Irreducible recurrence/transience dichotomy

If $E=\mathrm{\varnothing }$, there are no states to classify. Both universal conclusions in the disjunction hold vacuously.

Now suppose $E\mathrm{\ne }\mathrm{\varnothing }$ and fix one state $x\in E$. For every $y\in E$, irreducibility [F1] gives $x\leftrightarrow y$. This fixes one witness state only; no family of choices is made.

If $x$ is recurrent, [F2] and step 1.2 imply that every $y\in E$ is recurrent. Hence the first alternative holds.

If $x$ is not recurrent, consider any $y\in E$. If $y$ were recurrent, [F2] and step 1.2 would imply that $x$ is recurrent, a contradiction. Thus no state is recurrent. By [F3], every state is transient, so the second alternative holds.

By [F3], the fixed state $x$ is either recurrent or transient. Step 2.1 handles the recurrent case and step 2.2 handles the transient case. Therefore one of the two asserted universal alternatives always holds.

If $E$ has one state $x$, [F4] gives $p(x,x)=1$, so its positive-time return probability is one. If $E$ has more than one state, an absorbing row at any state would make the chain reducible; zero one-step weights are allowed, since irreducibility requires communication by some positive-probability finite path, not a positive one-step transition. The zero-step accessibility $x\to x$ alone does not establish recurrence, which requires a positive-time return [F3]. For a deterministic transition map on a nonempty irreducible state space with more than one state, fix $x$ and choose $y\mathrm{\ne }x$. The unique forward orbit from $x$ reaches $y$ and then returns to $x$ by irreducibility. It therefore contains a finite cycle through $x$; every state is reachable from $x$, so every state lies on this cycle and returns to itself. Steps 2.1 and 2.2 prove the forward and reverse uses of the class-property equivalence. AC [A1] is inherited by the canonical laws and class-property theorem; fixing one state in step 1.2 uses no choice principle.