Re: A different definition of MINUS, Part 3
From: Bob Badour <bbadour_at_pei.sympatico.ca>
Date: Sun, 21 Dec 2008 15:53:46 -0400
Message-ID: <494e9ecc$0$5459$9a566e8b_at_news.aliant.net>
>
> Actually, easy. The right side simplifies to
>
> x + y = (x' ^ y')'.
>
> Applying double negation (provable in RL), we get
>
> (x+y)' = x' ^ y'.
>
> De Morgan, which is again provable in RL. Amazing.
Date: Sun, 21 Dec 2008 15:53:46 -0400
Message-ID: <494e9ecc$0$5459$9a566e8b_at_news.aliant.net>
vadimtro_at_gmail.com wrote:
> On Dec 21, 11:36 am, vadim..._at_gmail.com wrote:
>
>>x + y = ((x ^ x)' ^ (y ^ y)')'. >>... is tough...
>
> Actually, easy. The right side simplifies to
>
> x + y = (x' ^ y')'.
>
> Applying double negation (provable in RL), we get
>
> (x+y)' = x' ^ y'.
>
> De Morgan, which is again provable in RL. Amazing.
Why are you amazed? Received on Sun Dec 21 2008 - 20:53:46 CET