Predicates and Quantifiers

Question

Question: Establish these logical equivalences, where x does not occur as a free variable in A. Assume that the domain is nonempty.

a) ∀x(A → P(x)) ≡ A → ∀xP(x)

b) ∃x(A → P(x)) ≡ A → ∃xP(x)

My Solution

a) Suppose A is false. Then A -> P(x) is trivially true because if hypothesis is false then conditional statement is trivially true. hence, both left-hand side and right-hand side are true.

Second case if A is true. Then there are two sub-cases.

(i) P(x) is true for every x, then left hand side is true, because if hypothesis and conclusion both are true then conditional proposition is true. Same reasoning can be given for right hand side also, right-hand side is also true as P(x) is true for every x.

(ii) P(x) is true for some x, left-hand side will be true for that x, as hypothesis is true and conclusion is also true. But for those x, where p(x) is false, Left hand side will be false as hypothesis is true and conclusion is false.

For right hand side it will always be false because A is true and ∀xP(x) is false

Hence, both propositions are not equivalent.

b) If A is false, then both left-hand and right-hand sides are trivially true as hypothesis is false.

If A is true, then there are two sub-cases.

i.P(x) is true for every x, then left-hand side is true, and same reasoning can be given for right hand-side, and right-hand side is also true.>

ii.If P(x) is true for some x, left hand side is true and right hand side is also true.

Hence, both propositions are equivalent.

Please validate if my solution is correct.

Answer 1

Your reasoning in a)ii) is not correct. IF $A$ is true, and some, but not all, of the objects in the domain have property $P$, then the left hand side, like the right hand side, is false, because for those objects that do not have property $P$, the conditional $A \rightarrow P(x)$ is false, and hence it is not true that for all objects in the domain $A \rightarrow P(x)$ is true. Since your reasoning for the other cases is correct, the two statements in ) are in fact equivalent.

Your reasoning for b) is all correct.

Answer 2

I believe we have an easier method for the case that you assume $A$ is True in part (a). We have the following formula $$p\rightarrow q\equiv q\vee \neg p.$$ Now since we assume $A$ is True we can use the above formula for each $x$, and we have
$$A\rightarrow P(x)\equiv P(x)\vee \neg A\equiv P(x)\vee F\equiv P(x).$$ Then right side is equivalent to $\forall x P(x)$.

If we use the formula on the left side, we have $$A\rightarrow \forall x P(x)\equiv \forall x P(x)\vee \neg A\equiv \forall x P(x)\vee F\equiv \forall x P(x).$$ Thus both sides are equivalent when $A$ is True since, as you mentioned, both sides are true when $A$ is False, then we are done.

Part (b) can be proven by using the same method.