Quantifiers implication question Let X be the domain of discourse.
Suppose $\forall a \in X$ $P(a)$ $\implies Q(a)$
How would I prove that the above implication implies $\forall a\in X$ $Q(a)$?
 A: What you are essentially trying to do is apply a rule of inference that looks like the following:
$p\rightarrow q$
$\therefore q$
However, no such inference rule exists because the above argument form is not valid. The single premise $\forall a \in X$ $P(a)$ $\implies Q(a)$ is not enough to prove $\forall a \in X$ $Q(a)$. You need more premises to work with.
A: The negation of your statement is satisfiable $\dots$
The correct version is similar to the rule MP or $\to$ Elim in PL:
$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}$
$$\fitch{~~1.~P\to Q\\~~2.~P}{~~3.~Q\hspace{10ex}{\to}\textsf{Elim}~1,2}$$
To conlude $Q$ is true also requires that $P$ is true, otherwise it's not sound.
In another word, $(P\to Q)\to Q$ will not hold when $P$ is false and $Q$ is false,
Since this will make $(P\to Q)\equiv\top$ and $Q\equiv\bot$ that $(\top\to\bot)\equiv\bot$ .
Therefore It's not a tautology.
In the case of FOL, Here is a proof of the correct version with natural deduction:

Note that $$\forall x\in X,P(x)$$
$$\text{ iff } \forall x,x\in X\to P(x)$$

We denote $x\in X$ as $X(x)$,
First we need to assume $\forall x(X(x)\to(P(x)\to Q(x)))$ and $\forall x(X(x)\to P(x))$:
$$\fitch{~~1.~\forall x(X(x)\to(P(x)\to Q(x)))\\~~2.~\forall x(X(x)\to P(x))}{\fitch{~~3.~\boxed{a}~X(a)}{~~4.~X(a)\to(P(a)\to Q(a))
\hspace{25ex}{\forall}~\textsf{Elim}~1\\~~5.~X(a)\to P(a)\hspace{35.5ex}{\forall}~\textsf{Elim}~2\\~~6.~P(a)\to Q(a)\hspace{34.5ex}{\to}~\textsf{Elim}~3,4\\~~7.~P(a)\hspace{43.3ex}{\to}~\textsf{Elim}~3,5\\~~8.~Q(a)\hspace{43.3ex}{\to}~\textsf{Elim}~6,7}\\~~9.~\forall x(X(x)\to Q(x))\hspace{34ex}{\forall}~\textsf{Intro}~3-8}$$
Finally, this proves $\forall x\in X,Q(x)$
