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  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 $p \rightarrow \neg p$?
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  • $\begingroup$ You jumped the gun at the last step. $\neg(\neg q \land \neg p) \lor \neg p \equiv q \lor p \lor \neg p = q\lor T \equiv T$. $\endgroup$
    – jl00
    Apr 4, 2022 at 4:40
  • $\begingroup$ @jl00 Oh I see. That is a fault. Thanks. $\endgroup$ Apr 4, 2022 at 4:43
  • $\begingroup$ For your second question, yes, and note that $p\implies\lnot p$ is equivalent to $\lnot p$. $\endgroup$
    – RobPratt
    Apr 4, 2022 at 4:46
  • $\begingroup$ Use truth Tables to verify your assumptions and Transformations. $\endgroup$
    – miracle173
    Apr 4, 2022 at 4:48
  • $\begingroup$ Thank you everyone. $\endgroup$ Apr 4, 2022 at 5:00

1 Answer 1

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For your second question, yes, that is absolutely correct.

...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 $p$?

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  • $\begingroup$ Is it true for the implication $p \rightarrow \neg p$? I am very confused... $\endgroup$ Apr 4, 2022 at 4:58
  • $\begingroup$ @filterhash $p \rightarrow \neg p$ is logically equivalent to $\neg p.$ $\endgroup$
    – ryang
    Apr 9, 2022 at 12:51

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