Can someone provide an interpretation to show that the following are not equivalent:
$$\forall x \in D, P(x) \vee Q(x)\;\;\text{vs.}\;\;(\forall x \in D, P(x)) \vee (\forall x \in D, Q(x))$$
They seem equivalent.
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Or; let $D=\mathbb Z$ and $$P(x): x~~~\text{is odd},~~Q(x): x~~~\text{is even}$$
Mikasa
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Great example! You had a great day today (yesterday)! – amWhy Jun 03 '13 at 01:35
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@amWhy: Thanks Amy. Yes I had if we can factor that terrible assignments. :D – Mikasa Jun 03 '13 at 04:04
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Let $D=\Bbb R$ and $P(x)$ mean $x\ge 0$. Can you find a suitable property $Q(x)$?
Git Gud
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@mathnoob Let's check your suggestion: does $(\forall x\in \Bbb R)(P(x)\lor Q(x))$ hold? – Git Gud Jun 02 '13 at 15:47
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@mathnoob Babak's suggestion also works. It's just a minor variant of the one we got. It's basically (but not exactly) letting $P$ mean something and $Q$ be mean $\neg P$. – Git Gud Jun 02 '13 at 15:52
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Yes the idea is to let $D$ be a domain and partition it into two disjoint halfs. Like let $D = R$. Then $P(x): x$ is rational and $Q(x): x$ is irrational also works. – mathnoob Jun 02 '13 at 15:55
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1@GitGud: It always takes time for you to learn someone a logical fact. And you won Git (+1). – Mikasa Jun 02 '13 at 16:03
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The first statement asserts that for all $x \in D, P$ is true or $Q$ is true. Perhaps both may be true, assuming the statement uses the inclusive OR. However, this statement does not claim that both $P \text{ and } Q$ must be true for any given $x$. It also does not claim that either (or both) must be true for all $x$.
The second statement on the other hand, claims that either $P$ must be true for all $x$ or $Q$ must be true for all $x$ or both are simultaneously true for all $x$. So for example, with this statement it may not be the case that $P$ is true for all $x$ and $Q$ is true for some $x$ but not others. However, this possibility is included by the first statement.
ThisIsNotAnId
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