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

On 5月28日, 下午5时37分, Rupert <rupertmccal...@xxxxxxxxx> wrote:
On May 28, 9:36 am, Conbra <s...@xxxxxxxxx> wrote:





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.- Hide quoted text -

- Show quoted text -

It looks like you are trying to re-invent the theory of Boolean
algebras. If your work were correct, your theory would be equi-
interpretable with the theory of Boolean algebras. I'm not sure that
this is actually the case; I tried a natural-seeming interpretation
and some of your axioms came out false. Anyway, you should read up on
the theory of Boolean algebras. It looks like what you are doing is
trying to find a new axiomatization for it.- 隐藏被引用文字 -

- 显示引用的文字 -

The concept algebra is a complete algebra (http://
conceptalgebra.blog.sohu.com/29782375.html ) and the Boolean algebra
is incomplete algebra. It is sub algebra of concept algebra. All of
the axioms of Boolean algebra could be reasoned from concept
algebra(see web: http://conceptalgebra.blog.sohu.com/43607593.html )

.

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