An invariant law need not be reversible

The matrix $p$ is a transition matrix, since each row has the single entry $1$ and all other entries $0$; and $\pi$ is invariant: for each $y$ there is exactly one predecessor $x=y-1$ with $p(x,y)=1$, so ${\sum }_x\pi (x)p(x,y)=\frac{1}{3}=\pi (y)$, which is the criterion of [F1].

Detailed balance fails for $\pi$: for the directed edge $0\to 1$, $\pi (0)p(0,1)=\frac{1}{3}\cdot 1=\frac{1}{3}$, while $\pi (1)p(1,0)=\frac{1}{3}\cdot 0=0$, so the two sides differ; by [F2] the invariant probability $\pi$ is not a reversible probability distribution.

The reverse kernel of [F3] is well defined because $E_{+}=\{x:\pi (x)>0\}=E$: $p_{\ast }(x,y)=\frac{\pi (y)p(y,x)}{\pi (x)}=p(y,x)$, so $p_{\ast }(i,i-1)=1$ for every $i$ and all other entries vanish; the reversed chain is the deterministic cycle running in the opposite direction.

The equivalence in [F3] gives a second proof of nonreversibility: $p_{\ast }(1,0)=1$ while $p(1,0)=0$, so $p_{\ast }\mathrm{\ne }p$ on $E_{+}$ and detailed balance fails; both computations agree.

Boundary and scope cases: the two-state deterministic cycle with $p(0,1)=p(1,0)=1$ is reversible, since $\frac{1}{2}\cdot 1=\frac{1}{2}\cdot 1$; hence three states is the minimal size for a deterministic cycle that refutes the implication, and the example is sharp in that respect; the uniform law remains invariant for the reversed kernel $p_{\ast }$, so reversing does not lose stationarity; the diagonal entries $p(i,i)=0$ satisfy detailed balance trivially in the sense $0=0$; steps 1.1–2.1 are finite computations on the given data and use no choice principle, while steps 2.2–3.1 spend the axiom [A1] exactly through the reverse-kernel theorem [F3], whose statement assumes Choice; and the example refutes only the implication "invariant $\Rightarrow$ reversible", not the converse, which is Detailed balance implies invariance.