# Re: Play "Stump the Intuitionists!"

*From*: Herbert Newman <nomail@invalid>*Date*: Sat, 10 Jan 2009 20:17:58 +0100

On 10 Jan 2009 05:40:31 -0800, Daryl McCullough wrote:

I guess, you meant it implies _|_, right? With other words,You can let P == Av~A. Then ~(Av~A) is a contradiction.

I'm not mistaken, (~P -> P) -> P is not int-valid.

~(A v ~A) -> _|_

or

~~(A v ~A)

is a theorem in int. logic. But how can we _prove_ that? (Well, Glivenko's

theorem comes to mind. Did you invoke it here?)

Anyway, then ~(Av~A) -> _|_ and _|_ -> Av~A would give ~(Av~A) -> Av~A.

So ~(Av~A) -> Av~A. So the [formula] (~P -> P) -> P impliesAh, nice.

excluded middle.

Herb

.

**Follow-Ups**:**Re: Play "Stump the Intuitionists!"***From:*Daryl McCullough

**Re: Play "Stump the Intuitionists!"***From:*Herbert Newman

**References**:**Play "Stump the Intuitionists!"***From:*MoeBlee

**Re: Play "Stump the Intuitionists!"***From:*MoeBlee

**Re: Play "Stump the Intuitionists!"***From:*Daryl McCullough

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