Corollary:
Under the assumptions of Theorem, a $(1-\alpha)100\%$ confidence interval for $\mu_x-\mu_y$ is $(\overline X -\overline Y)\pm t_{n+m-2}(\alpha/2)S_{\overline X -\overline Y}$.
Theorem:
How can I prove the Corollary using the Theorem? The difference in the Corollary is that $\sigma^2_x\neq\sigma^2_y$.
Thank you in advance for your help.
Since the Corollary says "under the assumptions of the Theorem," the two variances would still be equal in the Corollary.
So you have \begin{align} & -t < \frac{(\overline X - \overline Y) - (\mu_X-\mu_Y)}{s_p\sqrt{\frac 1 n + \frac 1 m}} < t \tag 1 \\[10pt] & -ts_p \sqrt{\frac 1 n + \frac 1 m} < (\overline X - \overline Y) - (\mu_X-\mu_Y) < ts_p\sqrt{\frac 1 n + \frac 1 m} \\[10pt] & \text{Then, multiplying all three expressions by $-1$, we get:} \\ &ts_p\sqrt{\frac 1 n + \frac 1 m} > (\mu_X-\mu_Y) - (\overline X - \overline Y) > - ts_p\sqrt{\frac 1 n + \frac 1 m} \\[10pt] & (\overline X-\overline Y) + ts_p\sqrt{\frac 1 n + \frac 1 m} > \mu_X-\mu_Y > -ts_p\sqrt{\frac 1 n + \frac 1 m}. \tag 2 \end{align} In the middle we have only the unobservable quantity to be estimated, and on the left and right we have only observable quantities. Since the probability of the event in line $(1)$ is $1-\alpha,$ so is the probability of the event in line $(2).$
If we take $S_{\overline X - \overline Y}$ to mean $s_p\sqrt{\frac 1 n + \frac 1 m},$ then that answers the question.