# Showing logical equivalence for predicates?

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.

• The basic is that R(x)∧Q(x))≡Q(x)∧R(x); details depend on the proof system. Commented Jun 19, 2023 at 7:55
• @MauroALLEGRANZA but $R(x)$ and $Q(x$ aren't propositions right? Can we still use logical equivalence on non-propositions? Commented Jun 19, 2023 at 8:00
• The predicates (Rx ∧ Qx)↔(Qx ∧ Rx) and x=x are certainly logical validities; as such, Rx ∧ Qx is logically equivalent to Qx ∧ Rx. However, depending on the textbook, it may or may not be legit to claim that these logically valid predicates are "true". Hope this makes sense, even as it sounds paradoxical. (On a related note: in real analysis, although the predicate $x^2\ge0$ is universally true, without implicit universal quantification, whether it is permissible to say that this predicate is "true" depends on how nitpicky we are being.) Commented Jun 19, 2023 at 12:13
• Commented Jun 19, 2023 at 12:16
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 ... $$\forall x (\ldots)$$ ...".
• I believe this means that we're using the fact that $\forall x, (Q(x) \wedge R(x)) \equiv \forall x, (R(x) \wedge Q(x))$? Commented Jun 20, 2023 at 7:58