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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.

  • The answer was that P->~Q and ~P v ~Q are equivalent, so the rule does apply. – Dennis Feb 13 '13 at 6:36
  • 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). – Dennis Feb 13 '13 at 6:54
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What do you mean does not apply?

  P   Q  ~P  ~Q  P->Q  ~P v Q   P->~Q  ~P v ~Q
 =================================================
  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.

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