I've spent a good 2-3 hours trying to look up examples online on how to do this exactly. I was astonished to find so few examples. I'm still trying to figure it out, so what I think I'm supposed to say is, if we assume $A$ is true. Then we can automatically eliminate the "$A$" from both sides of the equation since a proposition with any true proposition means it depends on whether the other proposition, $P(x)$ in this case is true or false. So then you can write it as $\forall x P(x) \equiv \forall x P(x)$ which is trivially true. I assume, if I am going at this with the right logic, you just do the same thing with the second part of this problem? I am a complete beginner and this is the first week of the semester, so I wouldn't be surprised if I am royally screwing this up.
Establish these logical equivalences, where $x$ does not occur as a free variable in $A$. Assume that the domain is nonempty.
a) $(\forall xP(x)) \land A \equiv \forall x(P (x) \land A)$
b) $(\exists xP(x)) \land A \equiv \exists x(P (x) \land A)$
Here's how to do a semantical proof. I'll do a) and leave b) to you
Take any interpretation $I$ with domain $D_I$
We have:
$I \models (\forall x P(x)) \land A$ iff (by semantics of $\land$)
$I \models (\forall x P(x))$ and $I \models A$ iff (by semantics of $\forall$)
It is true that for all objects $a$ in $D_I$: $I \models P[a]$ and it is also true that $I \models A$ iff (pure logic)
For all objects $a$ in $D_I$ it is true that $I \models P[a]$ and $I \models A$ iff (by semantics of $\land$)
For all objects $a$ in $D_I$: ($I \models P[a] \land A$) iff (by semantics of $\forall$)
$I \models \forall x (P(x) \land A)$
So, since for all interpretations $I$ we have $I \models (\forall x P(x)) \land A$ iff $I \models \forall x (P(x) \land A)$, we have by definition of logical equivalence proven that:
$(\forall x P(x)) \land A \equiv \forall x (P(x) \land A)$
Note: $I \models P[a]$ means that if we introduce a new constant symbol $c$ and extend interpretation $I$ to $I'$ that interprets $c$ as $a$, then $I' \models P(c)$