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Math 114L

Homework 2 Solutions

Spring 2011

1.7.2

Let be finitely satisfiable and complete, and let be as given in the problem.

We show by induction on that () = T iff .Base Case: If is a sentence symbol, () = () = T iff by the

definition of.Inductive Case for : If () = T then () = F, so by IH, , and

since is complete, we must have . If() = F then () = T, soby IH, ; if then {,} is a finite unsatisfiable subset of , andsince this is impossible, .

Inductive Case for : If ( ) = T then () = () = T, so by IH{, } ; if then ( ) , and so {,,( )} is afinite unsatisfiable subset of , and since this is impossible, . If () = F and () = F then by IH, , so . If we had then {,} would be a finite unsatisfiable subset of , and since there arenone, . If( ) = F and () = T then () = F, and a similarargument applies.

The other inductive cases are simialr.

1.7.3

Suppose Corollary 17A holds and that is not satisfiable. If were unsatis-fiable, we would have A1 A1. By the corollary, there must be a finite0 such that 0 A1 A1, and therefore 0 is a finite unsatisfiablesubset of , so is not finitely satisfiable. This is the contrapositive of thecompactness theorem, and therefore equivalent to the compactness theorem.

1.7.10

a

Let be the set of tautological consequences of . Consider the followingprocedure: For an expression , first check if is a wff. By Theorem 17B, thereis an effective procedure for this, so in a finite number of steps, this will produceyes if is a wff, no if is not. If no, we will output no. If yes, wewill continue as follows.

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is effectively enumerable, so we have an effective way to enumerate ,

{0

, 1

, 2

, . . .}. Let n

= {i

| i n}. Since each n

is finite, there is aneffective procedure for determining whether n . We successively check if0 , then if 0 , then if 1 , then if 1 , and so on. If we findsome n such that n , we output yes, if we find some n such that n ,we output no.

By assumption, either |= or |= but not both, so at some n, we willhave either n |= or n |= , so this procedure always eventually stops withthe correct answer.

b

Suppose that there is such that |= and |= . Then is unsatisfiable,so for every wff . Therefore we take the decision procedure which

outputs yes on every wff.Otherwise, there is no such , so we are in the case of part a.(Note that the procedure here is non-uniform, in the sense that we cant

decide, given a description of , which of the two prcedures to use. But thatdoesnt change the fact that the set is decidable, we just dont know how todecide it!)

1.7.11

a

Let A and B be effectively enumerable. By Theorem 17E, they are both semide-cidable. Let C= AB and let be an expression. We dovetail the two semide-

cision procedures: we first spend 1 minute checking if A, then 1 minutechecking if B, then 2 minutes checking if A, then 2 minutes checkingif B, and so on. If either A or B, this process will eventuallystop, and we output yes. If A B, this process runs forever. This is asemidecision procedure, so C is effectively enumerable.

b

Again, let A and B be effectively enumerable, and note that by Theorem 17Ethey are each semidecidable. Given an expression , first run the semidecisionprocedure checking if A. If the procedure runs forever, A, so AB,and we run forever.

Otherwise, the semidecision procedure eventually tells us A. Then we

run the semidecision procedure for B. If this procedure runs forever, B, so A B, and we run forever. If this procedure stops, we output yes, since A and B.

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1.7.12

a

= {A1,A1}

b

= {A1, A2,(A1 A2)}

c

= {A1, A2, A3,(A1 A2 A3)}

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