1
$\begingroup$

I'm trying to formalize and analyze the following reasoning:

  1. All equilateral triangles are isosceles triangles.
  2. Some triangles are not equilateral triangles.
  3. Therefore some triangles are not isosceles triangles.

The reasoning is obviously not valid. But I tried to find that out with Mathematica. So I typed what I would type in my math notebook:

Resolve[Implies[
  SubsetEqual[equilateralTriangles, isoscelesTriangles] &&
   Exists[triangle, NotElement[triangle, equilateralTriangles]]
  , Exists[triangle, NotElement[triangle, equilateralTriangles]]]]

and I got:

Resolve::elemc: Unable to resolve the domain or region membership condition triangle ∉ equilateralTriangles. >>

Resolve[equilateralTriangles \[SubsetEqual] isoscelesTriangles && \!\(
\*SubscriptBox[\(\[Exists]\), \(triangle\)]\(triangle \[NotElement] 
     equilateralTriangles\)\) \[Implies] \!\(
\*SubscriptBox[\(\[Exists]\), \(triangle\)]\(triangle \[NotElement] 
    equilateralTriangles\)\)]

Am I doing this the right way? Can one solve problems like this with Mathematica?

$\endgroup$
  • 1
    $\begingroup$ SubsetEqual has no meaning, for starters. It's there for output convenience, or can be overloaded with user-written functionality... $\endgroup$ – ciao Mar 7 '15 at 8:45
2
$\begingroup$

To answer your first question: your syllogism should probably be formulated as

Resolve[
  Implies[
    ForAll[triangle, 
      Implies[
        Element[triangle, equilateralTriangles], 
        Element[triangle, isoscelesTriangles]]] && 
      Exists[triangle, NotElement[triangle, equilateralTriangles]], 
    Exists[triangle, NotElement[triangle, isoscelesTriangles]]]]

but that still produces the Resolve::elemc message.

To answer your second question: I don't think so. Mathematica can only apply the predicate calculus to points in geometric regions or numerical domains, not totally abstract sets, and that is what the message Resolve::elemc is trying to tell you. See the documentation forExists

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.