Re: Answer "how to prove ~(P <--> Q) |- ~P <-->Q "
- From: "Jesse F. Hughes" <jesse@xxxxxxxxxxxxx>
- Date: Thu, 31 May 2007 07:38:04 -0400
herbzet <herbzet@xxxxxxxxx> writes:
If you enjoy this sort of thing, try
(P -> Q) & (R -> S) |- (P v R) -> (Q v S).
I've been too lazy to work it out Copi style.
Doesn't your copy of Copi have "Constructive Dilemma" (CD)?
1. (P -> Q) & (R -> S)
2. P v R (Assumption)
3. Q v S 1,2 CD
4. (P v R) -> (Q v S) 2-3 CP
--
"You are beneath contempt because you betray mathematics itself, and
spit upon the truth, spit upon decency, and spit upon the intelligence
of the world. You betrayed the world, and now it's time for the world
to notice." -- James S. Harris awaits Justice for crimes against Math.
.
- References:
- Answer "how to prove ~(P <--> Q) |- ~P <-->Q "
- From: Conbra
- Re: Answer "how to prove ~(P <--> Q) |- ~P <-->Q "
- From: George Dance
- Re: Answer "how to prove ~(P <--> Q) |- ~P <-->Q "
- From: herbzet
- Answer "how to prove ~(P <--> Q) |- ~P <-->Q "
- Prev by Date: Re: Answer "how to prove ~(P <--> Q) |- ~P <-->Q "
- Next by Date: Re: Formalizing the Fundamental Theorem of Arithmetic
- Previous by thread: Re: Answer "how to prove ~(P <--> Q) |- ~P <-->Q "
- Next by thread: question
- Index(es):
Relevant Pages
|
|
