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!

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- Sep 10th 2008, 05:11 AMNeedHelp18Logic -- 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 AMPlato
$\displaystyle \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}

$