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Assuming that $P \land \neg Q \leftrightarrow 0$, how would we go about proving $P \leftrightarrow Q$ using proof-by-cases?

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  • $\begingroup$ You have two different variables, $P$ and $Q$, so there are only four different cases in total. Just go through them one by one. $P$ is true (or $P=1$ in some notation) and $Q$ is false, etc... $\endgroup$ – Matti P. Feb 7 '18 at 8:46
  • $\begingroup$ $P \land \lnot Q$ is equivalent to $\lnot (P \to Q)$ $\endgroup$ – Mauro ALLEGRANZA Feb 7 '18 at 8:56
  • $\begingroup$ But note that $\lnot (P \land \lnot Q) \nvDash P \leftrightarrow Q$. $\endgroup$ – Mauro ALLEGRANZA Feb 7 '18 at 9:00
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Complete the table and compare the last two columns.

$$\begin{array}{l:l} P&Q&P\wedge\neg Q& P\wedge\neg Q\leftrightarrow 0 & P\leftrightarrow Q\\ \hline 1 & 1 & 0 & 1 & 1\\ 1 & 0 & & & \\ 0 & 1 & & & \\ 0 & 0 & & & \end{array}$$

Now use the results to guide you to the cases.

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