I've been reading Velleman's How to Prove It and it says that the $\forall x$ in$\forall x P(x) \rightarrow Q(x)$ only applies to $P(x)$ unless there is a parenthesis $\forall x (P(x) \rightarrow Q(x))$. But what is the difference $\forall x P(x) \rightarrow Q(x)$ and $\forall x (P(x) \rightarrow Q(x))$? Don't they mean the same thing?
2026-03-30 17:49:03.1774892943
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Question about Quantifiers
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Let's try writing out the formulae in words. For the first, $\forall x (P(x) \rightarrow Q(x))$ is the same as saying:
For any $x$, if $P(x)$ is true then $Q(x)$ is also true.
For the second, $\forall x P(x) \rightarrow Q(x)$ is the same as $\forall y P(y) \rightarrow Q(x)$, and this is the same as saying:
If for all $y$, P(y) is always true, then $Q(x)$ is true. Do you see the problem? The implication is that $Q(x)$ is always true, but what is $x$? So the second formula is not valid.
They are very different. In $(\forall x P(x))\implies Q(x)$ there are two different $x$'s. The first two are a dummy variable, and it is clearer to write it as $(\forall y P(y)) \implies Q(x)$. This is a predicate whose truth may depend on $x$. The other, $\forall x (P(x) \implies Q(x))$ is a sentence with no free variables. Its truth does not depend on $x$ as $x$ is a dummy. If $Q(x)$ is $x=1$ and $P(x)$ is $x$ is odd, the first is true for all $x$, because the antecedent is false, but the second is false as witnessed by $x=3$. If $P(x)$ is True, the first is true for $x=1$, and false otherwise.