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I am doing some simple Boolean algebra and wanted to perform some calculations in Mathematica but found it doesn't work.

Let us consider the expression:

expr1 = ((b < 0 && A <= AA) || A < AA);
expr2 = ((b < 0 && A == AA) || A < AA);

From our point of view they are the same. Or do we miss a special case?

FullSimplify[{expr1 == expr2}, #] & /@ {A < AA, A > AA, A == AA}

{{True}, {True}, {True}}

Now the problem is Mathematica does not see them as equivalent, in general:

FullSimplify[
expr1
 ==
expr2
,
A > 0 && AA > 0]

This does not yield True, but instead expr1 == expr2.

Do we have an error in reasoning?

How can I make Mathematica simplify expr1 to expr2?

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2 Answers 2

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Equivalent[expr1, expr2] // FullSimplify

(* True*)
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  • $\begingroup$ Whats the difference to my code? $\endgroup$
    – Philipp
    Feb 13, 2015 at 17:13
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    $\begingroup$ @Philipp Equivalent does not mean Equal.It only means that the truth tables for both are the same $\endgroup$ Feb 13, 2015 at 17:19
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Just to amplify belisarius comments, let b<0 be x, A < AA be y and A==AA be z,

then your expressions are:

e1 = (x && (y || z)) || z
e2 = (x && y) || z

BooleanMinimize applied to these yields: (x && y) || z

The truth tables can be shown:

Framed@TableForm[BooleanTable[{x, y, z, e1, e2}], 
  TableHeadings -> {None, {x, y, z, e1, e2}}]

enter image description here

And TautologyQ[Equivalent[e1, e2]] yields True.

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