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$\forall x [P(x) \rightarrow Q(x)] \Rightarrow [\forall x P(x) \rightarrow \forall x Q(x)]$

If the LHS is true, then Q(x) must be true for all values of x. Since Q(x) is true for any value, then Q(y) is always true. Thus the RHS is also always true.

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Why must $Q(x)$ be true for all $x$? Notive that "false implies false" is true. –  Gerry Myerson Oct 31 '12 at 0:12
If you mean to say $Q(x)$ is true for all $x$ because $\forall x P(x)$ you should say this, and explain exactly why. For instance for any $x$, we have $P(x) \rightarrow Q(x)$ and since $P(x)$ for every $x$ this means we have $Q(x)$ for every $x$. –  Deven Ware Oct 31 '12 at 0:13
hmm, but i don't think it's a tautology because Q() is true for any value of y and P() and Q() can have different values in the RHS; whereas P() and Q() must always have the same value in the LHS. Right? –  hjggh Oct 31 '12 at 0:24
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2 Answers

MJD has given the right response. But what an extraordinary thing to say:

If the LHS is true, then Q(x) must be true for all values of x

The LHS renders "Take anything $x$ you like, if (yes, IF) $Px$, then $Qx$". The 'if' is crucial. That you missed this suggests you badly need to get back to basics and have another look at a good exposition of the way we render statements of generality using quantifiers-and-variables in logic. I've mentioned it before here, but an excellent freely available resource is Paul Teller's A Modern Formal Logic Primer -- check out the first four chapters of Vol. II and you will never make a bad mistake like this again, promise! (Or of course get my Introduction to Formal Logic out of the library ...!)

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It is not a correct proof. Say that $P(x)$ is the statement "$x$ is a ruby" and $Q(x)$ is the statement "$x$ is red".

Your claim is:

If the LHS is true, then $Q(x)$ must be true for all values of $x$.

The LHS, $\forall x[P(x)\to Q(x)]$ is quite true, since for any $x$, if $x$ is a ruby then it is red. But contrary to your claim, $Q(x)$ is not true for all values of $x$, since not every $x$ is red—crows are not red, and neither is snow.

(However, the RHS is still true; it says that if every $x$ were a ruby, then every $x$ would be red, and this is correct, if the LHS is. But your argument is wrong.)

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