First, let $\mathcal{F}$ be a frame validating 5. Suppose $wRx$ and $wRy$. Let $M$ be a model based on $\mathcal{F}$, with $V(p)=\{x\}$. Since $\models_{x}p$, we have $\models_{w}\diamond p$, and so $\models_{w}\square\diamond p$, or $\models_{u}\diamond p$ for all $u$ such that $wRu$. In particular, $\models_{y}\diamond p$. So there is a $z$ such that $yRz$ and $\models_{z}p$. But this means $z=x$, whence $yRx$, meaning $R$ is Euclidean.

Conversely, suppose $\mathcal{F}$ is a Euclidean frame, and $M$ a model based on $\mathcal{F}$. Suppose $\models_{w}\diamond A$. Then there is a $v$ such that $wRv$ and $\models_{v}A$. Now, for any $u$ with $wRu$, we have $uRv$ since $R$ is Euclidean. So $\models_{u}\diamond A$. Since $u$ is arbitrary, $\models_{w}\square\diamond A$, and therefore $\models_{w}\diamond A\to\square\diamond A$. ∎

Suppose $R$ is both reflexive and Euclidean. If $aRb$, since $aRa$, $bRa$ so $R$ is symmetric. If $aRb$ and $bRc$, then $bRa$ since $R$ has just been proven symmetric, and therefore $aRc$, or $R$ is transitive. Conversely, suppose $R$ is an equivalence relation. If $aRb$ and $aRc$, then $bRa$ since $R$ is symmetric, so that $bRc$ since $R$ is transitive. Hence $R$ is Euclidean. ∎

Since any theorem $A$ in S5 is deducible from a finite sequence consisting of tautologies, which are valid in any frame, instances of T, which are valid in reflexive frames, instances of 5, which are valid in Euclidean frames by the proposition above, and applications of modus ponens and necessitation, both of which preserve validity in any frame, $A$ is valid in any frame which is both reflexive and Euclidean, and hence an equivalence frame. ∎

By the discussion above, it is enough to show that the canonical frame of S5 is reflexive, symmetric, and transitive. Since S5 contains T, B, and 4, $\mathcal{F}_{{\textbf{S5}}}$ is reflexive, symmetric, and transitive respectively, the proofs of which can be found in the corresponding entries on T, B, and S4. ∎

Let $\Lambda$ be such a logic. Suppose $uR_{{\Lambda}}v$ and $uR_{{\Lambda}}w$. We want to show that $vR_{{\Lambda}}w$, or $\Delta_{v}:=\{B\mid\square B\in v\}\subseteq w$. Let $A$ be any wff. If $A\notin w$, $A\notin\Delta_{u}$ since $uR_{{\Lambda}}v$, so $\square A\notin u$ by the definition of $\Delta_{u}$, or $\neg\square A\in u$ since $u$ is maximal, or $\diamond\neg A\in u$ by substitution theorem on $\neg\square A\leftrightarrow\diamond\neg A$, or $\square\diamond\neg A\in u$ by modus ponens on 5 and the fact that $u$ is closed under modus ponens. This means that $\diamond\neg A\in\Delta_{u}$ by the definition of $\Delta_{u}$, or $\diamond\neg A\in v$ since $uR_{{\Lambda}}v$, so that $\neg\square A\in v$ by the substitution theorem on $\diamond\neg A\leftrightarrow\neg\square A$, which means $\square A\notin v$ since $v$ is maximal, or $A\notin\Delta_{v}$ by the definition of $\Delta_{v}$. ∎