Metamath Proof Explorer

Theorem disjeqi

Description: Equality theorem for disjoints, inference version. (Contributed by Peter Mazsa, 22-Sep-2021)

Ref Expression
Hypothesis disjeqi.1 
|- A = B
Assertion disjeqi 
|- ( Disj A <-> Disj B )

Proof

Step Hyp Ref Expression
1 disjeqi.1 
 |-  A = B
2 disjeq 
 |-  ( A = B -> ( Disj A <-> Disj B ) )
3 1 2 ax-mp 
 |-  ( Disj A <-> Disj B )