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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$\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} $
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