I'm asked to show that the proof rule
\begin{equation} \dfrac{\varphi \to \psi}{\lnot \varphi \to \lnot \psi} \end{equation}
is not sound.
To show this would I just make the truth tables for the statement above the line and below the line and show that they are not equivalent?
I'm also asked to show $\vdash p \lor \lnot p$. I can have $\lnot (p \lor \lnot p) \to p \land \lnot p$ as an assumption. When I try to move from the conclusion upward I get
\begin{equation} \dfrac{\dfrac{p \land ¬p}{p}}{p \lor \lnot p} \end{equation} as I try to move toward the assumption, but I don't think that's right because $p \lor \lnot p$ should conclude $\bot$, not $p$. If I try to move from the assumption downward toward the conclusion I'm not sure what to do because for an implication elimination wouldn't I need to have
\begin{equation} \lnot(p \lor ¬p) \to p \land \lnot p \qquad\qquad \lnot (p \lor \lnot p) \end{equation}
as an assumption rather than just
\begin{equation} \lnot (p \lor \lnot p) \to p \land \lnot p \end{equation}
The first question is:
One would check that the premise implies the conclusion noting any line in the truth table where that is not true. From that line one can build a counter-example.
For the second question, the OP wants to show $⊢p∨¬p$. One is permitted to use this portion of the De Morgan rules: $¬(p∨¬p)→p∧¬p $.
Proceeding the way the OP starts to show this we could derive a proof using a Fitch-style proof checker as follows: