= Solution
Under the reference convention $\alpha_1=0$, the hypothesis becomes absence of the free group contrast $\alpha_2$. Introduce an inclusion indicator $J$: set $\alpha_2=0$ when $J=0$, and use its stated proper normal prior when $J=1$. Assign positive prior <probabilities> $q_0,q_1$ to those models, and retain $\mu,\beta_2$ in both.
A birth draws $u$ from a proposal <probability density function> $g(u)$ and appends it as $\alpha_2'=u$; the reverse death deletes it. The Jacobian is one. At fixed $\theta,\beta_2$, the <likelihood> ratio is
$$
\frac{L(\alpha_2=u)}{L(\alpha_2=0)}
=\exp\{Ru-I\theta(1+e^{\beta_2})(e^u-1)\}.
$$
Thus the birth acceptance ratio before capping is
$$
\frac{q_1p_\alpha(u)d_1}{q_0b_0g(u)}
\exp\{Ru-I\theta(1+e^{\beta_2})(e^u-1)\},
$$
where $p_\alpha$ is the normalized $N(0,\sigma_1^2)$ prior. Again, taking $g=p_\alpha$ cancels that factor. Use the reciprocal death rule and ordinary within-model parameter updates.
After warm-up, the <Markov chain Monte Carlo> occupation fraction estimates the desired <probability>:
$$
\boxed{\mathbb P(\alpha_1=\alpha_2=0\mid\mathbf x)
\approx\frac1N\sum_{s=1}^N\mathbf1_{\{J^{(s)}=0\}}.}
$$
This requires the chain to explore both models and satisfy its usual ergodic convergence conditions. Under the original continuous normal prior alone, the equality event has posterior <probability> zero; the explicit model indicator supplies the necessary prior atom. If intercept selection from part (v) is also performed, use two indicators and count all sampled states with $J=0$, regardless of the intercept indicator.
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