Clarification on theorem $(\forall x | R : P) \equiv (\forall x |: R \lor P \equiv P)$

Question

Given the theorem $(\forall x | R : P) \equiv (\forall x |: R \lor P \equiv P)$

Since in the second statement we have $R \lor P \equiv P$

Does that mean that I can simply go from the form $(\forall x | R : P)$ to either $(\forall x |: P)$ or $(\forall x | : R)$ else how should this theorem be understood?

Answer 1

No, it doesn't mean that. Think about this in plain language, first. The theorem says, intuitively, that the following statements are equivalent:

• $(i)$ "Every $x$ which satisfies $R$, satisfies $P$."

• $(ii)$ "For every $x$, we have $x$ satisfies $P$ iff $x$ satisfies $R$ or $P$."

Now it should be clear that $(i)$ doesn't imply "Every $x$ satisfies $P$" or "Every $x$ satisfies $R$." (For example, suppose we take $P=R$. Then $(i)$ is trivially true ...)

For a more concrete example of why this won't work, suppose $R$ is "$x$ is divisible by $4$" and $P$ is "$x$ is even." Then (in the usual context of the natural numbers) clearly $(i)$ holds, but also clearly neither "every $x$ satisfies $R$" nor "every $x$ satisfies $P$" holds.

The theorem can be used to do, well, exactly what it says on the tin: from $(i)$ we can deduce $(ii)$, and from $(ii)$ we can deduce $(i)$. Any usage of the theorem which doesn't exactly look like this has to be justified, even if it "sort of looks right," and the usages you've proposed cannot in fact be so justified.

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At this point it might be a good idea to remember why the theorem above is true in the first place.

• $(i)\implies(ii)$: Assuming $(i)$ holds, I want to prove $(ii)$. So let's suppose I have some element $x$. If $x$ satisfies $P$, then clearly $x$ satisfies ($R$ or $P$). Conversely, if $x$ satisfies ($R$ or $P$), there are two choices: $x$ satisfies $P$, in which case we're done, or $x$ satisfies $R$, in which case by $(i)$ we know $x$ also satisfies $P$, and so we're done. So from $(i)$, we've proved $(ii)$.

That is, $(i)$ lets us identify the (seemingly) stronger condition $P$ with the (seemingly) weaker condition $R\vee P$, by telling us that the "alternative to $P$" in the latter in fact still gives us $P$.

• $(ii)\implies (i)$: Assuming $(ii)$ holds, I want to prove $(i)$. So let's suppose I have some $x$ which satisfies $R$. Then clearly $x$ satisfies ($R$ or $P$). But by $(ii)$ this means $x$ satisfies $P$, so we're done.

This time, it's $(ii)$ which is letting us pass from a (seemingly) weaker claim to a (seemingly) stronger claim.