A logic assignment requires me to create a model in which most X's are smarter than most Y's, but most Y's are such that it is not the case that most X's are smarter than it.
It's easy to do this when we grant that $Sxy$ = "x smarter than y" is a non-commutative predicate eg. $Sxy => -Syx$.
However it seems impossible to do this if we assume that $Sxy$ is transitive ie. $(Sxy + Syz) => Sxz$
My question is regarding whether "smarter than" is, in reality, transitive.
My logic professor argued that it wasn't because we can imagine different types of smartness such that A > B, B > C and C > A. I disagree because:
Different types of smartness demand different predicates eg. Pxy = x is smarter than y at visual problems, Cxy = x is smarter than y at chess problems, etc
To try and rescue this idea, if we declared "smarter than" to mean "smarter than [in any capacity]" (ie. Sxy = Pxy or Cxy or ...), then we lose its non-commutativity and it becomes pretty meaningless.
He also suggested thinking about it in terms of a competition in which A usually beats B and B usually beats C and C usually beats A. There is no contradiction here, but again, "x beats y" is a different predicate with different semantics.
Therefore I continue to believe that "smarter than", reasonably, is a transitive relationship.
Who is correct??
Your argument is irrelevant. As you said, your professor argued that it wasn't because we can imagine non-transitive smartness. That is a red herring. There is a simple and transitive smartness relation that has the properties that were asked for. So it doesn't matter whether or not we are willing to consider nontransitive relations, because there is a transitive solution.