ψ) is a tautology. ϕ ψ ϕ ψ ϕ ϕ ψ (ϕ ψ) ( ϕ ψ) E1.6 Use the truth table method to show that the formula [ϕ (ψ χ)] [(ϕ ψ) (ϕ χ)] is a tautology. Let θ be the indicated formula. ϕ ψ χ ϕ ψ ϕ χ (ϕ ψ) (ϕ χ) ψ χ ϕ (ψ χ) θ E1.7 Use the truth table method to show that the formula (ϕ ψ) (ϕ ψ) is not a tautology. It is not necessary to work out the full truth table. ϕ ψ ϕ ψ ψ ϕ ψ (ϕ ψ) (ϕ ψ)

3 E1.8 Determine whether or not the following is a tautology: S 0 (S 1 (S 2 (S 3 S 1 ))). Suppose that f is an assignment making the indicated formula false; we work towards a contradiction. Thus (1) S 0 [f] = 1 and (2) (S 1 (S 2 (S 3 S 1 )))[f] = 0. From (2) we get (3) S 1 [f] = 1 and (4) (S 2 (S 3 S 1 ))[f] = 0. From (4) we get (5) S 2 [f] = 1 and (6) (S 3 S 1 )[f] = 0. From (6) we get S 1 [f] = 0, contradicting (3). E1.9 Determine whether or not the following is a tautology; an informal method is better than a truth table: ({[(ϕ ψ) ( χ θ)] χ} τ) [(τ ϕ) (θ ϕ)]. Suppose that f is an assignment which makes the given formula false; we want to get a contradiction. Thus we have (1) ({[(ϕ ψ) ( χ θ)] χ} τ)[f] = 1 and (2) [(τ ϕ) (θ ϕ)][f] = 0. By (2) we have (3) (τ ϕ)[f] = 1 and (4) (θ ϕ)[f] = 0. By (4) we have (5) θ[f] = 1 and (6) ϕ[f] = 0. By (3) and (6) we get (7) τ[f] = 0. By (1) and (7) we get (8) {[(ϕ ψ) ( χ θ)] χ}[f] = 0. 3

4 It follows that (9) [(ϕ ψ) ( χ θ)][f] = 1 and (10) χ[f] = 0. Now by (6) we have (11) (ϕ ψ)[f] = 1, and hence by (9), (12) ( χ θ)[f] = 1. By (5) we have (13) ( θ)[f] = 0, and hence by (12), (14) ( χ)[f] = 0. This contradicts (10). E1.10 Determine whether the following statements are logically consistent. If the contract is valid, then Horatio is liable. If Horation is liable, he will go bankrupt. Either Horatio will go bankrupt or the bank will lend him money. However, the bank will definitely not lend him money. Let S 0 correspond to the contract is valid, S 1 to Horatio is liable, S 2 to Horatio will go bankrupt, and S 3 to the bank will lend him money. Then we want to see if there is an assignment of values which makes the following sentence true: (S 0 S 1 ) (S 1 S 2 ) (S 2 S 3 ) S 3. We can let f(0) = f(1) = f(2) = 1 and f(3) = 0, and this gives the sentence the value 1. E1.11 Write out an actual proof for {ψ} ψ ϕ. This can be done by following the proof of Lemma 1.9, expanding it using the proof of the deduction theorem. Following the proof of Lemma 1.9, the following is a {ψ, ψ}-proof: (a) ψ (b) ψ ( ϕ ψ) (1) (c) ϕ ψ (a), (b), MP (d) ( ϕ ψ) (ψ ϕ) (3) (e) ψ ϕ (c), (d), MP (f) ψ (g) ϕ (e), (f), MP Now applying the proof of the deduction theorem, the following is a {ψ}-proof: (a) [ ψ [( ψ ψ) ψ]] [[ ψ ( ψ ψ)] ( ψ ψ)] (2) 4

5 (b) ψ [( ψ ψ) ψ] (1) (c) [ ψ ( ψ ψ)] ( ψ ψ) (a), (b), MP (d) ψ ( ψ ψ) (1) (e) ψ ψ (c), (d), MP (f) [ ψ ( ϕ ψ)] [ ψ [ ψ ( ϕ ψ)]] (1) (g) ψ ( ϕ ψ) (1) (h) ψ [ ψ ( ϕ ψ)]] (f), (g), MP (i) [( ϕ ψ) (ψ ϕ)] [ ψ [( ϕ ψ) (ψ ϕ)]] (1) (j) ( ϕ ψ) (ψ ϕ) (3) (k) ψ [( ϕ ψ) (ψ ϕ)] (i), (j), MP (l) [ ψ [( ϕ ψ) (ψ ϕ)]] [[ ψ ( ϕ ψ)] [ ψ (ψ ϕ)]] (2) (m) [ ψ ( ϕ ψ)] [ ψ (ψ ϕ)] (k), (l), MP (n) ψ (ψ ϕ) (g), (m), MP (o) ψ ( ψ ψ) (1) (p) ψ (q) ψ ψ (o), (p), MP (r) [ ψ (ψ ϕ)] [( ψ ψ) ( ψ ϕ)] (2) (s) ( ψ ψ) ( ψ ϕ) (n), (r), MP (t) ψ ϕ (q), (s), MP Solutions for exercises in Chapter 2 E2.1 Give an exact definition of a language for the structure (ω, <). The quadruple ({11},,, rnk), where rnk is the function with domain {11} such that rnk(11) = 2. E2.2 Give an exact definition of a language for the set A (no individual constants, function symbols, or relation symbols). The quadruple (,,, ). Note that the last is the empty function. E2.3 Describe a term construction sequence which shows that + v 0 v 0 v 1 is a term in the language for (R, +,, 0, 1, <). v 0, v 0 v 0, v 1,+ v 0 v 0 v 1. E2.4 In any first-order language, show that the sequence v 0, v 0 is not a term. Hint: use Proposition 2.2. Suppose that v 0, v 0 is a term. This contradicts Proposition 2.2(ii). E2.5 In the language for (ω, S, 0, +, ), show that the sequence +, v 0, v 1, v 2 is not a term. Here S(i) = i + 1 for any i ω. Hint: use Proposition 2.2. Suppose it is a term. By Proposition 2.2(ii)(c), there are terms σ, τ such that +, v 0, v 1, v 2 is + σ τ. Thus v 0, v 1, v 2 = σ τ. So the term v 0 is an initial segment of the term σ. By Proposition 2.2(iii) it follows that v 0 = σ. Hence v 1, v 2 = τ. This contradicts Proposition 2.2(ii). 5

