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Is it sufficient to show that since $A \Leftrightarrow B$ is equivalent to $(A \Rightarrow B) \land (B \Rightarrow A)$ and as conjunction cannot be expressed using conditional alone, neither can biconditional? I can't think of a convincing argument why this would hold. Maybe with some rearrangement it would still be possible?

I tried to get a contradiction using induction like this answer but could not find any.

antrep1234
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    The suggested proof wouldn't be enough. In general, we have to be careful when proving the negative of something: it's not enough to say "this is an approach to proving the positive statement, but that approach won't work"—even if true, conceivably some other approach might work. – Greg Martin Jul 25 '20 at 18:26

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If you write an expression just involving propositional variables and the connective $\implies$ focus on the right-most variable, $X$ say. If that takes the value $T$ (true) then the whole expression takes the value $T$. So the expression cannot be logically equivalent to $A\iff B$ since if $A$ and $B$ take distinct values the expression must evaluate to $F$ (false). If $A$ is $X$ then taking $A=T$ and $B=F$ gives a contradiction; likewise if $B$ is $X$. If neither $A$ nor $B$ is $X$, again take $A=T$ and $B=F$ and $X=T$ to give a contradiction.

Angina Seng
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