From set inclusion to set equality in lean

问题

Given a proof of set inclusion and its converse I'd like to be able to show that two sets are equal.

For example, I know how to prove the following statement, and its converse:

open set

universe u
variable elem_type : Type u
variable A : set elem_type
variable B : set elem_type

def set_deMorgan_incl : A ∩ B ⊆ set.compl ((set.compl A) ∪ (set.compl B)) :=
    sorry

Given these two inclusion proofs, how do I prove set equality, i.e.

def set_deMorgan_eq : A ∩ B = set.compl ((set.compl A) ∪ (set.compl B)) :=
    sorry

回答1:

You will want to use anti-symmetry of the subset relation, as proved in the stdlib package:

def set_deMorgan_eq : A ∩ B = set.compl ((set.compl A) ∪ (set.compl B)) :=
subset.antisymm (set_deMorgan_incl _ _ _) (set_deMorgan_incl_conv _ _ _)

As you can see in the proof of subset.antisymm, it combines both functional and propositional extensionality.

来源：https://stackoverflow.com/questions/45658919/from-set-inclusion-to-set-equality-in-lean