= Solution
Put $a=p(x)q(x,y)$ and $b=p(y)q(y,x)$. For $a+b>0$, the off-diagonal <probability> flux of the proposed rule is
$$
p(x)q(x,y)\beta(x,y)=a\frac{b}{a+b}=\frac{ab}{a+b}=b\frac{a}{a+b}=p(y)q(y,x)\beta(y,x).
$$
If $a=b=0$, define both acceptance <probabilities> to be zero; both fluxes are zero. The diagonal <detailed balance equations> hold automatically, including the <probability> of rejected proposals. Summing the <detailed balance equations> over $x$ gives <stationarity>:
$$
\sum_xp(x)P(x,y)=\sum_xp(y)P(y,x)=p(y).
$$
\b[The alternative rule therefore preserves $p$.] The unknown <normalizing constant> cancels after replacing $p$ by $r$. This <Barker acceptance probability> is no larger than the <Metropolis–Hastings acceptance probability> for the same proposed move: if $t=b/a>0$, then $t/(1+t)\leq\min(1,t)$. Preservation of $p$ still requires separate <Markov chain irreducibility> and convergence checks before it guarantees useful sampling.
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