I've got a quick question about Universal Quantifiers. Given the following:
$$ \forall x (p(x) \vee q(x)) $$
Can we do this: $$ \forall xp(x) \vee \forall xq(x) $$. i.e can we distribute the "for any" to the p and q?
I've got a quick question about Universal Quantifiers. Given the following:
$$ \forall x (p(x) \vee q(x)) $$
Can we do this: $$ \forall xp(x) \vee \forall xq(x) $$. i.e can we distribute the "for any" to the p and q?
On
Let be $A = \forall x (p(x)\lor q(x))$ and $B = \forall x p(x) \lor \forall x q(x)$. It holds that
$$B\Rightarrow A\text{.}$$
The opposite direction may not hold, which can be seen from the answer of user1331281. However, if $q$ does not depend on $x$, then the opposite direction holds also. In that case we have $$A\Leftrightarrow B$$ and this is known as Frobenius law.
No. Every integer is either even or odd, but it is not true that every integer is even or that every integer is odd.