Theorem eldisjeqd

Description: Equality theorem for disjoint elementhood, deduction version. (Contributed by Peter Mazsa, 23-Sep-2021)

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

Proof

Step	Hyp	Ref	Expression
1		eldisjeqd.1	|- ( ph -> A = B )
2		eldisjeq	|- ( A = B -> ( ElDisj A <-> ElDisj B ) )
3	1 2	syl	|- ( ph -> ( ElDisj A <-> ElDisj B ) )