6 E2.6 Prove Proposition 2.5. We show by complete induction on i that ϕ i Γ for all i < m. So, suppose that i < m and ϕ j Γ for all j < i. By the definition of formula construction sequence, we have the following cases. Case 1. ϕ i is an atomic formula. Then ϕ i Γ by (i). Case 2. There is a j < i such that ϕ i is ϕ j. By the inductive hypothesis, ϕ j Γ. Hence by (ii), ϕ i Γ. Case 2. There are j, k < i such that ϕ i is ϕ j ϕ k. By the inductive hypothesis, ϕ j Γ and ϕ k Γ. Hence by (iii), ϕ i Γ. Case 4. There exist j < i and k ω such that ϕ i is v k ϕ j. By the inductive hypothesis, ϕ j Γ. Hence by (iv), ϕ i Γ. This completes the inductive proof. E2.7 Show how the structure (ω, S, 0, +, ) can be put in the general framework of structures. (ω, S, 0, +, ) can be considered to be the structure (ω, Rel, Fcn, Cn ) where Rel =, Cn is the function 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) =. E2.8 Prove that in the language for the structure (ω, +), a term has length m iff m is odd. First we show by induction on terms that every term has odd length. This is true for variables. Suppose that it is true for terms σ and τ. Then also σ + τ has odd length. 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 σ of length 2m + 1. Then σ + v 0 has length 2m + 3. This finishes the inductive proof. E2.9 Give a formula ϕ in the language for (Q, +, ) such that for any a : ω Q, (Q, +, ) = ϕ[a] iff a 0 = 1. Let ϕ be the formula v 1 [v 0 v 1 = v 1 ]. E2.10 Give a formula ϕ which holds in a structure, under any assignment, iff the structure has at least 3 elements. v 0 v 1 v 2 ( (v 0 = v 1 ) (v 0 = v 2 ) (v 1 = v 2 )). E2.11 Give a formula ϕ which holds in a structure, under any assignment, iff the structure has exactly 4 elements. v 0 v 1 v 2 v 3 ( (v 0 = v 1 ) (v 0 = v 2 ) (v 0 = v 3 ) (v 1 = v 2 ) (v 1 = v 3 ) (v 2 = v 3 ) v 4 (v 0 = v 4 v 1 = v 4 v 2 = v 4 v 3 = v 4 )). E2.12 Write a formula ϕ in the language for (ω, <) such that for any assignment a, (ω, <) = ϕ[a] iff a 0 < a 1 and there are exactly two integers between a 0 and a 1. v 0 < v 1 v 2 v 3 [v 0 < v 2 v 2 < v 3 v 3 < v 1 v 4 [v 0 < v 4 v 4 < v 1 v 4 = v 2 v 4 = v 3 ]]. 6

7 E2.13 Prove that the formula v 0 = v 1 (Rv 0 v 2 Rv 1 v 2 ) is universally valid, where R is a binary relation symbol. Let A be a structure and a : ω A an assignment. Suppose that A = (v 0 = v 1 )[a]. Then a 0 = a 1. Also suppose that A = Rv 0 v 2 [a]. Then (a 0, a 2 ) R A. Hence (a 1, a 2 ) R A. Hence A = Rv 1 v 2 [a], as desired. E2.14 Give an example showing that the formula is not universally valid. v 0 = v 1 v 0 (v 0 = v 1 ) Consider the structure A def = (ω, <), and let a : ω ω be defined by a(i) = 0 for all i ω. Then A = (v 0 = v 1 )[a]. Now A = (v 0 = v 1 )[a 0 1 ] since 1 0, so A = v 0(v 0 = v 1 )[a]. Therefore A = (v 0 = v 1 v 0 (v 0 = v 1 ))[a]. E2.15 Prove that v 0 v 1 ϕ v 1 v 0 ϕ is universally valid. Assume that a : ω A and A = v 0 v 1 ϕ[a]. Choose u A so that A = v 1 ϕ[a 0 u ]. In order to show that A = v 1 v 0 ϕ[a], let w A be given. Then A = ϕ 01 uw. It follows that A = v 0 ϕ[u 1 w ]. Hence A = v 1 v 0 ϕ[a], as desired. Solutions to exercises in Chapter 3 E3.1 Do the case Rσ 0...σ m 1 for some m-ary relation symbol and terms σ 0,..., σ m 1 in the proof of Theorem 3.1, (L3). We are assuming that v i does not occur in Rσ 0...σ m 1 ; hence it does not occur in any term σ i. A = (Rσ 0...σ m 1 )[a] iff iff iff σ0 A (a),..., σr m 1 (a) RA σ0 A (b),..., σr m 1 (b) RA (by Proposition 2.4) A = (Rσ 0...σ m 1 )[b]. E3.2 Prove that (L6) is universally valid, in the proof of Theorem 3.1. Assume that A = (σ = τ)[a] and A = (ρ = σ)[a]. Then σ A (a) = τ A (a) and ρ A (a) = σ A (a), so ρ A (a) = τ A (a), hence A = (ρ = τ)[a]. E3.3 Prove that (L8) is universally valid, in the proof of Theorem 3.1. Assume that A = (σ = τ)[a]. Then σ A (a) = τ A (a). Assume that A = (Rξ 0...ξ i 1 σξ i+1...ξ m 1 )[a]; hence ξ A 0 (a),..., ξ A i 1(a), σ A (a), ξ A i+1(a),..., ξ A m 1(a) R A ; hence ξ A 0 (a),..., ξa i 1 (a), τa (a), ξ A i+1 (a),..., ξa m 1 (a) RA ; hence A = (Rξ 0...ξ i 1 τξ i+1...ξ m 1 )[a]; 7

8 hence (L8) is universally valid. E3.4 Finish the proof of Proposition 3.9. If i = 0 then ϕ itself is the desired segment, unique by Proposition 2.6(iii). If i > 0 then actually i > 1 so that ϕ i is within ψ, and the inductive hypothesis applies. E3.5 Finish the proof of Proposition Suppose inductively that ϕ is ψ. Thus ϕ is 1 ψ. It follows that i > 0, so that ϕ i appears in ψ; then the inductive hypothesis applies. Suppose inductively that ϕ is ψ χ. Thus ϕ is 2 ψ χ. It follows that i > 0, so that ϕ i appears in ψ or χ; then the inductive hypothesis applies. Finally, suppose that ϕ is v s ψ with ψ a formula and s ω. Thus ϕ is 4, 5(s+1) ψ. 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 ϕ i is an entry in ψ and hence by the inductive assumption, there is a segment ϕ i, ϕ i+1,...ϕ m which is a term; this is also a segment of ϕ, and it is unique by Proposition 2.6(iii). E3.6 Indicate which occurrences of the variables are bound and which ones free for the following formulas. v 0 (v 0 < v 1 ) v 1 (v 0 = v 1 ). v 4 + v 2 = v 0 v 3 (v 0 = v 1 ). v 2 (v 4 + v 2 = v 0 ). First formula: the first and second occurrences of v 0 are bound, and the third one is free. The first occurrence of v 1 is free, and the other two are bound. Second formula: the occurrence of v 3 is bound. All other occurrences of variables are free. Third formula: the two occurrences of v 2 are bound. The other occurrences of variables are free. E3.7 Prove Proposition Induction on ϕ. Suppose that ϕ is ρ = ξ. Then by Proposition 3.13, σ occurs in ρ or ξ. Suppose that it occurs in ρ. Let ρ be obtained from ρ by replacing that occurrence of σ by τ. Then ρ is a term by Proposition Since ψ is ρ = ξ, ψ is a formula. The case in which σ occurs in ξ is similar. Now suppose that ϕ is Rη 0...η m 1 with R an m-ary relation symbol and η 0,..., η m 1 are terms. Then the occurrence of σ is within some η i. Let η i be obtained from η i by replacing that occurrence by τ. Now ψ is Rη 0...η i 1 η i...η m 1, so ψ is a formula. Now suppose that the result holds for ϕ, and ϕ is ϕ. Then σ occurs in ϕ, so if ψ is obtained from ϕ by replacing the occurrence of σ by τ, then ψ is a formula by the inductive assumption. Since ψ is ψ also ψ is a formula. Next, suppose that the result holds for ϕ and ϕ, and ϕ is ϕ ϕ. Then the occurrence of σ is within ϕ or is within ϕ. If it is within ϕ, let ψ be obtained from ϕ by replacing that occurrence of σ by τ. Then ψ is a formula by the inductive hypothesis. Since ψ is ψ ϕ, also ψ is a formula. If the occurrence is within ϕ, let ψ be obtained 8

9 from ϕ by replacing that occurrence of σ by τ. Then ψ is a formula by the inductive hypothesis. Since ψ is ϕ ψ, also ψ is a formula. Finally, suppose that the result holds for ϕ, and ϕ is v k ϕ. If i = 1, then σ is v k, and by hypothesis τ is some variable v l. Then ψ is v l ϕ, which is a formula. If i > 1, then σ occurs in ϕ, so if ψ is obtained from ϕ by replacing the occurrence of σ by τ, then ψ is a formula by the inductive assumption. Since ψ is v k ψ also ψ is a formula. E3.8 Indicate all free and bound occurrences of terms in the formula v 0 = v 1 + v 1 v 2 (v 0 + v 2 = v 1 ). v 0 is free in both of its occcurrences. v 1 is free in all three of its occurrences. v 2 is bound in both of its occurrences. v 1 + v 1 is free in its occurrence. v 0 + v 2 is bound in its occurrence. E3.9 Prove Proposition 3.17 Induction on ϕ. If ϕ is atomic, then ψ is equal to ϕ, and θ is equal to χ and hence is a formula. Suppose the result is true for ϕ and ϕ is ϕ. If ψ = ϕ, again the desired conclusion is clear. Otherwise the occurrence of ψ is within the subformula ϕ. If θ is obtained from ϕ by replacing that occurrence by χ, then θ is a formula by the inductive hypothesis. Since θ is θ, also θ is a formula. Now suppose the result is true for ϕ and ϕ, and ϕ is ϕ ϕ. If ψ = ϕ, again the desired conclusion is clear. Otherwise the occurrence of ψ is within the subformula ϕ or is within the subformula ϕ. If it is within ϕ and θ is obtained from ϕ by replacing that occurrence by χ, then θ is a formula by the inductive hypothesis. Since θ is θ ϕ, also θ is a formula. If it is within ϕ and θ is obtained from ϕ by replacing that occurrence by χ, then θ is a formula by the inductive hypothesis. Since θ is ϕ θ, also θ is a formula. Finally, suppose the result is true for ϕ and ϕ is v i ϕ. If ψ = ϕ, again the desired conclusion is clear. Otherwise the occurrence of ψ is within the subformula ϕ. If θ is obtained from ϕ by replacing that occurrence by χ, then θ is a formula by the inductive hypothesis. Since θ is v i θ, also θ is a formula. E3.10 Show that the condition in Lemma 3.15 that the resulting occurrence of τ is free is necessary. Hint: use Theorem 3.2; describe a specific formula of the type in Proposition 3.15, but with τ not free, such that the formula is not universally valid. Consider the language for (ω, S), and the formula v 0 = v 1 ( v 1 (Sv 0 = v 1 ) v 1 (Sv 1 = v 1 )). Taking an assignment a : ω ω with a 0 = a 1 makes this sentence false; hence it is not provable, by Theorem 3.2. E3.11 Do the case of implication in the proof of Lemma Suppose inductively that ϕ is χ θ. 9

10 Case 1. The occurrence of σ in ϕ is within χ. Let χ be obtained from χ by replacing that occurrence by τ, such that that occurrence is free in ψ, hence free in χ. By the inductive hypothesis, σ = τ (χ χ ). Since ψ is χ θ, a tautology gives the desired result. Case 2. The occurrence of σ in ϕ is within θ. Let θ be obtained from θ by replacing that occurrence by τ, such that that occurrence is free in ψ, hence free in θ. By the inductive hypothesis, σ = τ (θ θ ). Since ψ is χ θ, a tautology gives the desired result. E3.12 Prove that the hypothesis of Theorem 3.25 is necessary. Consider the formula v 0 v 1 (v 0 < v 1 ) v 1 (v 1 < v 1 ). This formula is not universally valid; it fails to hold in (ω, <), for example. E3.13 Prove Proposition Proof. By definition, v i ϕ is v i ϕ. Now ϕ ϕ by a tautology. Hence using generalization and (L2) we get v i ϕ v i ϕ. Hence another tautology yields v i ϕ v i ϕ, i.e., v i ϕ v i ϕ. E3.14 Prove Proposition Proof. v i ϕ is the formula v i ϕ, so a simple tautology gives the result. E3.15 Prove Proposition Proof. By Theorem 3.25 we have v i ϕ Subf v i σ ( ϕ). Since clearly Subfv i σ ( ϕ) is the same as Subf v i σ ϕ, a tautology gives Subfv i σ ϕ v iϕ. E3.16 Prove Proposition By Corollary 3.26 and Corollary E3.17 Prove Proposition Proof. ϕ v i ϕ. Now use a tautology. E3.18 Prove Proposition Assume ϕ ψ. By a tautology, ϕ ψ. Hence by generalization and (L2), v i ϕ v i ψ. Similarly, v i ψ v i ϕ. The exercise follows by a tautology. E3.19 Prove Proposition 3.42 Assume ϕ ψ. By a tautology, ϕ ψ. Hence by exercise E3.18, v i ϕ v i ψ. Now a tautology gives the desired result. E3.20 Prove that v 0 v 1 (v 0 = v 1 ) v 0 (v 0 = v 1 v 0 = v 2 ). 10

11 v 0 v 1 (v 0 = v 1 ) v 0 = v 1 ; Cor twice, taut. (1) v 1 (v 0 = v 1 ) v 0 = v 2 ; Thm (2) v 0 v 1 (v 0 = v 1 ) v 0 = v 2 ; (2), Cor. 3.26, taut. (3) v 0 v 1 (v 0 = v 1 ) v 0 = v 1 v 0 = v 2 ; (1), (3), taut. (4) v 0 v 0 v 1 (v 0 = v 1 ) v 0 (v 0 = v 1 v 0 = v 2 ); (4), (L2), taut. (5) v 0 v 1 (v 0 = v 1 ) v 0 (v 0 = v 1 v 0 = v 2 ). (5), Prop. 3.27, taut. E3.21 Prove that v 0 ( v 0 = v 1 v 0 = v 2 ) v 0 v 1 ( v 0 = v 1 ). v 0 (v 0 = v 1 v 0 = v 2 ) v 0 v 1 (v 0 = v 1 ); E3.20, taut. (1) v 0 (v 0 = v 1 v 0 = v 2 ) v 0 (v 0 = v 1 v 0 = v 2 ); Prop (2) (v 0 = v 1 v 0 = v 2 ) ( (v 0 = v 1 ) (v 0 = v 2 )); taut. (3) v 0 (v 0 = v 1 v 0 = v 2 ) v 0 ( (v 0 = v 1 ) (v 0 = v 2 )); (3), Prop (4) v 0 (v 0 = v 1 v 0 = v 2 ) v 0 ( (v 0 = v 1 ) (v 0 = v 2 )); (2), (4), taut. (5) v 1 (v 0 = v 1 ) v 1 (v 0 = v 1 ); Prop (6) v 0 v 1 (v 0 = v 1 ) v 0 v 1 (v 0 = v 1 ); (6), Prop (7) v 0 v 1 (v 0 = v 1 ) v 0 v 1 (v 0 = v 1 ); Prop (8) v 0 v 1 (v 0 = v 1 ) v 0 v 1 (v 0 = v 1 ) (7), (8), taut. (9) v 0 ( v 0 = v 1 v 0 = v 2 ) v 0 v 1 ( v 0 = v 1 ). (1), (5), (9), taut. Solutions to exercises in Chapter 4 E4.1 Suppose that Γ ϕ ψ, Γ ϕ ψ, and Γ ϕ ϕ. Prove that Γ is inconsistent. The formula ( ϕ ϕ) ϕ is a tautology. Hence by Lemma 3.3, Γ ( ϕ ϕ) ϕ. Since also Γ ϕ ϕ, it follows that Γ ϕ. Hence Γ ψ and Γ ψ. Hence by Lemma 4.1, Γ is inconsistent. E4.2 Let L be a language with just one non-logical constant, a binary relation symbol R. Let Γ consist of all sentences of the form v 1 v 0 [Rv 0 v 1 ϕ] with ϕ a formula with only v 0 free. Show that Γ is inconsistent. Hint: take ϕ to be Rv 0 v 0. By Theorem 3.25 we have (1) Γ v 0 [Rv 0 v 1 Rv 0 v 0 ] [Rv 1 v 1 Rv 1 v 1 ]. Now [Rv 1 v 1 Rv 1 v 1 ] (v 0 = v 0 ) is a tautology, so from (1) we obtain Γ v 0 [Rv 0 v 1 Rv 0 v 0 ] (v 0 = v 0 ); 11

12 then generalization gives Then by Proposition 3.37 we get Γ v 1 [ v 0 [Rv 0 v 1 Rv 0 v 0 ] (v 0 = v 0 )]. Γ v 1 v 0 [Rv 0 v 1 Rv 0 v 0 ] (v 0 = v 0 ). But the hypothesis here is a member of Γ, so we get Γ (v 0 = v 0 ). Hence by Lemma 4.1, Γ is inconsistent. Alternate proof (due to a couple of students). Suppose that Γ is consistent. By the completeness theorem let A be a model of Γ. Taking ϕ to be Rv 0 v 0, we get A = v 1 v 0 [Rv 0 v 1 Rv 0 v 0 ]. Let a : ω A be any assignment. Then by Proposition 2.8(iv) there is a b A such that A = v 0 [Rv 0 v 1 Rv 0 v 0 ][a 1 b ]. By the definition of satisfaction of, it follows that for any c A we have A = [Rv 0 v 1 Rv 0 v 0 ][a 0 c b 1 ]. Hence (c, b) RA iff (c, b) / R A, contradiction. E4.3 Show that the first-order deduction theorem fails if the condition that ϕ is a sentence is omitted. Hint: take Γ =, let ϕ be the formula v 0 = v 1, and let ψ be the formula v 0 = v 2. {v 0 = v 1 } v 0 = v 1 {v 0 = v 1 } v 1 (v 0 = v 1 ) {v 0 = v 1 } v 1 (v 0 = v 1 ) v 0 = v 2 by Theorem 3.25 {v 0 = v 1 } v 0 = v 2. On the other hand, let A be the structure with universe ω and define a = 0, 0, 1, 1,.... Clearly A = [v 0 = v 1 v 0 = v 2 ][a]. Hence v 0 = v 1 v 0 = v 2 by Theorem 3.2. E4.4 In the language for A def = (ω, S, 0, +, ), let τ be the term v 0 + v 1 v 2 and ν the term v 0 + v 2. Let a be the sequence 0, 1, 2,.... Let ρ be obtained from τ by replacing the occurrence of v 1 by ν. (a) Describe ρ as a sequence of integers. (b) What is ρ A (a)? (c) What is ν A (a)? (d) Describe the sequence a 1 as a sequence of integers. ν A (a) (e) Verify that ρ A (a) = τ A (a 1 ) (cf. Lemma 4.4.) ν A (a) (a) ρ is v 0 + (v 0 + v 2 ) v 2 ; as a sequence of integers it is 7, 5, 9, 7, 5, 15, 15. (b) ρ A (a) = 0 + (0 + 2) 2 = 4. (c) ν A (a) = = 2. (d) a 1 = 0, 2, 2, 3,.... ν A (a) (e) ρ A (a) = 4, as above; τ A (a 1 ) = = 4. ν A (a) 12

13 E4.5 In the language for A def = (ω, S, 0, +, ), let ϕ be the formula v 0 (v 0 v 1 = v 1 ), let ν be the formula v 1 + v 1, and let a = 1, 0, 1, 0,.... (a) Describe Subf v 1 ν ϕ as a sequence of integers (b) What is ν A (a)? (c) Describe a 1 ν A (a) (d) Determine whether A = Subf v 1 ν (e) Determine whether A = ϕ[a 1 ν A (a) as a sequence of integers. ϕ[a] or not. ] or not. (a) Subf v 1 ν ϕ is v 0 (v 0 (v 1 + v 1 ) = v 1 + v 1 ; as a sequence of integers it is 4, 5, 3, 9, 5, 7, 10, 10, 7, 10, 10. (b) ν A (a) = (v 1 + v 1 ) A ( 1, 0, 1, 0,... ) = = 0. (c) a 1 = 1, 0, 1, 0,.... ν A (a) (d) A = Subf v 1 ν ϕ[a] iff A = [ v 0(v 0 (v 1 +v 1 ) = v 1 +v 1 ][ 1, 0, 1, 0,... ] iff for all a ω, a (0 + 0) = 0 + 0; this is true. (e) A = ϕ[a 1 ] iff A = [ v 0(v 0 v 1 = v 1 ][ 1, 0, 1, 0,... ] iff for all a ω, a 0 = 0; ν A (a) this is true. E4.6 Show that the condition in Lemma 4.6 that no free occurrence of v i in ϕ is within a subformula of the form v k µ with v k a variable occurring in ν is necessary for the conclusion of the lemma. In the language for A = (ω, S, 0, +, ), let ϕ be the formula v 1 [Sv 1 = v 0 ], ν = v 1, and a = 1, 1,.... Note that the condition on v 0 fails. Now Subf v 0 v 1 ϕ is the formula v 1 [Sv 1 = v 1 ], and there is no a ω such that Sa = a, and hence A = Subf v 0 v 1 ϕ[a]. Also, ν A (a) = v1 A (a) = a 1 = 1, and hence a 0 = 1, 1,.... Since S0 = 1, it follows that ν A (a) A = ϕ[a 0 ν A (a) ]. E4.7 Let A be an L -structure, with L arbitrary. Define Γ = {ϕ : ϕ is a sentence and A = ϕ[a] for some a : ω A}. Prove that Γ is complete and consistent. Note by Lemma 4.4 that A = ϕ[a] for some a : ω A iff A = ϕ[a] for every a : ω A. Let ϕ be any sentence. Take any a : ω A. If A = ϕ[a], then ϕ Γ and hence Γ ϕ. Suppose that A = ϕ[a]. Then A = ϕ[a], hence ϕ Γ, hence Γ ϕ. This shows that Γ is complete. Suppose that Γ is not consistent. Then Γ (v 0 = v 0 ) by Lemma 4.1. Then Γ = (v 0 = v 0 ) by Theorem 3.2. Since A is a model of Γ, it is also a model of (v 0 = v 0 ), contradiction. E4.8 Call a set Γ strongly complete iff for every formula ϕ, Γ ϕ or Γ ϕ. Prove that if Γ is strongly complete, then Γ v 0 v 1 (v 0 = v 1 ). Assume that Γ is strongly complete. Then Γ v 0 = v 1 or Γ (v 0 = v 1 ). If Γ v 0 = v 1, then by generalization, Γ v 0 v 1 (v 0 = v 1 ). Suppose that Γ (v 0 = v 1 ). Then by 13

14 generalization, Γ v 0 (v 0 = v 1 ). By Theorem 3.25, Γ v 0 (v 0 = v 1 ) (v 1 = v 1 ). Hence Γ (v 1 = v 1 ). But also Γ v 1 = v 1 by Proposition 3.4, so Γ is inconsistent by Lemma 4.1, and hence again Γ v 0 v 1 (v 0 = v 1 ). E4.9 Prove that if Γ is rich, then for every term σ with no variables occurring in σ there is an individual constant c such that Γ σ = c. By richness we have Γ v 0 (v 0 = σ) c = σ for some individual constant c. Then using (L4) it follows that Γ c = σ. E4.10 Prove that if Γ is rich, then for every sentence ϕ there is a sentence ψ with no quantifiers in it such that Γ ϕ ψ. We proceed by induction on the number m of symbols,, in ϕ. (More exactly, by the number of the integers 1,2,4 that occur in the sequence ϕ.) If m = 0, then ϕ is atomic and we can take ψ = ϕ. Assume the result for m and suppose that ϕ has m + 1 integers 1,2,4 in it. Then there are three possibilities. First, ϕ = ϕ. Let ψ be a quantifier-free sentence such that Γ ϕ ψ. Then Γ ϕ ψ. Second, ϕ = (ϕ ϕ ). Choose quantifier-free sentences ψ and ψ such that Γ ϕ ψ and Γ ϕ ψ. Then Γ ϕ (ψ ψ ). Third, ϕ = v i ϕ. By richness, let c be an individual constant such that Γ v i ϕ Subf v i c ϕ. Then by Theorem 3.31 we get (1) Γ v i ϕ Subf v i c ϕ. Now Subf v i c ϕ has only m integers 1,2,4 in it, so by the inductive hypothesis there is a sentence ψ with no quantifiers in it such that Γ Subf v i c ϕ ψ and hence (2) Γ Subf v i c ϕ ψ. From (1) and (2) and a tautology we get Γ v i ϕ ψ. Then by Proposition 3.31, Γ v i ϕ ψ, finishing the inductive proof. E4.11 Describe sentences in a language for ordering which say that < is a linear ordering and there are infinitely many elements. Prove that the resulting set Γ of sentences is not complete. Let Γ consist of the following sentences: v 0 (v 0 < v 0 ); v 0 v 1 v 2 [v 0 < v 1 v 1 < v 2 v 0 < v 2 ]; v 0 v 1 [v 0 < v 1 v 0 = v 1 v 1 < v 0 ]; (v i = v j ) for every positive integer n. i<j<n The following sentence ϕ holds in (Q, <) but not in (ω, <): v 0 v 1 [v 0 < v 1 v 2 (v 0 < v 2 v 2 < v 1 )]. Since ϕ does not hold in (ω, <), we have Γ ϕ, by Theorem 4.2. But since ϕ holds in (Q, <), we also have Γ ϕ by Theorem 4.2. So Γ is not complete. 14

15 E4.12 Prove that if a sentence ϕ holds in every infinite model of a set Γ of sentences, then there is an m ω such that it holds in every model of Γ with at least m elements. Suppose that ϕ holds in every infinite model of a set Γ of sentences, but for every m ω there is a model M of Γ with at least m elements such that ϕ does not hold in M. Let be the following set: Γ (v i = v j ) : n a positive integer { ϕ}. i<j<n Our hypothesis implies that every finite subset of has a model; for if m is the maximum of all n such that the above big conjunction is in, then the hypothesis yields a model of. By the compactness theorem we get a model N of. Thus N is an infinite model of Γ in which ϕ does not hold, contradiction. E4.13 Let L be the language of ordering. Prove that there is no set Γ of sentences whose models are exactly the well-ordering structures. Suppose there is such a set. Let us expand the language L to a new one L by adding an infinite sequence c m, m ω, of individual constants. Then consider the following set Θ of sentences: all members of Γ, plus all sentences c m+1 < c m for m ω. Clearly every finite subset of Θ has a model, so let A = (A, <, a i ) i<ω be a model of Θ itself. (Here a i is the 0-ary function, i.e., element of A, corresponding to c i.) Then a 0 > a 1 > ; so {a i : i ω} is a nonempty subset of A with no least element, contradiction. E4.14 Suppose that Γ is a set of sentences, and ϕ is a sentence. Prove that if Γ = ϕ, then = ϕ for some finite Γ. We prove the contrapositive: Suppose that for every finite subset of Γ, = ϕ. Thus every finite subset of Γ { ϕ} has a model, so Γ { ϕ} has a model, proving that Γ = ϕ. E4.15 Suppose that f is a function mapping a set M into a set N. Let R = {(a, b) : a, b M and f(a) = f(b)}. Prove that R is an equivalence relation on M. If a M, then f(a) = f(a), so (a, a) R. Thus R is reflexive on M. Suppose that (a, b) R. Then f(a) = f(b), so f(b) = f(a) and hence (b, a) R. Thus R is symmetric. Suppose that (a, b) R and (b, c) R. Then f(a) = f(b) and f(b) = f(c), so f(a) = f(c) and hence (a, c) R. E4.16 Suppose that R is an equivalence relation on a set M. Prove that there is a function f mapping M into some set N such that R = {(a, b) : a, b M and f(a) = f(b)}. Let N be the collection of all equivalence classes under R. For each a M let f(a) = [a] R. Then (a, b) R iff a, b M and [a] R = [b] R iff a, b M and f(a) = f(b). E4.17 Let Γ be a set of sentences in a first-order language, and let be the collection of all sentences holding in every model of Γ. Prove that = {ϕ : ϕ is a sentence and Γ ϕ}. For, suppose that ϕ. To prove that Γ ϕ we use the compactness theorem, proving that Γ = ϕ. Let A be any model of Γ. Since ϕ, it follows that A is a model of Γ, as desired. 15

16 For, suppose that ϕ is a sentence and Γ ϕ. Then by the easy direction of the completeness theorem, Γ = ϕ. That is, every model of Γ is a model of ϕ. Hence ϕ. Solutions, Chapter 6 E6.1 Prove that if f : A B and C i : i I is a system of subsets of A, then f [ i I C ] i = i I f[c i]. [ ] x f C i i I iff x rng ( f i I C i ) iff y i I C i [f(y) = x] iff i I y C i [f(y) = x] iff i I[x rng(f C i )] iff i I[x f[c i ]] iff x i I f[c i ]. E6.2 Prove that if f : A B and C, D A, then f[c D] f[c] f[d]. Give an example showing that equality does not hold in general. Take any x f[c D]. Choose y C D such that x = f(y). Since y C, we have x f[c]. Similarly, x f[d]. So x f[c] f[d]. Since x is arbitrary, this shows that f[c D] f[c] f[d]. For the required example, let dmn(f) = {a, b} with a b and with f(a) = a = f(b). Let C = {a} and D = {b}. Then C D =, so f[c D] =, while f[c] = {a} = f[d] and hence f[c] f[d] = {a}. So f[c D] f[c] f[d]. E6.3 Given f : A B and C, D A, compare f[c\d] and f[c]\f[d]: prove the inclusions (if any) which hold, and give counterexamples for the inclusions that fail to hold. We claim that f[c]\f[d] f[c\d]. For, suppose that x f[c]\f[d]. Choose c C such that x = f(c). Since x / f[d], we have c / D. So c C\D and hence x f[c\d], proving the claim. The other inclusion does not hold. For, take the same f, C, D as for exercise E6.2. Then C\D = {a} and so f[c\d]. But f[c] = {a} = f[d], so f[c]\f[d] =. E6.4 Prove that if f : A B and C i : i I is a system of subsets of B, then f [ 1 i I C ] i = i I f 1 [C i ]. For any b B we have b f 1 [ i I C i ] iff f(b) i I C i iff i I[f(b) C i ] 16

17 iff i I[b f 1 [C i ]] iff b i I f 1 [C i ]. E6.5 Prove that if f : A B and C i : i I is a system of subsets of B, then f [ 1 i I C ] i = i I f 1 [C i ]. For any a, a f 1 [ i I C i ] iff f(a) i I C i iff i I[f(a) C i ] iff i I[a f 1 [C i ]] iff a i I f 1 [C i ]. E6.6 Prove that if f : A B and C, D B, then f 1 [C\D] = f 1 [C]\f 1 [D]. For any a, a f 1 [C\D] iff f(a) C\D E6.7 Prove that if f : A B and C A, then iff iff iff f(a) C and f(a) / D a f 1 [C] and a / f 1 [D] a f 1 [C]\f 1 [D]. {b B : f 1 [{b}] C} = B\f[A\C]. First suppose that b is in the left side; but suppose also, aiming for a contradiction, that b f[a\c]. Say b = f(a), with a A\C. Then a f 1 [{b}], so a C, contradiction. Second, suppose that b is in the right side. Take any a f 1 [{b}]. Then f(a) = b, and it follows that a C, as desired. E6.8 For any sets A, B define A B = (A\B) (B\A); this is called the symmetric difference of A and B. Prove that if A, B, C are given sets, then A (B C) = (A B) C. Let D = A B C, A = D\A, B = D\B, and C = D\C. Then A B = (A B ) (B A ); (A B) = ((A B ) (B A )) = (A B ) (B A ) = (A B) (B A) = (A B ) (A B). 17

18 These equations hold for any sets A, B. Now A (B C) = (A (B C) ) ((B C) A = (A ((B C ) (B C))) (((B C ) (C B )) A ) = (A B C ) (A B C) (A B C ) (A B C). This holds for any sets A, B, C. Hence (A B) C = C (A B) = (C A B ) (C A B) (C A B ) (C A B) = A (B C). E6.9 For any set A let Id A = { x, x : x A}. Justify this definition on the basis of the axioms. Id A = {y A A : x A[y = x, x ]}. E6.10 Suppose that f : A B. Prove that f is surjective iff there is a g : B A such that f g = Id B. Note: the axiom of choice might be needed. : given b B, we have b = (f g)(b) = f(g(b)); so f is surjective. : Assume that f is surjective. Let A = {{(b, a) : a A, f(a) = b} : b B}. Each member of A is nonempty; for let x A. Choose b B such that x = {(b, a) : a A, f(a) = b}. Choose a A such that f(a) = b. So (b, a) x. The members of A are pairwise disjoint: suppose x, y A with x y. Choose b, c so that x = {(b, a) : a A, f(a) = b} and y = {(b, a) : a A, f(a) = c}. If u x y, then there exist a, a A such that u = (b, a), f(a) = b, and also (u = (c, a ), f(a ) = c. So by Theorem 6.3, b = c. But then x = y, contradiction. Now by the axiom of choice, let C have exactly one element in common with each member of A. Then define g = {(b, a) C : a A, f(a) = b}. Now g is a function. For, suppose that (b, a), (b, a ) g. Let x = {(b, a ) : a A, f(a ) = b}. Then (b, a), (b, a ) C x, so (b, a) = (b, a ). Hence a = a. Clearly g B A. Next, dmn(g) = B, for suppose that b B. Choose x C {(b, a ) : a A, f(a ) = b}; say x = (b, a) with a A, f(a) = b. Then x g and so b dmn(g). Thus g : B A. Take any b B, and let g(b) = a. So (b, a) g and hence f(a) = b. So f g = Id B. 18

19 E6.11 Let A be a nonempty set. Suppose that f : A B. Prove that f is injective iff there is a g : B A such that g f = Id A. First suppose that f is injective. Fix a A, and let g = f 1 {(b, a) : b B\rng(f)}. Then g is a function. In fact, suppose that (b, c), (b, d) g. If both are in f 1, then (c, b(, (d, b) f, so f(c) = b = f(d) and hence c = d since f is injective. If (b, c) f 1 and b B\rng(f), the (c, b) f, so b rng(f), contradiction. If (b, c), (b, d) / f 1, then c = d = a. Clearly then g : B A. For any a A we have (a, f(a)) f, hence (f(a), a) f 1 g, and so g(f(a)) = a. Second, suppose that g : B A and g f = Id A. Suppose that f(a) = f(a ). Then a = (g f)(a) = g(f(a)) = g(f(a )) = (g f)(a ) = a. E6.12 Suppose that f : A B. Prove that f is a bijection iff there is a g : B A such that f g = Id B and g f = Id A. Prove this without using the axiom of choice. : Assume that f is a bijection. By E6.11 there is a g : B A such that g f = Id A. We claim that f g = Id B. Since f is a bijection, the relation f 1 is also a bijection. Now for any b B, (f g)(b) = f(g(b)) = f(g(f(f 1 (b)))) = f((g f)(f 1 (b))) = f(f 1 (b)) = b. So f g = Id B, as desired. : Assume that g is as indicated. Then f is injective, since f(a) = f(b) implies that a = g(f(a)) = g(f(a )) = a. And f is surjective, since for a given b B we have f(g(b)) = b. E6.13 For any sets R, S define R S = {(x, z) : y((x, y) R (y, z) S)}. Justify this definition on the basis of the axioms. R S = {(x, z) dmn(r) rng(s) : y((x, y) R (y, z) S)}. E6.14 Suppose that f, g : A A. Prove that is a function. (A A)\[((A A)\f) ((A A)\g)] Suppose that (x, y), (x, z) are in the indicated set, with y z. By symmetry say f(x) y. Then (x, y) [(A A)\f], so it follows that (y, z) g, as otherwise (x, z) [((A A)\f) ((A A)\g)]. Hence (y, y) / g, so (x, y) [((A A)\f) ((A A)\g)], contradiction. 19

20 E6.15 Suppose that f : A B is a surjection, g : A C, and x, y A[f(x) = f(y) g(x) = g(y)]. Prove that there is a function h : B C such that h f = g. Define h as a set of ordered pairs. Let h = {(f(a), g(a)) : a A}. Then h is a function, for suppose that (x, y), (x, z) h. Choose a, a A so that x = f(a), y = g(a), x = f(a ), and y = g(a ). Thus f(a) = f(a ), so g(a) = g(a ), as desired. Since f is a surjection it is clear that dmn(h) = B. Clearly rng(h) C. So h : B C. If a A, then (f(a), b(a)) h, hence h(f(a)) = g(a). This shows that h f = g. E6.16 The statement A A B B(A B) implies that A B is slightly wrong. Fix it, and prove the result. If A has a nonempty member and B is empty, the implication does not hold. Add the hypothesis B. Suppose that a A and B B; we want to show that a B. Choose A A such that a A. Since A B, we have a B. E6.17 Suppose that A A B B(A B). Prove that A B. Suppose that a A ; we want to show that a B. Choose A A such that a A. Then choose B B such that A B. Then a B. Hence a B. E6.18 The statement A A B B(B A) implies that B A. is slightly wrong. Fix it, and prove the result. If A is empty and B is nonempty, the statement is false. Fix it by adding the hypothesis that A is nonempty. Suppose that b B and A A ; we want to show that b A. Choose B B such that B A. Now b B since b B, so b A. Solutions, exercises in Chapter 7 E7.1 Prove that if x is an ordinal, then x is transitive and (x, {(y, z) x x : y z}) is a well-ordered set. By definition, x is transitive. Let R = {(y, z) x x : y z}). Obviously R is a relation. By definition, R x x. R is irreflexive on x by Theorem 7.5. R is transitive since x is transitive. R is linear on x by Theorem 7.7. The final well-ordering property follows from Theorem E7.2 Assume that x is transitive and (x, {(y, z) x x : y z}) is a well-ordered set. Prove that for all y, z x, either y = z or y z or z y. This is obvious. E7.3 Assume that x is transitve and for all y, z x, either y = z or y z or z y. Prove that for all y, if y x and y is transitive, then y x. Hint: apply the foundation axiom to x\y. 20

