# Logic -- have no idea how to finish this!

• Sep 10th 2008, 05:11 AM
NeedHelp18
Logic -- have no idea how to finish this!
Demonstrate that p ->q is equivalent to (p^q')'.

This is what i got so far: T = true, F= false

p q p -> q q'
T T T F
T F F T
F T T F
F F T T

now what?? grrr
any help is appreciated thanks!
• Sep 10th 2008, 08:21 AM
Plato
$\begin{array}{ccccc}
p & q &\vline & {p \to q} & {\neg \left( {p \wedge \neg q} \right)} \\
\hline
t & t &\vline & t & t \\
t & f &\vline & f & f \\
f & t &\vline & t & t \\
f & f &\vline & t & t \\
\end{array}
$