(1)$\exists x \in X \ ( P(x) \implies \forall y \in X P(y) )$
(2) $\exists x\in X P(x) \implies \forall y \in X P(y)$
What’s the difference? And are they both always true?
It seems to me that the second is false, but I read that the first is true. May someone please clarify the difference? And tell me why that is the case, please?
The first one is equivalent to : $∀xPx → ∀yPy$, which is always true.
The second one is not equivalent to the first one, and is not always true. Consider the following counter-example : "if there is a number that is even, then every number is even".
To have an insight about the difference, consider what happens to the first one with the same interpretation used above : domain $\mathbb N$ and predicate symbol $P(x)$ interpreted with "$x$ is Even".
We have that $\forall y P(y)$ is False (because it is not true that every natural is Even).
But also $P(1)$ is False.
Thus, $P(1) \to \forall y P(y)$ is True (because $\text F \to \text F$ is $\text T$) and thus :
$\exists x (Px \to \forall y P(y))$
is True.
See the so-called Drinker paradox.
And see also this post for proofs of the validity of the formula.
The distinction is in the placement of the brackets, and the operator precedence.
$\exists x\in X~\Big(P(x)\to\forall y\in X~P(y)\Big)$
This states: "There is something in $X$ where if that thing satisfies $P$, then everything in $X$ satisfies $P$". Now, implications are only false when their antecedent is true and consequent false. However, when the consequent is false, then there is something in $X$ that makes the antecedent false too.
So you can always find something in $X$ that makes the implication hold (well, unless there is no things in $X$).
So, as long as $X$ is not empty, this existential statement is true.
Because implication has precedance over quantification, there is implicit brackettng around the existential in the antecedant.
$\Big(\exists x\in X~P(x)\Big)\to\forall y\in X~P(y)$
This states "If there is something in $X$ that satisfies $P$, then everything in $X$ will satisfy $P$."
It is possible to have an $X$ and $P$ that make this implication false.
