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

On 5月28日, 上午5时14分, translogi <wilem...@xxxxxxxxxxxxxx> wrote:
On May 26, 6:37 am, 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. 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

says

The procedure of proving is as follows:

~(P Q) / (~P Q)

= ~((P / Q) * (Q / P)) / ((~P / Q) * (Q / ~P))

= (( P - Q) + (Q - P)) / ((~P / Q) * (Q / ~P))

= (( P - Q) + (Q - P)) / ((~P + ~Q) * (Q + P))

= (( P - Q) + (Q - P)) / (( P - Q) + (Q - P))

= υ

but the importand bit is how do you get from line 1 to line 2 and from
line 2 to line 3
ect.

Also a bit a problem with your website, my computer doesn't like
chineese characters.

~(P <--> Q) / (~P <--> Q)


= ~((P / Q) * (Q / P)) / ((~P / Q) * (Q / ~P)) Axiom GW11


= (( P - Q) + (Q - P)) / ((~P / Q) * (Q / ~P)) Theorem A


= (( P - Q) + (Q - P)) / ((~P + ~Q) * (Q + P)) Theorem B


= (( P - Q) + (Q - P)) / (( P - Q) + (Q - P))


= υ


If you'd like I will prove the related theorem in detail to you.


.