Theorem eldisjeqd

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

	Ref	Expression
Hypothesis	eldisjeqd.1	$$⊢\phi \to A=B$$
Assertion	eldisjeqd	$$⊢\phi \to \left(ElDisjA\leftrightarrow ElDisjB\right)$$

Proof

Step	Hyp	Ref	Expression
1		eldisjeqd.1	$$⊢\phi \to A=B$$
2		eldisjeq	$$⊢A=B\to \left(ElDisjA\leftrightarrow ElDisjB\right)$$
3	1 2	syl	$$⊢\phi \to \left(ElDisjA\leftrightarrow ElDisjB\right)$$