Metamath Proof Explorer


Theorem disjss

Description: Subclass theorem for disjoints. (Contributed by Peter Mazsa, 28-Oct-2020) (Revised by Peter Mazsa, 22-Sep-2021)

Ref Expression
Assertion disjss A B Disj B Disj A

Proof

Step Hyp Ref Expression
1 cnvss A B A -1 B -1
2 funALTVss A -1 B -1 FunALTV B -1 FunALTV A -1
3 1 2 syl A B FunALTV B -1 FunALTV A -1
4 relss A B Rel B Rel A
5 3 4 anim12d A B FunALTV B -1 Rel B FunALTV A -1 Rel A
6 dfdisjALTV Disj B FunALTV B -1 Rel B
7 dfdisjALTV Disj A FunALTV A -1 Rel A
8 5 6 7 3imtr4g A B Disj B Disj A