Re: Answer "how to prove ~(P <--> Q) |- ~P <-->Q "

On May 25, 10:37 pm, Conbra <s...@xxxxxxxxx> wrote:
Answer one question

At google group sci.logic one person asks a question "How do you prove
~(P <--> Q) |- ~P <--> Q using natural deduction?". Not only the
relation [~(P <--> Q) |- ~P <--> Q] could be proved on concept
algebra, but also the relation [~P <--> Q |-~(P <--> Q)] could be
proved.


"how to prove ~(P<->Q) |- ~P<-> Q "?

Proof in a Copi-style system:

1. ~(P <-> Q)
2. ~((P -> Q) & (Q -> P)) 1 Equiv1
3. ~(P -> Q) v ~(Q -> P) 2 DeM
4. ~(~P v Q) v ~(~Q v P) 3 Impl
5. (~~P & ~Q) v (~~Q & ~P) 4 DeM
6. (~~P & ~Q) v (Q & ~P) 5 DN
7. (Q & ~P) v (~~P & ~Q) 6 Comm
8. (~P & Q) v (~~P & ~Q) 7 Comm
9. ~P <-> Q 8 Equiv2



That is to say the relation [~(P <--> Q) = ~P <--> Q] could be
proved on concept algebra. At this article we are going to prove the
late relation on concept algebra.

Full text see following webhttp://blog.tom.com/blog/read.php?blogid=62728&bloggerid=762122


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