The validity of $∀x(Bx↔∃yRxy)↔\big(∀x∃y(Bx→ Rxy)∧ ∀x∀y(Rxy→Bx)\big)$

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$(\forall x.B(x) \leftrightarrow \exists y. R(xy)) \leftrightarrow (\forall x \exists y.B(x)\rightarrow R(xy)) \land \forall xy. (R(x,y) \rightarrow B(x))$ is a valid formula.

My attempt to prove the validity of RHS→LHS

RHS: $B(x)$ is true iff there exists a $R$-successor to $x$.

LHS: if B(x) is true and there exists an $R$-successor to $x$, then the both conjuncts seem to make sense and are right. If $B(x)$ and there exists no R-successor to $x$, then both conjuncts are trivially satisfied.

Hence, RHS→LHS.

The other direction is not so clear. Is there a simple-intuitive way to prove it?