Prove $(s \to p) \to (s \to q)$ using propositional logic

Question

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 $s \to q$.

1. $(s \to p) \lor (t \to q)$ (Promise)
2. $s \to p \ \ \ \ \ \ \ \ \ $ (Assumption)
3. ...
4. $s \to q$
5. $(s \to q) \lor (t \to p)$

and then

1. $t \to q \ \ \ \ \ \ \ \ \ $ (Assumption)
2. ...
3. $t \to p$
4. $(s \to q) \lor (t \to p)$

Any hint is very well appreciated. Thanks.

Answer 1

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) } $

Answer 2

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}