Period is constant on communicating classes

If $x=y$, the conclusion is the identity $d(x)=d(x)$, whether or not $R_x$ is empty. Suppose $x\mathrm{\ne }y$. By [F1], choose route lengths $r,s\in {\mathbb{N}}_0$ with $p^{(r)}(x,y)>0$ and $p^{(s)}(y,x)>0$. By [F4], neither length is zero, so $r,s\ge 1$. Twice applying [F3] and retaining the route terms gives $p^{(r+s)}(x,x)\ge p^{(r)}(x,y)p^{(s)}(y,x)>0$ and $p^{(r+s)}(y,y)\ge p^{(s)}(y,x)p^{(r)}(x,y)>0$. Thus both return-time sets are nonempty and both periods are positive.

Fix any $t\in R_x$. By [F3], the route from $y$ to $x$, a $t$-step return at $x$, and the route from $x$ to $y$ give $p^{(s+t+r)}(y,y)\ge p^{(s)}(y,x)p^{(t)}(x,x)p^{(r)}(x,y)>0$. Step 1.1 also shows $r+s\in R_y$. Hence the positive integer $d(y)$ divides both $r+s$ and $r+t+s$, so it divides their difference $t$. As this holds for every $t\in R_x$, $d(y)$ is a common positive divisor of $R_x$ and therefore $d(y)\le d(x)$ by [F2].

Interchanging $x$ and $y$ in step 2.1 shows that every $t\in R_y$ is divisible by $d(x)$; hence $d(x)$ is a common positive divisor of $R_y$ and $d(x)\le d(y)$. Together with step 2.1 this proves equality for distinct communicating states. The case $x=y$ was settled in step 1.1.

If $E=\mathrm{\varnothing }$, there are no communicating states and the assertion is vacuous. If $x=y$ and there is no positive return, both sides equal the stipulated zero; if $x\mathrm{\ne }y$ communicate, step 1.1 proves positive returns exist, so neither period is zero. One-state, deterministic, and absorbing cases are covered by the same alternatives. The accessibility witnesses are positive for distinct states because [F4] makes a zero-step transition possible only from a state to itself. The proof chooses routes only for this fixed pair, so it uses no choice function. This one-way equality statement is not an iff.