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I have been trying to solve this but I got stuck at the end.

$$\begin{align} \neg q \to \neg (q \land (p \to \neg q)) &\equiv \neg \neg q\lor \neg (q \land ( \neg p\lor \neg q)) \\& \equiv q\lor \neg q \land \neg ( \neg p\lor \neg q) \\&\equiv T \land \neg ( \neg p\lor \neg q)\end{align}$$

Aaron Maroja
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Fasil
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  • I'd suggest you think about what $q\to (p\to q)$ means and then you think again about the claim in question. – Nikolaj-K Apr 21 '15 at 12:14
  • $(\neg q)\to\neg(q\land\phi)$ is a tautology for any $\phi$, and you might find it easier to prove, since you won't be tempted to dig into the details of the $(p\to\neg q)$ part of your formula. – Andreas Blass Apr 21 '15 at 19:28

2 Answers2

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In the second to last step, you commited an error \begin{align*} ¬q→¬(q∧(p→¬q)) &\equiv ¬¬q∨¬(q∧(¬p∨¬q))\\ &\equiv q∨¬q\color{red}\lor¬(¬p∨¬q)\\ &\equiv T \color{red}\lor ¬(¬p∨¬q)\\ &\equiv T. \end{align*}

martini
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2

A slightly different way to see this

$$\begin{align}\neg q \to \neg (q \wedge (p \to \neg q)) &\color{red}\equiv (q \wedge (p \to \neg q)) \to q \\&\equiv (\neg q \vee q )\vee \neg( p\to \neg q)\\&\equiv T \vee \neg (\neg p \vee\neg q) \\&\equiv \color {red} T\end{align}$$

Aaron Maroja
  • 17,571