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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: [mg62042] Re: [mg62015] ((a&&b)||c)==((a||c)&&(b||c))
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 10 Nov 2005 02:50:25 -0500 (EST)
  • References: <200511090845.DAA17387@smc.vnet.net> <B544096B-72F4-4119-86F0-42F575461D2F@mimuw.edu.pl>
  • Sender: owner-wri-mathgroup at wolfram.com

On 9 Nov 2005, at 19:09, 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))
>>
>> This little function shows that the lhs  and rhs have the same  
>> truth tables,
>> and are therefore equivalent:
>>
>> TruthTable[s_, argc_] := Module[
>>     {tt = Tuples[{True, False}, argc]},
>>     {#, s @@ #} & /@ tt // TableForm
>>     ]
>>
>> -- 
>> The Mathematica Wiki: http://www.mathematica-users.org/
>> Math for Comp Sci http://www.ifi.unizh.ch/math/bmwcs/master.html
>> Math for the WWW: http://www.w3.org/Math/
>>
>
> 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:
>
>
> Implies[(a â?¨ c) â?§ (b â?¨ c),(a â?§ b) â?¨ c]//FullSimplify
>
>
> True
>
>
> Implies[(a â?§ b) â?¨ c,(a â?¨ c) â?§ (b â?¨ c)]//FullSimplify
>
>
> True
>
> Andrzej Kozlowski
>
>


Actually, there is at least one more way to do this, which you may  
like better:


LogicalExpand[(a â?§ b) â?¨ c]==LogicalExpand[(a â?¨ c) â?§ (b â?¨ c)]


True

But this one is actually less reliable.

Andrzej Kozlowski


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