Is it always a tautology?

Question

If any two compound propositions $P$ and $Q$ are equivalent, then is the proposition formed from their biconditional $P \leftrightarrow Q$ always a tautology?

Answer 1

Yes; this can be proved using the deduction theorem.

TFAE.

1. $(P \vdash Q) \wedge (Q \vdash P)$
2. $(\vdash P \rightarrow Q) \wedge (\vdash Q \rightarrow P)$
3. $\vdash (P \rightarrow Q) \wedge (Q \rightarrow P)$
4. $\vdash (P \leftrightarrow Q)$

Answer 2

No.

I'll define "equivalence" as meaning that two propositions have the same truth-value.

↔ is defined by this table where $F$ indicates falsity, $T$ indicates truth, and $U$ indicates that the truth-value of a proposition takes on some other value than truth or falsity:

↔  F  U  T
F  T  U  F
U  U  U  U
T  F  U  T

If $P$ has truth value of $U$, and so does $Q$, then $P$ and $Q$ end up equivalent. $P$ could come as the compound proposition (a↔b) and $Q$ as the compound proposition (a↔b). However, ($P$↔$Q$) is not a tautology.