Re: ((a&&b)||c)==((a||c)&&(b||c))
- To: mathgroup at smc.vnet.net
- Subject: [mg62142] Re: ((a&&b)||c)==((a||c)&&(b||c))
- From: "Steven T. Hatton" <hattons at globalsymmetry.com>
- Date: Sat, 12 Nov 2005 03:32:39 -0500 (EST)
- References: <200511090845.DAA17387@smc.vnet.net> <43721D94.9050307@wolfram.com> <dkv013$7ts$1@smc.vnet.net> <dl1juo$25$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
John Doty wrote:
> Steven T. Hatton wrote:
>
>> On Wednesday 09 November 2005 11:02 am, Daniel Lichtblau wrote:
>
>>>Equal does not do logical manipulations on its operands.
>>
>>
>> That's part of what I'm trying to understand. Mathamatica AFAIK takes
>> expressions and transforms them into the simplest form possible, and then
>> does pattern patching.
>
> No, that's not what it does. There are a few transformations it does
> automatically, but they don't add up to anything like a transformation
> to "simplest form possible".
I should have said "most specific form available". That is, a symbol has
downvalues, and upvalues associated with it. These are listed in
descending order of specificity according to some mechanism for determining
specificity which I do not understand. The user can override this ordering
and assert the order of transformation rules if desired. I acknowledge
that specificity and simplicity are not necessarily the same thing.
> The main difference is that Simplify[] won't apply LogicalExpand[]
> automatically. I suppose this would be hazardous if some symbols didn't
> represent booleans. However:
>
> In[10]:= Simplify[((a && b) || c) == ((a || c) && (b || c)),
> TransformationFunctions -> {Automatic, LogicalExpand}]
>
> Out[10]= True
>
> -jpd
That looks useful, but following what Daniel Lichtblau posted, it looks like
LogicalExpand isn't the whole answer either. I believe I need a function
that performs Quine-McCluskey minimization in order to get a
'deterministic' biconditional.
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- References:
- ((a&&b)||c)==((a||c)&&(b||c))
- From: "Steven T. Hatton" <hattons@globalsymmetry.com>
- ((a&&b)||c)==((a||c)&&(b||c))