Let $P$, $Q$, and $R$ be predicate symbols and $x$ be the predicate variable.
Do I use the fact that $$\forall x, Q(x)\wedge R(x)\equiv \forall x, R(x)\wedge Q(x)$$ or the fact that $$\exists x, Q(x)\wedge R(x)\equiv \exists x, R(x)\wedge Q(x)$$ to show that $$\forall x,(P(x) \leftrightarrow (R(x)\wedge Q(x)) \equiv \forall x,(P(x) \leftrightarrow (Q(x)\wedge R(x)) \;?$$
For propositions $p, q,$ and $r$, we can use the fact that $q\wedge r \equiv r\wedge q$ to show that $$p \leftrightarrow (q \wedge r) \equiv p \leftrightarrow (r \wedge q)$$
We can "substitute" $q \wedge r$ for $r \wedge q$ since they're logically equivalent. But since predicates aren't propositions, they don't have a truth value until concrete values are substituted in place for the predicate variable(s). So I'm not sure if we can say that $$Q(x) \wedge R(x) \equiv R(x) \wedge Q(x)$$ since they're not propositions. From what I know, predicates only become propositions when we substitute the predicate variable(s) for some concrete value(s) or use quantifiers on the predicates. That's why I was confused on whether to use the fact that $\forall x, Q(x)\wedge R(x)\equiv \forall x, R(x)\wedge Q(x)$ or the fact that $\exists x, Q(x)\wedge R(x)\equiv \exists x, R(x)\wedge Q(x)$ to show that $$\forall x,(P(x) \leftrightarrow (R(x)\wedge Q(x)) \equiv \forall x,(P(x) \leftrightarrow (Q(x)\wedge R(x))$$
I want to "substitute" $R(x)\wedge Q(x)$ with $Q(x)\wedge R(x)$ but since neither $R(x)$ nor $Q(x)$ are propositions, I turn them into propositions using quantifiers which gave me the two quantified statements that I showed earlier and I want to "substitute" either one with $R(x)\wedge Q(x)$, but I'm not sure which one to use.
It could be that I'm getting this whole idea incorrect. I'd appreciate to be corrected. Thanks.
Since the relevant logically equivalent part is inside the scope of a quantifier, you would use the fact that $Q(x) \land R(x)$ and $R(x) \land Q(x)$ are equivalent for any assignment of $x$, and hence also under the universal quantification. Something along the lines of "Since $v(x)$ was arbitrary, the above holds for all assignments, therefore ... $\forall x (\ldots) $ ...".