Prove $(s \to p) \to (s \to q)$ using propositional logic I need to demonstrate
$(s \to p) \lor (t \to q) \vdash (s \to q) \lor (t \to p)$
I know that, if I can do something like the following, I can succesfully demonstrate the validity of this logical sentence. However, I have no idea how to get from $s \to p$ to $s \to q$.

*

*$(s \to p) \lor (t \to q)$ (Promise)

*$s \to p \ \ \ \ \ \ \ \ \ $ (Assumption)

*...

*$s \to q$

*$(s \to q) \lor (t \to p)$
and then

*

*$t \to q \ \ \ \ \ \ \ \ \ $ (Assumption)

*...

*$t \to p$

*$(s \to q) \lor (t \to p)$
Any hint is very well appreciated.
Thanks.
 A: Your proof skeleton seems sensible. However, I do not think it is going to lead you to the conclusion, in this case. I think a proof by contradiction is needed.
$
\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\def\Ae#1{\qquad\mathbf{\forall\,Elim}\colon #1 \\}
\def\Ai#1{\qquad\mathbf{\forall\,Intro}\colon #1 \\}
\def\Ee#1{\qquad\mathbf{\exists\,E}\colon #1 \\}
\def\Ei#1{\qquad\mathbf{\exists\,Intro}\colon #1 \\}
\def\R#1{\qquad\mathbf{R}\colon #1 \\}
\def\ci#1{\qquad\mathbf{\land\,I}\colon #1 \\}
\def\ce#1{\qquad\mathbf{\land\,E}\colon #1 \\}
\def\oe#1{\qquad\mathbf{\lor\,E}\colon #1 \\}
\def\ii#1{\qquad\mathbf{\to I}\colon #1 \\}
\def\ie#1{\qquad\mathbf{\to E}\colon #1 \\}
\def\be#1{\qquad\mathbf{\leftrightarrow E}\colon #1 \\}
\def\bi#1{\qquad\mathbf{\leftrightarrow I}\colon #1 \\}
\def\qi#1{\qquad\mathbf{=I}\\}
\def\qe#1{\qquad\mathbf{=E}\colon #1 \\}
\def\ne#1{\qquad\mathbf{\neg E}\colon #1 \\}
\def\ni#1{\qquad\mathbf{\neg I}\colon #1 \\}
\def\IP#1{\qquad\mathbf{IP}\colon #1 \\}
\def\X#1{\qquad\mathbf{\bot\,Elim}\colon #1 \\}
\def\DNE#1{\qquad\mathbf{DNE}\colon #1 \\}
$
$
\fitch{(s \to p) \lor (t \to q)}{
  \fitch{\lnot((s \to q) \lor (t \to p))}{
   \fitch{s \to p}{
     \fitch{s}{
       \vdots\\
       \bot\\
       q
}\\
s \to q\\
(s \to q) \lor (t \to p)\\
}\\
\fitch{t \to q}{
 \vdots\\
 (s \to q) \lor (t \to p)
}\\
(s \to q) \lor (t \to p)\\
\bot
}\\
(s \to q) \lor (t \to p)
}
$
A: We need to prove
\begin{eqnarray}
(s \to p) \lor (t \to q) \equiv (s \to q) \lor (t \to p)
\end{eqnarray}
Note that
\begin{eqnarray}
(s \to p ) \lor (t \to q) &\equiv& (s \lor \lnot p) \lor (t \lor \lnot q)\\
&\equiv& (s\lor \lnot q) \lor (t \lor \lnot p)\\
&\equiv& (s \to q) \lor (t \to p)
\end{eqnarray}
