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:
- $\land$ – Introduction and $\land$ – Elimination
- Modus Ponens and Modus Tollens
- Conditional Proof
- Double Negation Elimination and Double Negation Introduction
- $\Leftrightarrow$ – Introduction and $\Leftrightarrow$ – Elimination
- 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.
Best Answer
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}