# Show that the proof rule is not sound and proof question

I'm asked to show that the proof rule

$$$$\dfrac{\varphi \to \psi}{\lnot \varphi \to \lnot \psi}$$$$

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

$$$$\dfrac{\dfrac{p \land ¬p}{p}}{p \lor \lnot p}$$$$ as I try to move toward the assumption, but I don't think that's right because $$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

$$$$\lnot(p \lor ¬p) \to p \land \lnot p \qquad\qquad \lnot (p \lor \lnot p)$$$$

as an assumption rather than just

$$$$\lnot (p \lor \lnot p) \to p \land \lnot p$$$$

• For your first question, find a pair of values for $\phi$ and $\psi$ such that the formula above the line is true but the formula below is false. For your second question, see Mauro's answer. – palmpo Jan 6 at 19:37

For the second question, the OP wants to show $$⊢p∨¬p$$. One is permitted to use this portion of the De Morgan rules: $$¬(p∨¬p)→p∧¬p$$.