Theorem rabeqd

Description: Deduction form of rabeq . Note that contrary to rabeq it has no disjoint variable condition. (Contributed by BJ, 27-Apr-2019)

	Ref	Expression
Hypotheses	rabeqd.nf	⊢ Ⅎ 𝑥 𝜑
	rabeqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	rabeqd	⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜓 } )

Proof

Step	Hyp	Ref	Expression
1		rabeqd.nf	⊢ Ⅎ 𝑥 𝜑
2		rabeqd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
3		eleq2	⊢ ( 𝐴 = 𝐵 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
4	3	anbi1d	⊢ ( 𝐴 = 𝐵 → ( ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ↔ ( 𝑥 ∈ 𝐵 ∧ 𝜓 ) ) )
5	2 4	syl	⊢ ( 𝜑 → ( ( 𝑥 ∈ 𝐴 ∧ 𝜓 ) ↔ ( 𝑥 ∈ 𝐵 ∧ 𝜓 ) ) )
6	1 5	rabbida4	⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜓 } )