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Metamath Proof Explorer

Theorem disjex

Description: Two ways to say that two classes are disjoint (or equal). (Contributed by Thierry Arnoux, 4-Oct-2016)

Ref Expression
Assertion disjex 
|- ( ( E. z ( z e. A /\ z e. B ) -> A = B ) <-> ( A = B \/ ( A i^i B ) = (/) ) )

Proof

Step Hyp Ref Expression
1 orcom 
 |-  ( ( A = B \/ -. E. z ( z e. A /\ z e. B ) ) <-> ( -. E. z ( z e. A /\ z e. B ) \/ A = B ) )
2 df-in 
 |-  ( A i^i B ) = { z | ( z e. A /\ z e. B ) }
3 2 neeq1i 
 |-  ( ( A i^i B ) =/= (/) <-> { z | ( z e. A /\ z e. B ) } =/= (/) )
4 abn0 
 |-  ( { z | ( z e. A /\ z e. B ) } =/= (/) <-> E. z ( z e. A /\ z e. B ) )
5 3 4 bitr2i 
 |-  ( E. z ( z e. A /\ z e. B ) <-> ( A i^i B ) =/= (/) )
6 5 necon2bbii 
 |-  ( ( A i^i B ) = (/) <-> -. E. z ( z e. A /\ z e. B ) )
7 6 orbi2i 
 |-  ( ( A = B \/ ( A i^i B ) = (/) ) <-> ( A = B \/ -. E. z ( z e. A /\ z e. B ) ) )
8 imor 
 |-  ( ( E. z ( z e. A /\ z e. B ) -> A = B ) <-> ( -. E. z ( z e. A /\ z e. B ) \/ A = B ) )
9 1 7 8 3bitr4ri 
 |-  ( ( E. z ( z e. A /\ z e. B ) -> A = B ) <-> ( A = B \/ ( A i^i B ) = (/) ) )