Proving that universal quantification distributes over conjunction Show that $$\vdash [\forall x(P(x))\wedge \forall x(Q(x))]\to \forall x[P(x)\wedge Q(x)]$$
My answer: by Q_{1}, it is the case that $\forall x \phi\to\phi_{t}^{x}$
so we have
$$\forall x  P(x)\to P(t)$$
We also have
$$\forall x  Q(x)\to Q(t)$$
Then $$ (\forall x)P(x)\wedge(\forall x)Q(x)\to P(t)\wedge Q(t)$$
and by QR  $ P(x)\wedge Q(x)\to (\forall x)P(x)\wedge Q(x)$
so we get
$$\vdash [(\forall x)(P(x))\wedge(\forall x)Q(x))]\to(\forall x)[P(x)\wedge Q(x)]$$
it is correct?

A formal proof, according to 


*

*Christopher Leary, A Friendly Introduction to Mathematical Logic (2000) 


is : 
1) $\forall xP(x) \land \forall xQ(x)$ --- premise
2) $\forall xP(x)$ --- from 1) by (PC) : $p \land q \vDash p$
3) $\forall xQ(x)$ --- from 1) by (PC) : $p \land q \vDash q$
4) $P(x)$ --- from quantifier axiom (Q1) and 2) by modus ponens
5) $Q(x)$ --- from quantifier axiom (Q1) and 3) by modus ponens
6) $P(x) \land Q(x)$ --- from 4) and 5) by (PC) : $p,q \vDash p \land q$
7) $(\forall xP(x) \land \forall xQ(x)) \rightarrow (P(x) \land Q(x))$ --- from 1) and 7) by Deduction Theorem

8)  $(\forall xP(x) \land \forall xQ(x)) \rightarrow \forall x(P(x) \land Q(x))$ --- from 7) by (QR) : $\psi \rightarrow \phi \vdash \psi \rightarrow \forall x \phi$, $x$ not free in $\psi$ [in our case, $\psi$ is : $\forall xP(x) \land \forall xQ(x)$].

