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  The following content is from a book: Rosen discrete mathematics and I understand most of it.   But, why can we conclude that $P(a)$ and $Q(a)$ is true for every element in the domain ?   Why every element?   Because I always have this intuition that "a" is a single element.   Is it because "a" is a variable?   Can someone explain the reasoning behind?

Content: suppose that $\forall x~(P(x) \wedge Q(x))$ is true.   This means that if $a$ is in the domain, then $P(a)\wedge Q(a)$ is true.   Hence, $P(a)$ is true and $Q(a)$ is true.   Because $P(a)$ is true and $Q(a)$ is true for every element in the domain, we can conclude that $\forall x~P(x)$ and $\forall x~Q(x)$ are both true.   This means that $\forall x~P(x)~\wedge~\forall x~Q(x)$ is true.

Graham Kemp
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Snailwalker
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You can conclude that because $a$ is an arbitrary element of the domain. Because of the universal quantifier in the first statement, $P(a)$ is true for any element $a$ of the domain and we can deduce $\forall x P(x).$

The formal rule is called universal generalization: From $\phi(x),$ conclude $\forall x\phi(x).$

spaceisdarkgreen
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    Thank you for leading me to the right point. I found this discussion particularly helpful ! https://math.stackexchange.com/questions/1889107/why-does-universal-generalization-work-the-rule-of-inference – Snailwalker Jun 13 '17 at 04:21