# Logic Question

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• Feb 25th 2008, 05:00 PM
brand_182
Logic Question
I'm trying to show that this implication is true:

~P, P v Q --> Q

I can show it simply by writing out my thought process and showing that a contradiction occurs if I make ~P true and try to make the conditional on the right false.

But, when I do a truth table to check my work, I see that the two sentences do not have equivalent truth columns. Does this mean I've done something wrong in my analysis or does it sometimes not work with the truth table?
• Feb 25th 2008, 05:26 PM
Soroban
Hello, brand_182!

You are misreading the problem . . .

Quote:

I'm trying to show that this implication is true:

~P, P v Q --> Q

When I do a truth table to check my work,
the two sentences do not have equivalent truth columns.
. . There is only ONE sentence.

The sentence is: . $[\sim p \text{ and }(p \vee q)] \text{ implies }q$

The truth table looks like this:

. . $\begin{array}{c|c|ccccccc}
p & q & \sim p & \wedge & (p & \vee & q) & \to & q \\ \hline
T & T & F & F & T & T & T & {\color{blue}T} & T \\
T & F & F & F & T & T & F & {\color{blue}T} & F \\
F & T & T & T & F & T & T & {\color{blue}T} & T \\
F & F & T & F & F & F & F & {\color{blue}T} & F \\ \hline
& & _1 & _3 & _1 & _2 & _1 & _4 & _1\end{array}$