21 Assume the hypothesis, and suppose that y x and y is transitive. Choose z x\y such that z (x\y) =. If u y, then u x since y x. So u, z x, so by hypothesis we have u z, u = z, or z u. Now u z since z / y and u y. And z / u, since z u would imply, because y is transitive and u y, that z y, which is not true. Hence u z. This is true for any u y. So y z. Clearly also z y, so y = z x. E7.4 Assume that x is transitive and for all y, if y x and y is transitive, then y x. Show that x is an ordinal. Hint: let y = {z x : z is an ordinal}, and get a contradiction from the assumption that y x. Assume the hypothesis. Let y = {z x : z is an ordinal}. So y x. Suppose that y x. Now y is transitive, for assume that z y. Thus z x and z is an ordinal. Suppose that w z. Then w x since x is transitive, and w is an ordinal since z is an ordinal. So w y. Thus, indeed, y is transitive. So by assumption y x. Now y is a transitive set of transitive sets, so y is an ordinal. It follows that y y, contradiction. This proves that x = y. So x is a transitive set of transitive sets, and hence x is an ordinal. E7.5 Show that if x is an ordinal, then the following two conditions hold: (i) For all y x, either y {y} = x or y {y} x. (ii) For all y x, either y = x or y x. Assume that x is an ordinal. Then (i) holds by Proposition Now suppose that y x. If z y, choose w y such that z w. Then also w x, so z x since x is transitive. This shows that y x. By Proposition 7.3, y is an ordinal. Hence by Proposition 7.8, y x. E7.6 Assume the two conditions of exercise E7.5. Show that x is an ordinal. Hint: Show that there is an ordinal α not in x. Taking such an ordinal α, show that there is a least β α {α} such that β / x. Work with such a β to show that x is an ordinal. By Theorem 7.6 there is an ordinal α not in x. Then by Theorem 7.13 there is a least β α {α} such that β / x. Now we have two possibilities: Case 1. β = β. Now β x, so by (ii) second clause, since β = β / x we have x = β, hence x is an ordinal, as desired. Case 2. β = ( β) + 1. Thus β is an ordinal smaller than β, so it is in x. By (i), since β = β + 1 / x we have x = ( β) + 1, hence x is an ordinal. Solutions, exercises in Chapter 8 E8.1 Give an example of A,R such that R is not well-founded on A and is not set-like on A. We take On and R, where R = {(α, β) : α > β}. As shown after 8.2, R is not set-like on On. It is also not well-founded on On, since ω is a nonempty set of ordinals, but if m ω then (m + 1, m) R, so that ω does not have an R-minimal element. E8.2 Give an example of A,R such that R is not well-founded on A but is set-like on A. Give one example with R and A are proper classes, and one example where they are sets. 21

22 Both are sets: let A = ω and R = {(m, n) : m, n ω and m > n. Then ω does not have an R-minimal element, since for any m ω we have (m + 1, m) R. Both are proper classes: let A = On and let R = {(m, n) : m, n ω and m > n} {(α, β) : α < β}. E8.3 Give an example of A,R such that R is well-founded on A but is not set-like on A. Let A = V and R = {(a, ) : a V, a }. Thus pred AR ( ) = V, so R is not set-like on V. Now let X be a nonempty set. If X = { }, then X and a X[(a, ) / R]. If X { }, take any a X\{ }. Then b X[(b, a) / R]. E8.4 Suppose that R is a class relation contained in A A, x A, and v pred AR (x). Prove by induction on n that if n ω\1, f is a function with domain n + 1, i < n[(f(i), f(i + 1)) R] and f(n) = v, then f(0) pred AR (x). Suppose that R is a class relation contained in A A, x A, and v pred AR (x). We take n = 1 in the condition to be proved. So, suppose that f is a function with domain 2 such that i < 1[(f(i), f(i + 1)) R] and f(1) = v. Thus (f(0), v) R, so f(0) pred AR (v). By Lemma 8.3(ii), f(0) pred AR (x). Now suppose that if n ω\1, f is a function with domain n + 1, i < n[(f(i), f(i + 1)) R] and f(n) = v, then f(0) pred AR (x). Suppose also now that f is a function with domain n + 2, i < n + 1[(f(i), f(i + 1)) R] and f(n + 1) = v. Define g with domain n + 1 by setting g(i) = f(i + 1) for all i < n + 1. Then i < n[(g(i), g(i + 1)) = (f(i + 1), f(i + 2)) R] and g(n) = f(n + 1) = v. Hence by the inductive assumption, f(1) = g(0) pred AR (x). We also have (f(0), f(1)) R, so by Lemma 8.3(ii), f(0) pred AR (x). E8.5 Suppose that R is a class relation contained in A A, (u, v) R, and (v, w) R. Show that (u, w) R. Assume that R is a class relation contained in A A, (u, v) R, and (v, w) R. Since (u, v) R, there exist n ω\1 and a function f with domain n + 1 such that i < n[(f(i), f(i + 1)) R, f(0) = u, and f(n) = v. From exercise E8.4 it follows that (u, w) R. E8.6 Give an example of a proper class X which has a proper class of -minimal elements. Let X = {{α} : α 2}. We claim that all elements of X are -minimal. Suppose that α, β 2 and {α} {β}. Then {α} = β, Since β 2 we have 0, 1 β, so 0 = α = 1, contradiction. E8.7 Give an example of a proper class relation R contained in A A for some proper class A, and a class function G mapping A V into V such that R is set-like on A but not well-founded on A and there is no class function F mapping A into V such that F(a) = G(a,F pred AR (a)) for all a A. Let A = On and R = {(m, n) : m, n ω and m > n} {(α, β) : ω α < β}. 22

23 Thus R is a proper class relation contained in A A. Clearly R is set-like on On but it is not well-founded on On. Define G : On V V by setting { {a(α + G(α, a) = 1)} if α ω and a is a function with domain {m ω : m > α}, otherwise. Suppose that F : A V is such that F(α) = G(α,F pred OnR (α)) for all α On. Let f = F ω. Choose b rng(f) such that b rng(f) =. Say b = f(m) with m ω. Now f(m) = F(m) = G(m,F pred OnR (m)) = G(m,F {n : n ω, n > m}) = {F(m + 1)} = {f(m + 1)}, so that f(m + 1) f(m) rng(f), contradiction. E8.8 Give an example of a proper class relation R contained in some A A for some proper class A and a class function G mapping A V into V such that R is set-like on A but not well-founded on A but still there is a class function F mapping A into V such that F(a) = G(a,F pred AR (a)) for all a A. Let A and R be as in exercise E8.7, but define G(α, a) = α for all α On and all a V. Then the function F : On V such that F(α) = α for all α On is as desired. Solutions to exercises in chapter 9 E9.1 Let (A, <) be a well order. Suppose that B A and b B a A[a < b a B]. Prove that there is an element a A such that B = {b A : b < a}. Let a be the least element of A\B. We claim that a is as desired. For, if b B, then it cannot happen that a b, since this would imply that a B; so b < a. And if b < a, then b B by the minimality of a. E9.2 Let (A, <) be a well order. Suppose that B A and b B a A[a < b a B]. Prove that (A, <) is not isomorphic to (B, <). Suppose that f is such an isomorphism from (A, <) onto (B, <). By exercise E9.1, let a A be such that B = {x A : x < a}. By Proposition 9.11, a f(a), contradicting the assumption that f maps into B. E9.3 Suppose that f is a one-one function mapping an ordinal α onto a set A. Define a relation which is a subset of A A such that (A, <) is a well-order and f is an isomorphism of (α, <) onto (A, ). Define = {(a, b) A A : f 1 (a) < f 1 (b)}. We check that (A, <) is a well-order. If a A and a a, then f 1 (a) < f a (a), contradiction. So is irreflexive. Suppose that a b c. Then f 1 (a) < f 1 (b) < f 1 (c), so f 1 (a) < f 1 (c) and hence a c. So is transitive. Now given a, b A, either f 1 (a) < f 1 (b) or f 1 (a) = f 1 (b) or f 1 (b) < f 1 (a), so a b or a = b or b a. Thus (A, ) is a linear order. Finally, suppose that X A. Then f 1 [X], so let ξ be the least element of f 1 [X]. Then f(ξ) X. Suppose that b X. Then f 1 (b) f 1 [X], so ξ f 1 (b). Hence 23

