Derivation of the posterior distribution based on a "complete sample analysis" Let $y$ denote the i.i.d. data, and suppose it is splitted into two parts $y = \{y_1, y_2\}$.
Upon a Bayesian analysis of the first part of the data, let's assume that the posterior distribution of $\theta$ (the parameter of interest) is normal:
$$
\theta | y_1 \sim N(\mu, \sigma^2)
$$
Finally, let's suppose that the analysis of the second part of the data yields the exact same posterior:
$$
\theta | y_2 \sim N(\mu, \sigma^2)
$$
Is it possible to derive the posterior distribution based on the complete sample,
$$
\theta | \{y_1, y_2\} \sim ?
$$

Given the two partial posteriors
$$\pi_1(\theta|y_1)\quad\text{and}\quad\pi_2(\theta|y_2)$$
Bayes formula applied to both shows that
$$\pi_1(\theta|y_1)\times\pi_2(\theta|y_2)=\frac{\pi(\theta)f_1(y_1|\theta)}{m_1(y_1)}\times\frac{\pi(\theta)f_2(y_2|\theta)}{m_2(y_2)}\\=\pi(\theta)\times \pi(\theta)f_1(y_1|\theta)f_2(y_2|\theta)\times\frac{1}{m_1(y_1)m_2(y_2)}\\=\pi(\theta)\times\frac{\pi(\theta)f_1(y_1|\theta)f_2(y_2|\theta)}{m(y_1,y_2)}\times\frac{m(y_1,y_2)}{m_1(y_1)m_2(y_2)}\\=\pi(\theta)\times\pi(\theta|y_1,y_2)\times\frac{m(y_1,y_2)}{m_1(y_1)m_2(y_2)}$$
Therefore,
$$\dfrac{\pi_1(\theta|y_1)\times\pi_2(\theta|y_2)}{\pi(\theta)} \propto \pi(\theta|y_1,y_2)$$
