# Logic: What are the faults in my solution?

1. Show that $$(\neg q \wedge (p \rightarrow q)) \rightarrow \neg p$$ is a tautology.

The solution is: \begin{align} (\neg q \wedge (p \rightarrow q)) \rightarrow \neg p &\equiv \neg q \wedge (\neg p \vee q) \rightarrow \neg p \\ &\equiv ((\neg q \wedge \neg p) \vee (\neg q \wedge q)) \rightarrow \neg p \\ &\equiv \neg (((\neg q \wedge \neg p) \vee (\neg q \wedge q))) \vee p \\ &\equiv \neg(\neg q \wedge \neg p) \vee \neg p \equiv q \end{align} As a result, it is NOT a tautology. What is the fault in this proof?

1. (edit) I am somewhat confused of syllogism. Premise: $$p \rightarrow q$$, $$q \rightarrow \neg p$$. Then, can we conclude $$p \rightarrow \neg p$$?
• You jumped the gun at the last step. $\neg(\neg q \land \neg p) \lor \neg p \equiv q \lor p \lor \neg p = q\lor T \equiv T$.
– jl00
Apr 4, 2022 at 4:40
• @jl00 Oh I see. That is a fault. Thanks. Apr 4, 2022 at 4:43
• For your second question, yes, and note that $p\implies\lnot p$ is equivalent to $\lnot p$. Apr 4, 2022 at 4:46
• Use truth Tables to verify your assumptions and Transformations. Apr 4, 2022 at 4:48
• Thank you everyone. Apr 4, 2022 at 5:00

...and I suspect that that will prompt another question, so I am going to get in first. Given $$p\to(\neg p)$$, what can you deduce about $$p$$?
• Is it true for the implication $p \rightarrow \neg p$? I am very confused... Apr 4, 2022 at 4:58
• @filterhash $p \rightarrow \neg p$ is logically equivalent to $\neg p.$ Apr 9, 2022 at 12:51