Theorem raleqbidvv

Description: Version of raleqbidv with additional disjoint variable conditions, not requiring ax-8 nor df-clel . (Contributed by BJ, 22-Sep-2024)

	Ref	Expression
Hypotheses	raleqbidvv.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
	raleqbidvv.2	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
Assertion	raleqbidvv	⊢ ( 𝜑 → ( ∀ 𝑥 ∈ 𝐴 𝜓 ↔ ∀ 𝑥 ∈ 𝐵 𝜒 ) )

Proof

Step	Hyp	Ref	Expression
1		raleqbidvv.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		raleqbidvv.2	⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
3	2	alrimiv	⊢ ( 𝜑 → ∀ 𝑥 ( 𝜓 ↔ 𝜒 ) )
4		dfcleq	⊢ ( 𝐴 = 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
5	1 4	sylib	⊢ ( 𝜑 → ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) )
6		19.26	⊢ ( ∀ 𝑥 ( ( 𝜓 ↔ 𝜒 ) ∧ ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) ↔ ( ∀ 𝑥 ( 𝜓 ↔ 𝜒 ) ∧ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) )
7	3 5 6	sylanbrc	⊢ ( 𝜑 → ∀ 𝑥 ( ( 𝜓 ↔ 𝜒 ) ∧ ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) )
8		imbi12	⊢ ( ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) → ( ( 𝜓 ↔ 𝜒 ) → ( ( 𝑥 ∈ 𝐴 → 𝜓 ) ↔ ( 𝑥 ∈ 𝐵 → 𝜒 ) ) ) )
9	8	impcom	⊢ ( ( ( 𝜓 ↔ 𝜒 ) ∧ ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵 ) ) → ( ( 𝑥 ∈ 𝐴 → 𝜓 ) ↔ ( 𝑥 ∈ 𝐵 → 𝜒 ) ) )
10	7 9	sylg	⊢ ( 𝜑 → ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 → 𝜓 ) ↔ ( 𝑥 ∈ 𝐵 → 𝜒 ) ) )
11		albi	⊢ ( ∀ 𝑥 ( ( 𝑥 ∈ 𝐴 → 𝜓 ) ↔ ( 𝑥 ∈ 𝐵 → 𝜒 ) ) → ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜓 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝜒 ) ) )
12	10 11	syl	⊢ ( 𝜑 → ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜓 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝜒 ) ) )
13		df-ral	⊢ ( ∀ 𝑥 ∈ 𝐴 𝜓 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝜓 ) )
14		df-ral	⊢ ( ∀ 𝑥 ∈ 𝐵 𝜒 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐵 → 𝜒 ) )
15	12 13 14	3bitr4g	⊢ ( 𝜑 → ( ∀ 𝑥 ∈ 𝐴 𝜓 ↔ ∀ 𝑥 ∈ 𝐵 𝜒 ) )