# Thread: Please check my Truth Table 2!

1. ## SOLVED!! Please check my Truth Table 2!

( 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.

2. Originally Posted by aeubz
( P AND Q ) AND ( ~ P OR ~ Q ) AND ( P AND Q ) AND ( ~ P OR Q )
Is this a contradiction?
I think that you mised a 'not'.
]( P AND Q ) AND ( ~ P OR ~ Q ) AND ( P AND ~ Q ) AND ( ~ P OR Q )
Did you?

3. nope its typed correctly

4. If this is it: ( P AND Q ) AND ( ~ P OR ~ Q ) AND ( P AND Q ) AND ( ~ P OR Q )

Then drop a redundancy ( ~ P OR ~ Q ) AND ( P AND Q ) AND ( ~ P OR Q )

5. it is a contradiction, is it not?

6. Hello, aeubz!

If it is typed correctly, it should read: . $(P \wedge Q) \wedge (\sim\!P \:\vee \sim\!Q) \wedge (\sim\!P \vee Q)$

And we need a few theorems:
. . $\begin{array}{cc}(1) & R \:\wedge \sim\!R \:=\:f \\ (2) & R \vee f \:=\:R \\ (3) & R \wedge f \:=\:f \end{array}$

I would prove it like this:

$\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}$

7. TY All for your responses !! it is contradiction.. by what Soroban wrote. even though i did it the long way by using truth tables but his way is much neater and much better