Theorem disjeqd

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

	Ref	Expression
Hypothesis	disjeqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	disjeqd	⊢ ( 𝜑 → ( Disj 𝐴 ↔ Disj 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		disjeqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		disjeq	⊢ ( 𝐴 = 𝐵 → ( Disj 𝐴 ↔ Disj 𝐵 ) )
3	1 2	syl	⊢ ( 𝜑 → ( Disj 𝐴 ↔ Disj 𝐵 ) )