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Failure to simplify logical expressions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg27951] Failure to simplify logical expressions
  • From: Ignacio Rodriguez <ignacio at sgirmn.pluri.ucm.es>
  • Date: Sat, 24 Mar 2001 15:15:53 -0500 (EST)
  • Organization: UCM
  • Sender: owner-wri-mathgroup at wolfram.com

I am using Mathematica 3.0 for Win 2k and I want to prove the following
theorem:

((a=>b)&&(!a=>c))<=>((a&&b)||(!a&&c))

so let us try the following code

Iff[a_,b_]:=And[Implies[a,b],Implies[b,a]]
f=FullSimplify[LogicalExpand[And[Implies[a,b],Implies[Not[a],c]]]]
g=FullSimplify[LogicalExpand[Or[And[a,b],And[Not[a],c]]]]
t=Iff[f,g]
FullSimplify[t]
FullSimplify[LogicalExpand[t]]

clearly, f is the first member of the theorem, g is the second and t is
the theorem. The answer from Mathematica is

Out[2] = a&&b||b&&c||c&&!a
Out[3] = a&&b||c&&!a
Out[4] = Implies[a&&b||b&&c||c&&!a,a&&b||c&&!a] &&
Implies[a&&b||c&&!a,a&&b||b&&c||c&&!a]
Out[5] = Implies[a&&b||b&&c||c&&!a,a&&b||c&&!a] &&
Implies[a&&b||c&&!a,a&&b||b&&c||c&&!a]
Out[6] = True

so the theorem is clearly proved. However, I have the following
concerns:

If expressions 2 and 3 are equivalent, why does FullSimplify fail to
give the same results?
If expression 5 is true, why does FullSimplify fail to show it?
Is it possible to do anything with Equal? I had no luck.

--

Ignacio Rodriguez Ramirez de Arellano
Unidad de RMN
Universidad Complutense
Paseo Juan XXIII, 1
Madrid 28040, Spain

Tel. 34-91-394-3288
Fax  34-91-394-3245
e-mail: ignacio at sgirmn.pluri.ucm.es





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