Create a Truth Table: (p ^ q) ->(p ▼ ~q)

[rymatson406]

(p ^ q) ->(p ▼ ~q)

Need help creating a truth table (6 columns) for the above.Thanks

 

[Evgeny.Makarov]

I assume the formula is $(p\land q)\to(p\lor \neg q)$.

Have you seen examples of creating a truth table? Can you write a truth table for $p\land q$, which is a part of the required truth table? What exactly is your difficulty?

 

[Deveno]

Presumably your first two columns are for $p$ and $q$, which have two possible values each, making the table of necessity four rows "deep". I would go with:

T T
T F
F T
F F

for the first two columns (but other row-orders are possible).

It's hard to say what the next four columns are supposed to be, my guess is:

$\neg q,\ p \wedge q,\ p \vee \neg q$ and $(p \wedge q) \rightarrow (p \vee \neg q)$.

Really, only the last one is "necessary", but it's easier on the ol' noggin to include the other 3.

 

[soroban]

Hello, rymatson406!

Create a truth table for: .(p \wedge q)\;\to\;(p \,\vee \sim q)

. . . . \begin{array}{|c|c|c|c|c|c|c|c|c|} <br /> p &amp; q &amp; (p &amp; \wedge &amp; q) &amp; \to &amp; (p &amp; \vee &amp; \sim q) \\ \hline<br /> T&amp;T &amp; T&amp;T&amp;T &amp;T&amp; T&amp;T&amp;F \\<br /> T&amp;F &amp;T&amp;F&amp;F &amp;T&amp; T&amp;T&amp;T \\<br /> F&amp;T &amp;F&amp;F&amp;T &amp;T&amp; F&amp;F&amp;F \\<br /> F&amp;F &amp; F&amp;F&amp;F &amp;T&amp; F&amp;T&amp;T \\ \hline<br /> &amp;&amp; 1&amp;2&amp;1&amp;3&amp;1&amp;2&amp;1 \\ \hline \end{array}**