It is possible to have this logical consequence? $$ \forall x (p(x) \vee q(x)) \models \forall x p(x) \vee \forall x q(x) $$
$\begingroup$
$\endgroup$
1
-
2$\begingroup$ Does "every number is even or odd" imply "every number is even, or every number is odd"? $\endgroup$– Andrej BauerCommented Apr 25, 2022 at 17:47
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
The statement $\forall x (p(x) \vee q(x))$ does not imply $\forall x p(x) \vee \forall x q(x)$ because there exist a domain, a propositional function p(x) and a propositional function q(x) for which $\forall x (p(x) \vee q(x))$ is true and $\forall x p(x) \vee \forall x q(x)$ is false.
For an example, let the domain be the set of all integers, let $p(x)$ be the propositional function "$x$ is even'' and let $q(x)$ be the propositional function "$x$ is odd''. Then, $\forall x (p(x) \vee q(x))$ is true because every integer is either even or odd, but $\forall x p(x) \vee \forall x q(x)$ is false because it's not the case that ever integer is even and it's not the case that every integer is odd.
Of course, you can find other domains and propositional functions $p(x)$ and $q(x)$ such that both expressions have the same truth value, but the implication does not hold because it does not hold in general.
$\begingroup$
$\endgroup$
Take the structure $\mathfrak{A} = (\{a, b\}, =)$ and $p(x) \equiv x = a$, $q(x) \equiv x = b$. Then $\mathfrak{A} \models \forall x (p(x) \vee q(x))$ but $\mathfrak{A} \not\models \forall x p(x) \vee \forall x q(x)$.
