Metamath Proof Explorer

Theorem disjeqd

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

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

Proof

Step Hyp Ref Expression
1 disjeqd.1 
 |-  ( ph -> A = B )
2 disjeq 
 |-  ( A = B -> ( Disj A <-> Disj B ) )
3 1 2 syl 
 |-  ( ph -> ( Disj A <-> Disj B ) )