How can $\exists x \in X P(x) \rightarrow ((P(c) \rightarrow Q) → Q)$ be true? $x$ is a natural number
$X$ is a set of all natural numbers
$P$ is a one-place predicate
For context, this statement is found here. If the proposition of a material conditional can be true while both of the propositions it joins are false (i.e $P \rightarrow Q$ is true if both $P$ and $Q$ are false) then how can the proposition in the title be true?
If we deduct the proposition $P(c) \rightarrow Q$ and assume both $P(c)$ and $Q$ are false, the proposition would still be true. And if we assume both $P(c)$ and $Q$ are false then how can $((P(c) \rightarrow Q) \rightarrow Q)$ be valid at all? If $Q$ is false and $P(c) \rightarrow Q$ is true then the proposition $((P(c) \rightarrow Q) → Q)$ is self-contradictory.
What am I missing here?
The only thing I can think of is that the proposition in the title implies that the predicate, $P(c)$, is true if the proposition $\exists x \in X P(x)$ is true because $P(c)$ is just a different representation of $P(x)$. But I cannot confirm this and the phrasing in that paragraph is vague to me. 
Also, as a side question, have I used terminology appropriately here?
 A: $c$ is an object that we named but that is otherwise explicitly unspecified. Is $P(c)$ true then? Well, maybe, maybe not, we cannot know. There are only two situations where we can conclude that $P(c)$ must be true or must be false: If $\forall x P(x)$, then necessarily $P(c)$. If $\forall x\neg P(x)$, the necessarily $\neg P(x)$. Here, we are but interested in $\exists xP(x)$. Then it is certainly possible that $P(c)$ is true, whereas we have no idea whether $\neg P(c)$ can be true. Now if we can derive $P(c)\to Q$, then $Q$ (which does not depend on $c$) cannot be false, for it may (and we are given that it can) happen that $P(c)$ is true. As even in that case $P(c)\to Q$ must hold, it must be the case that $Q$ is true.
A: The Wikipedia page you cite is wrong. It is a valid inference rule to infer $Q$ from $\exists x.P(x)$ and $P(c) \to Q$, where $c$ is some constant that does not appear in $Q$:
$$
{{{\exists x.P(x)} \quad P(c) \to Q} \over Q} \quad\mbox{[where $c$ does not appear in $Q$]}
$$
However, that does not mean that the implication $\exists x.P(x) \to (P(c) \to Q) \to Q$ is provable.
The rule is valid, because if you can prove $P(c) \to Q$ and $\exists x. P(x)$ then $P(c) \to Q$ holds in any model including the model in which $c$ is interpreted as the  witness to $\exists x. P(x)$ (which exists in every model since $\exists x. P(x)$ is provable), but that means that $Q$ holds in any model, since the truth value of $Q$ does not depend on the value assigned to $c$. The implication  $\exists x.P(x) \to (P(c) \to Q) \to Q$ need not hold in every model. E.g., take $P(x) \equiv \exists y.x = 2y$ (i.e., $P(x)$ says "$x$ is even") and take $Q \equiv 1 = 2$. Then $\exists x.P(x) \to (P(c) \to Q) \to Q$ does not hold if $c$ is interpreted as $1$ (since it reduces to $\mathsf{true}\to\mathsf{true} \to \mathsf{false}$).
A way of seeing this thinking about proofs rather than models is that $c$ in the inference rule can be considered to be universally quantified (because it only appears in the deduction of $P(c) \to Q$ and not in $Q$ itself). So the rule works like:
$$
{{{\exists x.P(x)} \quad  \forall y(P(y) \to Q)} \over Q} \quad\mbox{[where $y$ does not appear in $Q$]}
$$
