Can this proposition be used to prove an iff statement? 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.
 A: 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 $\lnot q$.
A: *

*

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.
