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Determine whether $\forall x (P(x) \rightarrow Q(x))$ and $\forall x P(x) \rightarrow \forall xQ(x)$ are logically equivalent . Justify your answer.

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4 votes
4 votes
$∀x(P(x)→Q(x)) \ and \ ∀xP(x)→∀xQ(x)$ are logically equivalent if and only if ,

$(∀x(P(x)→Q(x)) \ → \ ∀xP(x)→∀xQ(x)) \ and \ ((∀xP(x)→∀xQ(x)) \ → \ ∀x(P(x)→Q(x)))$.

Let's prove the implications in parts.

$(∀x(P(x)→Q(x)) \ → \ ∀xP(x)→∀xQ(x))$

We need to prove that this implication is Valid.

The implication is not valid if we can find a True -> False case.

RHS is false if $∀xP(x)→∀xQ(x)$ is false thus $∀xP(x)$ must be True , and for some $x \ Q(x)$ must be false.

Since $P(x)$ is true for every $x$ of the domain , and $Q(x)$ must be false for some $x$ , thus LHS can never be true since there will be atleast one case of $True->False$ on LHS , thus making the LHS completely false.

Thus we couldn't prove the implication to be invalid , thus the implication is valid.

Now let's prove the other part of the implication.

$((∀xP(x)→∀xQ(x)) \ → \ ∀x(P(x)→Q(x)))$.

Here too we need to show atleast one $True->False$ case.

RHS if false if $∀x(P(x)→Q(x)))$ is false , thus there must exist one such $x$ for which $P(x)$ is true and $Q(x)$ is false.

LHS is true , if $((∀xP(x)→∀xQ(x))$ is true . Let there be some $x$ for which both $P(x) \ and \ Q(x)$ is false , this will make LHS True .

Thus we were able to find a $True->False$ situation , making the implication Invalid.

 

Thus both are $NOT \ EQUIVALENT$
4 votes
4 votes

Bot are not equivelent .

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