$\forall$-distributive rule I would state the axioms of a commutative monoid as follows:


*

*for all $a, b, c$: $\quad a(bc)=(ab)c$;

*for all $a, b$: $\quad ab = ba$.


But I also read texts in which the author dragged the for-all out:
for all $a, b, c$:


*

*$\quad a(bc)=(ab)c$;

*$\quad ab = ba$.


How can one see that the two are equivalent? Why does $\forall$ distribute over a conjunction?
 A: Assume $$\tag1\forall x\colon( \phi(x)\land \psi(x)).$$
We want to show $\forall x\colon \phi(x)$.  By specializationm of $(1)$, we obtain 
$$ \tag2\phi(x)\land \psi(x).$$
By conjuction elimination, $$\tag3\phi(x).$$
By universal generalizaion
$$\tag4\forall x\colon \phi(x).$$
Similarly, we find 
$$\tag5\forall x\colon \psi(x).$$
Finally, by adjunction fro $(4)$ and $(5)$
$$\tag6\forall x\colon \phi(x)\,\land\,\forall x\colon \psi(x).$$
In summary, $$\tag7 \forall x\colon( \phi(x)\land \psi(x))\implies \forall x\colon \phi(x)\,\land\,\forall x\colon \psi(x).$$

Now assume $$\tag8\forall x\colon \phi(x)\,\land\,\forall x\colon \psi(x).$$
By conjunction elimination
$$\tag9 \forall x\colon \phi(x)$$
and by specialization
$$\tag{10} \phi(x).$$
Similarly, show 
$$\tag{11} \phi(x).$$
By adjunction from $(10)$, and $(11)$
$$\tag{12} \phi(x)\land \psi(x)$$
and then by generalizaion
$$\tag{13}\forall x\colon( \phi(x)\land \psi(x)).$$
In summry, 
$$\tag{14} \forall x\colon \phi(x)\,\land\,\forall x\colon \psi(x)\implies \forall x\colon( \phi(x)\land \psi(x)).$$
A: "Every dog has a tail and every dog has a nose" is the same thing as "every dog has a tail and a nose".
Or formally,
$$(\forall x~P(x)) \land (\forall x~Q(x))$$
is equivalent to
$$\forall x ~ (P(x) \land Q(x))$$
