Re: A different definition of MINUS, Part 3
From: <vadimtro_at_gmail.com>
Date: Sun, 21 Dec 2008 11:49:56 -0800 (PST)
Message-ID: <9d1d537d-cb01-4af6-ab57-5f88c3290d5a_at_g39g2000pri.googlegroups.com>
Date: Sun, 21 Dec 2008 11:49:56 -0800 (PST)
Message-ID: <9d1d537d-cb01-4af6-ab57-5f88c3290d5a_at_g39g2000pri.googlegroups.com>
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.