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$$\forall x (P(x))\vee \forall x (Q(x))\\ \forall x (P(x)\vee Q(x))\\ $$

Why would a truth value of True for the first proposition mean that the second proposition is True as well?

Do both expressions not just mean the same thing?

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  • $\begingroup$ I don't get your notation. By $\forall P(x)$, do you mean $\forall x,P(x)$ ? $\endgroup$ – Hippalectryon Nov 4 '15 at 12:44
  • $\begingroup$ I apologize for the typo. I fixed it now. $\endgroup$ – nikolita Nov 4 '15 at 12:54
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The first implies the second but not the other way around (so the answer to the question in the title is yes, but the answer to the question of whether they are the same is no).

The first statement says that (at least) one of the two following statements is true: "for all $x$, $P(x)$ is true" and "for all $x$, $Q(x)$ is true". We can take for example $P(x)$ to be the statement "$x$ is even" and $Q(x)$ to be the statement "$2x$ is even" (where in both cases $x$ is an integer), then $Q(x)$ is true for all $x$. We must then answer the question of whether, for any given integer $x$, it is true that either $P(x)$ or $Q(x)$ is true. The answer is yes, because we know that $Q(x)$ is true for all $x$. (In other cases, it may be $P(x)$ which is true for all $x$, but either way one of $P(x)$ or $Q(x)$ is true.)

However, the two statements are not the same (i.e, they are not equivalent), because the second does not imply the first. That is, even if we know that, for any given $x$, one of $P(x)$ or $Q(x)$ is true, we cannot say that either $P(x)$ or $Q(x)$ is true for all $x$, because it can be the case that, for some $x$, only $P(x)$ is true, and for others, only $Q(x)$ is. For example, let $P(x)$ be the statement "$x$ is even", and $Q(x)$ be the statement "$x$ is odd". Then every integer is either even or odd, so for every $x$ it is true that either $P(x)$ or $Q(x)$ is true. However, it is not true that every integer is even or that every integer is odd, so it is not true that $P(x)$ holds for all $x$ nor that $Q(x)$ does.

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  • $\begingroup$ So, by using$ P(x)$ to denote the even integers and $Q(x)$ to denote the odd integers, how can I think about these two statements? I have a hard time with reading the notation. The first one is: all x in the integers they are even for all x in the integers are odd? $\endgroup$ – nikolita Nov 4 '15 at 13:03
  • $\begingroup$ @nikolita $P(x)$ is a statement about $x$, not a set itself. So $P(x)$ could be "x is an even integer." $\endgroup$ – Cheerful Parsnip Nov 4 '15 at 13:05
  • $\begingroup$ More precisely, $P(x)$ (resp. $Q(x)$) would denote the statement "$x$ is even" (resp. odd). Then the first statement in your question means "either every integer is even or every integer is odd" (which is false) and the second means "every integer is either even or odd" (which is true). $\endgroup$ – fkraiem Nov 4 '15 at 13:06
  • $\begingroup$ @fkraiem So basically, because $F$ IMPLIES $T$ is $T$. The second proposition is $T$, so it doesn't matter what the true value of the first is? $\endgroup$ – nikolita Nov 4 '15 at 13:13
  • $\begingroup$ No, what I gave is a counterexample showing that the second does not imply the first (to answer your question of whether they are the same). Tell me if you would like me to expand of why the first implies the second. $\endgroup$ – fkraiem Nov 4 '15 at 13:15

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