I know this statement is not always true, but I'm having a hard time proving it.
I'm also wondering what the difference between:
$[\exists x \in U, P(x)] \implies [\forall x \in U, P(x)]$
and
$[\exists x \in U, P(x)] \implies [\forall y \in U, P(y)]$
would be?
For the first one, when I try to negate it and show that the negation is true, I end up with something like:
$[\exists x \in U, P(x)] \land [\exists x \in U, \lnot P(x)]$,
but if it was written as the second one I could just provide some counterexample y...?
Or am I just not parsing the statement correctly?
Thanks in advance!
Unless you restrict the set $U$ to having one element, then your statement is not a theorem: you basically state: if there is one element that satisfies property $P$, then all elements satisfy property $P$. Like Bryan said, a counterexample, i.e., a world/interpretation in which your premises are not mapped to truths. Consider a world with two elements {$a,b$}, and just declare $P(a) $ $\land \neg P(B) $ . For a more substantial example, let $P(x):=$ x is even, be a property defined on the universe of all numbers, then your statement would be interpreted as: "if there is an even number, then all numbers are even." Or consider the example with $P(x)$:= "x is male" , defined in the world of all people. Then, you interpret this as: If there is a male, all people are males.
And, actually $ A \implies B $ is equivalent to $\neg A \lor B$, so that its negation is $A \land \neg B $, i.e., your statement $$[\exists x \in U, P(x)] \implies [\forall x \in U, P(x)]$$ negates to:
$$ [\exists x \in U, P(x)] \land \neg [\forall x \in U, P(x)]$$ negates to: