Proving that $(p \to \lnot p) \vdash \lnot p$

Question

I'm working through Tomassi's Logic book and I've come across an exercise that challenges you to find a proof for $(p \to \lnot p) \vdash \lnot p$ in 11 steps (p. 82, Exercise 3.2.1.10). Only the following strategies have been mentioned so far in the book:

1. $\land$ - Introduction and $\land$ - Elimination
2. Modus Ponens and Modus Tollens
3. Conditional Proof
4. Double Negation Elimination and Double Negation Introduction
5. $\Leftrightarrow$ - Introduction and $\Leftrightarrow$ - Elimination
6. Deduction Theorem

Other than stating the premise the only strategy I can think of from here is to assume $p$ and then free it from its assumption, but I'm not sure where else to go from here. Any help would be appreciated on what the next steps should be.

Answer 1

I can do it in 9 steps:

\begin{array}{llll} \{1\}&1.&P \to \neg P&Premise\\ \{2\}&2.&P&Premise\\ \{2\}&3.&\neg \neg P&2 \ DNI\\ \{1,2\}&4.&\neg P&1,2 \ MP\\ \{2\}&5.&(P\to \neg P) \to \neg P&1,4\ CP\\ \{2\}&6.&\neg (P \to \neg P)&3,5\ MT\\ \{ \}&7.&P\to \neg(P \to \neg P)&2,6 \ CP\\ \{1\}&8.&\neg \neg (P \to \neg P)&1 \ DNI\\ \{1\}&9.&\neg P&7,8\ MT\\ \end{array}

Answer 2

Hint

$$(p \to \lnot p) \vdash \lnot p$$

By conditional exchange

$\neg p \lor \neg p$

If the conditional exchange hasn't been introduced yet (which I see now that it hasn't), you can use a conditional proof and assume $p$. Then, by modus tollens, you can derive $\neg p$