Theorem eldisjeqi

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

	Ref	Expression
Hypothesis	eldisjeqi.1	$$⊢A=B$$
Assertion	eldisjeqi	$$⊢ElDisjA\leftrightarrow ElDisjB$$

Proof

Step	Hyp	Ref	Expression
1		eldisjeqi.1	$$⊢A=B$$
2		eldisjeq	$$⊢A=B\to \left(ElDisjA\leftrightarrow ElDisjB\right)$$
3	1 2	ax-mp	$$⊢ElDisjA\leftrightarrow ElDisjB$$