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: 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 : $\forall xP(x) \land \forall xQ(x)$].

A: The first step of your proof is not very clear, and in fact appears to be using what you are trying to prove to justify the step.
Lets try a simple proof by contradiction for some variety.


*

*Assume $\neg ([\forall x(P(x))\wedge \forall x(Q(x))]\to \forall x[P(x)\wedge Q(x)])$

*$[\forall x(P(x))\wedge \forall x(Q(x))] \wedge \neg \forall x[P(x)\wedge Q(x)]$ by the definition of $\to$ and DeMorgan the $\neg$ over the $\lor$.

*$[\forall x(P(x))\wedge \forall x(Q(x))] \wedge \exists x \neg[P(x)\wedge Q(x)]$ by the definition of $\neg \forall$

*$[\forall x(P(x))\wedge \forall x(Q(x))] \wedge \neg[P(t)\wedge Q(t)]$ by existential instantiation.

*$[\forall x(P(x))\wedge \forall x(Q(x))]\wedge [\neg P(t) \lor \neg Q(t)]$ by DeMorgan

*$ (\forall x(P(x))\wedge \forall x(Q(x)) \wedge \neg P(t)) \lor (\forall x(P(x))\wedge \forall x(Q(x)) \wedge \neg Q(t)))$ Distribute $\wedge$ over $\lor$

*Since $t$ is some object in the domain of discourse, $\forall x(P(x)) \land \neg P(t)$ is a contradiction, and by the same argument $\forall x(Q(x)) \land \neg Q(t)$.
