Show that $(\forall x)\alpha(x) \lor (\forall x)\beta(x)\rightarrow (\forall x)(\alpha(x)\lor \beta(x)) $ So here is what I have:
Assume $(\forall x)\alpha(x) \lor (\forall x)\beta(x) $
Then by universal instantiation, $\alpha(t) \lor \beta(s)$.
Right here is where I am stuck. I'm stuck because the variables for $\alpha$ and $\beta$ are $t$ and $s$ respectively, so what allows me to use universal generalization to conclude $(\forall x)(\alpha(x)\lor \beta(x))$? I'm confused because the variables are different due to them being in different quantifer scopes.
 A: See Prenex Normal Form for justification
\begin{align}
& (\forall x)\alpha(x) \lor (\forall x)\beta(x) \\
\iff & (\forall x)\alpha(x) \lor (\forall y)\beta(y) \\
\iff & (\forall y)((\forall x)\alpha(x) \lor \beta(y)) \\
\iff & (\forall x)(\forall y)(\alpha(x) \lor \beta(y))
\end{align}
In the last line if one takes the special case $x = y$, it implies the required formula.
A: Hint: The most straightforward proof (in a natural deduction system) will use ${\lor}E$ near the top level:
\begin{align*}
1.~ & (\forall x)\alpha(x) \lor (\forall x)\beta(x) & & \mathrm{assumption} \\
2.~ & \quad (\forall x)\alpha(x) & & \mathrm{assumption} \\
& \quad\quad \vdots \\
N.~ & \quad\quad (\forall x) (\alpha(x) \lor \beta(x)) \\
N+1.~& \quad (\forall x)\beta(x) & & \mathrm{assumption} \\
& \quad\quad \vdots \\
M.~& \quad\quad (\forall x) (\alpha(x) \lor \beta(x)) \\
M+1.~& \quad(\forall x) (\alpha(x) \lor \beta(x)) & & {\lor}E ~ 1, 2-N, (N+1)-M \\
M+2.~& ((\forall x) \alpha(x) \lor (\forall x) \beta(x)) \rightarrow (\forall x) (\alpha(x) \lor \beta(x)) & & {\rightarrow}I ~ 1-(M+1)
\end{align*}
Now it's just left to you to fill in how you get from the assumption $(\forall x)\alpha(x)$ to the conclusion $(\forall x) (\alpha(x)\lor \beta(x))$, and similarly for the other case.
A: Disassemble the formula into the the truth conditional clauses for each connective step by step.
As you correctly figured out, to show the implication $(\forall x)\alpha(x) \lor (\forall x)\beta(x)\rightarrow (\forall x)(\alpha(x)\lor \beta(x))$ we need to show that $(\forall x)(\alpha(x)\lor \beta(x))$ is true under the assumption that $(\forall x)\alpha(x) \lor (\forall x)\beta(x)$ is true.
Now what is the case if $(\forall x)\alpha(x) \lor (\forall x)\beta(x)$ is true? By the semantics of $\lor$, either $(\forall x)\alpha(x)$ is true or $(\forall x)\beta(x)$ is true. So we do a proof by cases:

*

*Assume $(\forall x)\alpha(x)$. That means for any term $t$, $\alpha(t)$ is true. With $\alpha(t)$ follows $\alpha(t) \lor \beta(t)$. Since the above  holds for any $t$, we have by universal generalization $(\forall x)(\alpha(x)\lor \beta(x))$.

*Analogous (for you to fill in).

Since the succendent follows in any of the two cases, the implication holds.
