Originally Posted by

**andrei** $\displaystyle

\begin{array}{rcl}

\forall x(P(x)\to Q(x))&\vdash&\forall x(P(x)\to Q(x)).\\

\forall xP(x),\,\forall x(P(x)\to Q(x))&\vdash&\forall xP(x).\\

\forall xP(x),\,\forall x(P(x)\to Q(x))&\vdash&P(x)\to Q(x)\text{ --- }\forall\text{-elimination, 1}.\\

\forall xP(x),\,\forall x(P(x)\to Q(x))&\vdash&P(x).\\

\forall xP(x),\,\forall x(P(x)\to Q(x))&\vdash&Q(x).\\

\forall xP(x),\,\forall x(P(x)\to Q(x))&\vdash&\forall xQ(x).\\

\end{array}

$

(I cannot finish the proof because of latex restrictions.) Now by using twice $\displaystyle \forall$-introduction to the 6th deduction, we get above formula.