Theorem eldisjeqd

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

	Ref	Expression
Hypothesis	eldisjeqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	eldisjeqd	⊢ ( 𝜑 → ( ElDisj 𝐴 ↔ ElDisj 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		eldisjeqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		eldisjeq	⊢ ( 𝐴 = 𝐵 → ( ElDisj 𝐴 ↔ ElDisj 𝐵 ) )
3	1 2	syl	⊢ ( 𝜑 → ( ElDisj 𝐴 ↔ ElDisj 𝐵 ) )