Basically my question is why does ∃x[P(x)∨Q(x)] ≡ ∃xP(x)∨∃xQ(x)?
I have a case that's confusing me.
Say P(x): x = 2 and Q(x) = x=5 and the domain is D = {2,5}
So let's say for ∃x[P(x) ∨ Q(x)], x = 5. That means that 5=5 or 5=2, which basically says true.
but for ∃xP(x) ∨ ∃xQ(x) we can say that Q(2) and P(5), which would make the entire case false.
If one of them is true and one of them is false, how can both of them equal? Sorry if this is a stupid question but I don't know if im missing a crucial concept or what?
Can someone help me out? Thanks
$\exists xP(x)\vee \exists xQ(x)$ is not false. Remember that "$\exists x A$" is true iff there is some $x$ making $A$ true, and that "$\varphi \vee \psi$" is true iff at least one of $\varphi, \psi$ is true.
Now, both $\exists xP(x)$ and $\exists xQ(x)$ are true (take $x=2$ and $x=5$, respectively), so $\exists xP(x)\vee \exists xQ(x)$ is true too.
When we plug in $x=5$ to $P(x)$, the result is indeed false; but this doesn't disprove $\exists xP(x)$! Instead, it disproves $\forall xP(x)$. Similarly, noticing that $x=2$ doesn't satisfy $Q(x)$ doesn't disprove $\exists xQ(x)$.
Another way to put this: "$\exists x A$" and "$\exists x\neg A$" are not contradictory! Indeed, in the example you give, both "$\exists xP(x)$" and "$\exists x\neg P(x)$" are true - take $x=2$ and $x=5$ respectively.