24 f(ξ) b. This shows that f(ξ) is the -least element of X. We have shown that (A, ) is a well-order. We are given that f is a bijection from α onto A. If ξ, η α and ξ < η, then f(ξ) f(η). If f(ξ) f(η), then ξ < η. Thus f is an isomorphism. E9.4 Prove that 1 + m = m + 1 for any m ω. (Ordinary) induction on m = 1 = using Theorem 9.21(vi). Assume that 1 + m = m + 1. Then 1 + (m + 1) = (1 + m) + 1 = (m + 1) + 1. E9.5 Prove that m + n = n + m for any m, n ω. With m fixed, induction on n. 0 + m = m = m + 0 using Theorem 9.21(vi). Assume that m + n = n + m. Then (n + 1) + m = n + (1 + m) = n + (m + 1) (by exercise E9.4) = (n + m) + 1 = (m + n) + 1 = m + (n + 1). E9.6 Prove that ω α iff 1 + α = α. First note that 1 + ω = m ω (1 + m) = m ω (m + 1) = ω, using Theorem 9.21(vi). : Assume that ω α. By Theorem 9.21(vii) let δ be such that ω + δ = α. Then 1 + α = 1 + (ω + δ) = (1 + ω) + δ = ω + δ = α. : It suffices to show that if m < ω then 1 + m m. This is true by Theorem 9.21(vi). E9.7 For any ordinals α, β let α β = (α {0}) (β {1}). We define a relation as follows. For any x, y α β, x y iff one of the following three conditions holds: (i) There are ξ, η < α such that x = (ξ, 0), y = (η, 0), and ξ < η. (ii) There are ξ, η < β such that x = (ξ, 1), y = (η, 1), and ξ < η. (ii) There are ξ < α and η < β such that x = (ξ, 0) and y = (η, 1). Prove that (α β, ) is a well order which is isomorphic to α + β. Clearly is a well-order. We show by transfinite induction on β, with α fixed, that (α β, ) is order isomorphic to α + β. For β = 0 we have α + β = α + 0 = α, while α β = α 0 = α {0}. Clearly ξ (ξ, 0) defines an order-isomorphism from α onto (α {0}, ). So our result holds for β = 0. Assume it for β, and suppose that f is an order-isomorphism from α + β onto (α β, ). Now the last element of α (β + 1) is (β, 1), and the last element of α + (β + 1) is α + β, so the function f {(α + β, (β, 1))} is an order-isomorphism from α + (β + 1) onto α (β + 1). Now assume that β is a limit ordinal, and for each γ < β, the ordinal α + γ is isomorphic to α γ. For each such γ let f γ be the unique isomorphism from α + γ onto 24

25 α γ. Note that if γ < δ < β, then f δ γ is an isomorphism from α +γ onto α γ; hence f δ γ = f γ. It follows that is an isomorphism from α + β onto α β, finishing the inductive proof. E9.8 Given ordinals α, β, we define the following relation on α β: γ<β (ξ, η) (ξ, η ) iff ((ξ, η) and (ξ, η ) are in α β and: f γ η < η, or (η = η and ξ < ξ ). We may say that this is the anti-dictionary or anti-lexicographic order. Show that the set α β under the anti-lexicographic order is a well order which is isomorphic to α β. We may assume that α 0. It is straightforward to check that is a well-order. Now we define, for any (ξ, η) α β, f(ξ, η) = α η + ξ. We claim that f is the desired order-isomorphism from α β onto α β. If (ξ, η) α β, then f(ξ, η) = α η + ξ < α η + α = α (η + 1) α β. Thus f maps into α β. To show that f is one-one, suppose that (ξ, η), (ξ, η ) α β and f(ξ, η) = f(ξ, η ). Then by Theorem 9.26, (ξ, η) = (ξ, η ). So f is one-one. To show that f maps onto α β, let γ < α β. Choose ξ and η so that γ = α η + ξ with ξ < α. Now η < β, as otherwise γ = α η + ξ α η α β. It follows that f(ξ, η) = α η + ξ = γ. so f is onto. Finally, we show that the order is preserved. Suppose that (ξ, η) (ξ, η ). Then one of these cases holds: Case 1. η < η. Then f(ξ, η) = α η + ξ < α η + α = α (η + 1) α η α η + ξ = f(ξ, η ), as desired. Case 2. η = η and ξ < ξ. Then f(ξ, η) < f(ξ, η ). Now it follows that f is the desired isomorphism. E9.9 Suppose that α and β are ordinals, with β 0. We define α β w = {f α β : {ξ < α : f(ξ) 0} is finite}. 25

26 For f, g α β w we write f g iff f g and f(ξ) < g(ξ) for the greatest ξ < α for which f(ξ) g(ξ). Prove that ( α β w, ) is a well-order which is order-isomorphic to the ordinal exponent β α. (A set X is finite iff there is a bijection from some natural number onto X.) If α = 0, then β α = 1, and α β w also has only one element, the empty function (= the emptyset). So, assume that α 0. If β = 1, then α β w has only one member, namely the function with domain α whose value is always 0. This is clearly order-isomorphic to 1, as desired. So, suppose that β > 1. Now we define a function f mapping β α into α β w. Let f(0) be the member of α β w which takes only the value 0. Now suppose that 0 < ε < β α. By Theorem 9.29 write ε = β γ(0) δ(0) + β γ(1) δ(1) + + β γ(m 1) δ(m 1), where ε γ(0) > γ(1) > > γ(m 1) and 0 < δ(i) < β for each i < m. Note that β γ(0) ε < β α, so γ(0) < α. Then we define, for any ζ < α, (f(ε))(ζ) = { 0 if ζ / {γ(0),..., γ(m 1)}, δ(i) if ζ = γ(i) with i < m. Clearly f(ε) α β w. To see that f maps onto α β w, suppose that x α β w. If x takes only the value 0, then f(0) = x. Suppose that x takes on some nonzero value. Let {ξ < α : x(ξ) 0} = {γ(0), γ(1),..., γ(m 1)}, where γ(0) > γ(1) > > γ(m 1). Let δ(i) = x(γ(i)) for each i < m, and let ε = β γ(0) δ(0) + β γ(1) δ(1) + + β γ(m 1) δ(m 1). Clearly then f(ε) = x. Now we complete the proof by showing that for any ε, θ < β α, ε < θ iff f(ε) < f(θ). This equivalence is clear if one of ε, θ is 0, so suppose that both are nonzero. Write ε = β γ(0) δ(0) + β γ(1) δ(1) + + β γ(m 1) δ(m 1), where α γ(0) > γ(1) > > γ(m 1) and 0 < δ(i) < β for each i < m, and θ = β γ (0) δ (0) + β γ (1) δ (1) + + β γ (n 1) δ (n 1), where α γ (0) > γ (1) > > γ (n 1) and 0 < δ (i) < β for each i < n. By symmetry we may suppose that m n. Note that N(β, m, γ, δ), k(β, m, γ, δ) = ε, N(β, n, γ, δ ), and k(β, n, γ, δ ) = θ. We now consider several possibilities. Case 1. ε = θ. Then clearly f(ε) = f(θ). Case 2. γ γ, δ δ, and m < n. Thus ε < θ. Also, γ (m) is the largest ξ < α such that (f(ε))(ξ) (f(θ))(ξ), and (f(ε))(ξ) = 0 < δ (m) = (f(θ))(γ (m)), so f(ε) < f(θ). 26