You used \[Epsilon]. It is just a symbol, you are looking for MemberQ
you used == (i.e., Equal) when you put seta == setb automatically
the operation will be evaluated as False. So just make some symbolic function that will denotes the two sets are identical or equal (eg., eq[x,y])
Last one, Implies[False, True] returns True meaning that if there is some z that belongs to set y but not to set x the relation is still True… so the “axiom” is practically ForAll[{x,y}, eq[x,y]] regardless of z, so ANY two sets are defined equal by the presented logic of this expression
Thank you…I’m not sure why, but that doesn’t work when I try it. Perhaps that’s because the lines are broken to begin with. Maybe I just need to restart Mathematica. Would you like to shed some light on that?
Trying to correct code with the suggestions. I didn’t see how I could use MemberQ; so I used Element. How would you tweak the code below to get a proof that {1,2,3} = {3,2,1}.
It seems that you will not be able to do that. Element[1,{1,2,3}] does not return either True or False for relations with sets, the expression is just returned unevaluated.
Within programs MemberQ[{1, 2, 3}, 1] returns True but it is not useful for logical proofs.
From what I noticed in the documentation it seems that proofs in set theory are not supported by the theorem proving tool.
I hope if someone more involved in this area would be able to comment further.