Communication is an equivalence relation

By [F3], $p^{(0)}(x,x)=1$, hence $x\to x$ and $x\leftrightarrow x$ for every $x\in E$. The definition in [F2] is symmetric in $x,y$, so communication is symmetric.

Suppose $x\to y$ and $y\to z$. By [F1], choose $m,n\in {\mathbb{N}}_0$ with $p^{(m)}(x,y)>0$ and $p^{(n)}(y,z)>0$. The nonnegative series in [F4] contains the term at $w=y$, so $p^{(m+n)}(x,z)\ge p^{(m)}(x,y)p^{(n)}(y,z)>0$. Thus $x\to z$. The argument permits either witness length to be zero.

If $x\leftrightarrow y$ and $y\leftrightarrow z$, then $x\to y\to z$ gives $x\to z$ by step 1.2, and $z\to y\to x$ gives $z\to x$ by the same step. Therefore $x\leftrightarrow z$, proving transitivity of communication.

Define $[x]:=\{y\in E:x\leftrightarrow y\}$. Reflexivity makes each $[x]$ contain $x$, so these classes cover $E$. If $c\in [x]\cap [z]$, symmetry and transitivity give $x\leftrightarrow z$; then every member of either class belongs to the other, so $[x]=[z]$. Thus distinct classes are disjoint and the classes partition $E$.

If $E=\mathrm{\varnothing }$, there are no states or classes and the assertion is vacuous. If $E$ has one state, step 1.1 gives its sole class. Absorbing or otherwise degenerate rows cause no exception: the proof uses only zero-step identity and positive accessibility witnesses. No global choice is made; for each fixed triple in step 1.2, the two existential witnesses are used locally.