The last step introducing y is not the bad one; that step is fine. The bad one is the second from last one, concluding Q(x0) by x e and violating the side condition that x0 may not leave the scope of its box. You can try a few other ways of ‘proving’ this sequent, but none of them should work (assuming that our proof system is sound with respect to semantic entailment, which we define in the next section). Without this side condition, we would also be able to prove that ‘all x satisfy the property P as soon as one of them does so,’ a semantic disaster of biblical proportions!

2.3.2 Quantifier equivalences

We have already hinted at semantic equivalences between certain forms of quantification. Now we want to provide formal proofs for some of the most commonly used quantifier equivalences. Quite a few of them involve several quantifications over more than just one variable. Thus, this topic is also good practice for using the proof rules for quantifiers in a nested fashion.

For example, the formula x y φ should be equivalent to y x φ since both say that φ should hold for all values of x and y. What about ( x φ) ( x ψ) versus x (φ ψ)? A moment’s thought reveals that they should have the same meaning as well. But what if the second conjunct does not start with x? So what if we are looking at ( x φ) ψ in general and want to compare it with x (φ ψ)? Here we need to be careful, since x might be free in ψ and would then become bound in the formula x (φ ψ).

Example 2.12 We may specify ‘Not all birds can fly.’ as ¬x (B(x) → F (x)) or as x (B(x) ¬F (x)). The former formal specification is closer to the structure of the English specification, but the latter is logically equivalent to the former. Quantifier equivalences help us in establishing that specifications that ‘look’ di erent are really saying the same thing.

Here are some quantifier equivalences which you should become familiar with. As in Chapter 1, we write φ1 φ2 as an abbreviation for the validity of φ1 φ2 and φ2 φ1.

Theorem 2.13 Let φ and ψ be formulas of predicate logic. Then we have the following equivalences:

1.(a) ¬x φ x ¬φ

(b)¬x φ x ¬φ.

2.Assuming that x is not free in ψ:

You have seen this sort of proof before, in Chapter 1. It is an example of something which requires proof by contradiction, or ¬¬e, or LEM (meaning that it simply cannot be proved in the reduced natural deduction system which discards these three rules) – in fact, the proof above used the rule PBC three times.

Now we prove the validity of ¬x P (x) x ¬P (x) similarly, except that where the rules for and were used we now use those for and :

You will really benefit by spending time understanding the way this proof mimics the one above it. This insight is very useful for constructing predicate logic proofs: you first construct a similar propositional proof and then mimic it.

Next we prove that ¬x φ x ¬φ is valid:

2¬x ¬φ assumption

3x0

120 2 Predicate logic

Proving that the reverse x ¬φ ¬x φ is valid is more straightforward, for it does not involve proof by contradiction, ¬¬e, or LEM. Unlike its converse, it has a constructive proof which the intuitionists do accept. We could again prove the corresponding propositional sequent, but we leave that as an exercise.

2. (a) Validity of x φ ψ x (φ ψ) can be proved thus:

4x0

7(φ ψ)[x0/x] identical to 6, since x not free in ψ

The converse sequent has x (φ ψ) as premise, so its proof has to use x e as its last rule; for that rule, we need φ ψ as a temporary assumption and need to conclude ( x φ) ( x ψ) from those data; of course, the assumption φ ψ requires the usual case analysis:

4.(b) Given the premise x y φ, we have to nest x e and y e to conclude y x φ. Of course, we have to obey the format of these elimination rules as done below: