P -> Q is equivalent to ~P v Q, so why isn't P -> ~Q equivalent to ~P v ~Q?
I can't figure out why the rule for P -> Q does not apply to P -> ~Q.
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The answer was that P->~Q and ~P v ~Q are equivalent, so the rule does apply.– DennisFeb 13, 2013 at 6:36
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Yea, I was more just putting a succinct, direct answer to your question since SF.'s answer is only clear to someone who understands basic logic already (i.e., it was more for future readers of this post than you).– DennisFeb 13, 2013 at 6:54
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1 Answer
What do you mean does not apply?
P Q ~P ~Q P->Q ~P v Q P->~Q ~P v ~Q
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0 0 1 1 1 = 1 1 = 1
0 1 1 0 1 = 1 1 = 1
1 0 0 1 0 = 0 1 = 1
1 1 0 0 1 = 1 0 = 0
As for me, everything matches.