( P AND Q ) AND ( ~ P OR ~ Q ) AND ( P AND Q ) AND ( ~ P OR Q )
Is this a contradiction?
Work: Well, lol, I got it is a contradiction. There is always one false in each row. So I assume it is a contradiction.
( P AND Q ) AND ( ~ P OR ~ Q ) AND ( P AND Q ) AND ( ~ P OR Q )
Is this a contradiction?
Work: Well, lol, I got it is a contradiction. There is always one false in each row. So I assume it is a contradiction.
Hello, aeubz!
If it is typed correctly, it should read: .$\displaystyle (P \wedge Q) \wedge (\sim\!P \:\vee \sim\!Q) \wedge (\sim\!P \vee Q) $
And we need a few theorems:
. . $\displaystyle \begin{array}{cc}(1) & R \:\wedge \sim\!R \:=\:f \\ (2) & R \vee f \:=\:R \\ (3) & R \wedge f \:=\:f \end{array}$
I would prove it like this:
$\displaystyle \begin{array}{ccc}(P \wedge Q) \wedge (\sim\!P\:\vee \sim\!Q) \wedge (\sim\!P \vee Q) & \text{given} \\
(P \wedge Q) \wedge \left[\sim\!P \vee (\sim\!Q \wedge Q)\right] & \text{distr.} \\
(P \wedge Q) \wedge \left[\sim\!P \vee f\right] & (1) \\
(P \wedge Q) \:\wedge \sim\!P & (2) \\
(P \:\wedge \sim\!P) \wedge Q & \text{comm/assoc.} \\
f \wedge P & (1) \\
f & (3)
\end{array}$