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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)

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 is false, because for those objects that do not have property P, the conditional A→P(x) is false, and hence it is not true that for all objects in the domain A→P(x) is true.

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

Hence, both propositions are 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.