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To me, it seems like the following two statements are equivalent

x ∈ Arbitrary set P(x)→ Q(x) =

∀x ∈ Arbitrary set (P(x)→ Q(x))

For example, if x ∈ all people, P(x) stands for "x is a man", and Q(x) stands for "x isn't a female", then ∀x(P(x)→ Q(x)) can be thought of as saying "For any person x, if x is a man, then x isn't a female". But isn't that what P(x)→ Q(x) is already saying, just without the explicit "For any person x" and instead with the implications from x ∈ all people? I mean, if we are assuming P(x)→ Q(x) is a valid premise, then shouldn't x be allowed to range over all possible x, i.e, "any person"?

Sorry if I'm not making too much sense and using incorrect notation; I've just started this stuff, probably obviously so lol.

GetDownJam
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    What is your source? Usually, we distinguish between variables and closed terms. That is, if we want $Pa \to Pb$ to be meaningful, we need to stipulate what elements of the universe $a$ and $b$ denote. If $x$ is a variable, then it is somewhat / usually regarded as ill-posed to ask what is the truth value of say $Px$, for $x$ does not denote anything. I would prefer more background before explaining more carefully, however, as I don't want to jump ahead of whatever your source says — different authors deal differently with free variables; this is a common source of confusion, BTW. – wet Nov 18 '16 at 09:38
  • Probably your question refers to so-called universal closure when you take a formula with free variable(s) and add $\forall$ (variable) to the beginning. Usually when the first statement is a theorem, the second statement will be as well, but they are not the same. You may want to read: http://math.andrej.com/2012/12/25/free-variables-are-not-implicitly-universally-quantified/ – Abstraction Nov 18 '16 at 09:42
  • As a matter of general problem solving technique, there is no reason to add the $\to Q(x)$. Your question is one of whether $P(x)$ is equivalent to $\forall x ~ P(x)$. Adding the $\to Q(x)$ is just unnecessary complication, and should really be kept out of the question unless it is helpful. – DanielV Nov 18 '16 at 09:49
  • @DanielV Well, I know P(x) and ∀x P(x) are not equivalent. I'm just getting confused I guess because I'm thinking that if x ∈ arbitrary set and P(x) →Q(x) describes a true relationship of that set, then I don't see how it's not equal to ∀x(P(x)→Q(x)). Maybe the translation of a conditional statement into words is tripping me up. I'm seeing the "if" in the conditional statement as the glue that's keeping them the same. – GetDownJam Nov 18 '16 at 11:48

1 Answers1

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No, P(x) and $\forall x ~ P(x)$ are not always the same thing. For example, $\pi > 7$ is not the same thing as $\forall \pi~\pi > 7$.

Free variables are implicitly quantified over their entire proof, but not necessarily any individual theorem. Imagine the following proof of the statement "if $x < 5$, then $x < 7$" (assume we have already established $y < n \implies y < n + 1$ , and our universe is only natural numbers):

$$\begin{array} {r|l|l} (1) & \quad x < 5 & \text{assumption} \\ (2) & \quad x < 6 & \text{From 1} \\ (3) & \quad x < 7 & \text{From 2} \\ (4) & x < 5 \implies x < 7 & \text{From 1 through 3} \\ \end{array}$$

Notice that from line (4), it would be appropriate to infer $\forall x ~ x < 5 \implies x < 7$. However, from line (2), it would not be correct to infer $\forall x ~ x < 6$. That is because there is an assumption made about the $x$ in line (2): the assumption is that $x < 5$. In general, the meaning of a free variable is that it is universally quantified over the entire proof:

$$\forall x ~ \left(\begin{array} {r|l|l} (1) & \quad x < 5 & \text{assumption} \\ (2) & \quad x < 6 & \text{From 1} \\ (3) & \quad x < 7 & \text{From 2} \\ (4) & x < 5 \implies x < 7 & \text{From 1 through 3} \\ \end{array}\right)$$

Due to the fact that $\forall x ~ (B \implies C(x))$ is equivalent to $B \implies \forall x ~ C(x)$ when $x$ is not a free variable in $B$, you can infer from line (4) $\forall x ~ x < 5 \implies x < 7$.

DanielV
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