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MathGroup Archive 2005

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Re: ((a&&b)||c)==((a||c)&&(b||c))

  • To: mathgroup at smc.vnet.net
  • Subject: [mg62057] Re: [mg62015] ((a&&b)||c)==((a||c)&&(b||c))
  • From: "Steven T. Hatton" <hattons at globalsymmetry.com>
  • Date: Thu, 10 Nov 2005 02:50:43 -0500 (EST)
  • References: <200511090845.DAA17387@smc.vnet.net> <B544096B-72F4-4119-86F0-42F575461D2F@mimuw.edu.pl> <DEC0DD7D-F78B-4A06-8C11-E0D72D49F9C8@mimuw.edu.pl>
  • Sender: owner-wri-mathgroup at wolfram.com

Andrzej Kozlowski wrote:

> On 9 Nov 2005, at 17:45, Steven T. Hatton wrote:
> 
>> Why does Mathematica not determine that the following is true?
>>
>> ((a \[And] b) \[Or] c) == ((a \[Or] c) \[And] (b \[Or] c))

> Because, (as I tried to explain in a reply to a recent posting of
> yours) this is not the way == is used in Mathematica. The correct way
> to show this is:

The rules of logic are very tricky.  For example:

Tautology <=> True
P && Tautology <=> P
P || Tautology <=> Tautology
!Tautology <=> Contradiction


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