$\forall$-distributive rule

Question

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?

Answer 1

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, $$\tag7 \forall x\colon( \phi(x)\land \psi(x))\implies \forall x\colon \phi(x)\,\land\,\forall x\colon \psi(x).$$

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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, $$\tag{14} \forall x\colon \phi(x)\,\land\,\forall x\colon \psi(x)\implies \forall x\colon( \phi(x)\land \psi(x)).$$

Answer 2

"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

$$\forall x ~ (P(x) \land Q(x))$$