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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: [mg62034] Re: ((a&&b)||c)==((a||c)&&(b||c))
  • From: Peter Pein <petsie at dordos.net>
  • Date: Wed, 9 Nov 2005 05:05:53 -0500 (EST)
  • References: <dksdln$h87$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Steven T. Hatton schrieb:
> 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
>     ]
> 
But Mathematica does:

In[1]:= expr = ((a && b) || c) == ((a || c) && (b || c))
Out[1]= ((a && b) || c) == ((a || c) && (b || c))
In[2]:= LogicalExpand /@ expr
Out[2]= True


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