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

```

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