Are the following equivalent:
(1) $((p\implies q) \implies r)$ $\wedge$ $((r\implies q) \implies p)$
(2) $p \iff q \iff r$?
I thought of writing a truth table, but it'll take a while, so I was wondering if there's a slicker way to prove or disprove the above. All I really care about is if they're equivalent or not.
(2) needs bracketing to be well-formed.
However, suppose $p$ is true, $q$ false, $r$ true.
Then (1) is true (two conditionals with true consequents).
And (2) is false, whether (i) bracketed as $(p \iff q) \iff r)$ or as (ii) $p \iff (q \iff r)$ or treated as (iii) abbreviating $(p \iff q) \land (q \iff r)$