Let's say I want to prove that $p\Leftrightarrow q$. My proof looks like this: $p\Leftrightarrow a\Leftrightarrow q$. I have used two "theorems", namely, $p\Leftrightarrow a$ and $a\Leftrightarrow q$. The problem is that the theorem $p\Leftrightarrow a$ has the requirement that $p$ or $q$ should be true. Can I still use this theorem in my proof? This is not a made up case, by the way. I am actually trying to find a proof for something.
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To prove that $p\implies q$, suppose $p$ is true and then show that $q$ is true. So it seems like you can use it? – Keshav Oct 05 '21 at 17:43
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Yes you can. If $p$ or $q$ hold then your theorem states that $p\Leftrightarrow a$ and your other theorem states that $a\Leftrightarrow q$. On the other hand, if neither $p$ nor $q$ holds, then this creates no problem for $p\Leftrightarrow q$. Basically you have:
If we have $p$ then we have $a$ and thus also $q$. If we have $\lnot p$ then we either have $\lnot q$, or we have $\lnot a$ and thus $\lnot q$.
Lazy
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The problem is that the theorem $p\Leftrightarrow a$ has the requirement that $p$ or $q$ should be true.
- The theorem $$p\Rightarrow a$$ is indeed not useful when $p$ is false, for then no conclusion can be derived;
- however, the theorem $$p\Leftrightarrow a\\\equiv\;\; p\Rightarrow a \;\text{ and }\; a\Rightarrow p \;\text{ and }\; \lnot p\Rightarrow \lnot a \;\text{ and }\; \lnot a\Rightarrow \lnot p,$$ is effectively saying that $p$ and $a$ have the same truth value, and so is useful even when $p$—and consequently, $a$—is false.
So, yes, proving $$p\Leftrightarrow q$$ is equivalent to proving $$(p \Leftrightarrow a) \;\text{ and }\; (a \Leftrightarrow q).$$
A technical note: a strict logician would construe the statement $$p\Leftrightarrow a\Leftrightarrow q$$ to mean $$p \Leftrightarrow (a \Leftrightarrow q),$$ which is not equivalent to $$(p \Leftrightarrow a) \;\text{ and }\; (a \Leftrightarrow q),$$ which is what you really mean.
ryang
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