Show that $\phi \equiv \psi$ if and only if $\models (\phi \leftrightarrow \psi)$.

Question

This is in the predicate calculus. I know that I must show that this is true for any structure with any given interpretation. However, I'm having trouble proceeding and actually writing a proof. I hope someone can help?

Answer 1

This is just the matter of definitions and formal proof writing. Let's first unfold the definitions.

$\models \phi \leftrightarrow \psi$ means for all structure like $\mathcal{M}$, $\mathcal{M}\models \phi \leftrightarrow \psi$, which is $\mathcal{M} \models (\phi \rightarrow \psi) \wedge (\psi \rightarrow \phi)$ which is $\mathcal{M} \models (\phi \rightarrow \psi)$ and $\mathcal{M} \models (\psi \rightarrow \phi)$. Also by the definition we we can finally say

• $\models \phi \leftrightarrow \psi$ : (If $\mathcal{M} \models \phi$ then $\mathcal{M} \models \psi$) and (If $\mathcal{M} \models \psi$ then $\mathcal{M} \models \phi$).

and for the $\equiv$ we have the following.

• $\phi \equiv \psi$: For all structure like $\mathcal{M}$, $\mathcal{M}\models \phi$ iff $\mathcal{M}\models \psi$.

but we know $P$ iff $Q$ means: If $P$ then $Q$ and if $Q$ then $P$. so the two definitions become equivalent.