Okay, for my first question, I dont understand how the following statement is equal to the other:

the statement : (P-->Q)---> ~(~Q-->P)

is equal to: (P^Q) v (~Q ^~P)

How?

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- Oct 8th 2008, 10:08 AMdoublecheckmyanswerspleasCan you assist me with a few logic questions?
Okay, for my first question, I dont understand how the following statement is equal to the other:

the statement : (P-->Q)---> ~(~Q-->P)

is equal to: (P^Q) v (~Q ^~P)

How? - Oct 8th 2008, 10:59 AMSoroban
Hello, doublecheckmyanswerspleas!

I suspect there's a typo . . .

Quote:

I dont understand how the following statement is equal to the other:

The statement: .$\displaystyle (P \to \:{\color{red}\sim}\,Q) \to \;\sim(\sim\!Q \to P)$

is equal to: .$\displaystyle (P \wedge Q) \vee (\sim\!Q \:\wedge \sim\!P)$

$\displaystyle p \to q \;\equiv \;\sim p \vee q$ . . I call it ADI (alternate definition of implication).

$\displaystyle \begin{array}{ccccc}

(P \to \:\sim Q) \; \to \; \sim(\sim\!Q \to P) & & \text{Given} \\ \\

(\sim\!P \:\vee \sim\!Q) \;\to \;\sim(Q \vee P) & & \text{ADI} \\ \\

(\sim\!P \:\vee \sim\!Q) \;\to \;(\sim\!Q \:\wedge \sim\!P) & & \text{DeMorgan} \\ \\

\sim(\sim\!P\:\vee \sim\!Q) \;\vee \;(\sim\!Q \:\wedge \sim\!P) & & \text{ADI} \\ \\

(P \wedge Q) \;\vee \;(\sim\!Q \:\wedge \sim\!P)& & \text{DeMorgan}

\end{array}$