Theorem raleqbidvvOLD 3335

Description: Obsolete version of raleqbidvv 3334 as of 9-Mar-2025. (Contributed by BJ, 22-Sep-2024.) (New usage is discouraged.) (Proof modification is discouraged.)

Hypotheses

Ref	Expression
raleqbidvv.1	⊢ (𝜑 → 𝐴 = 𝐵)
raleqbidvv.2	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
raleqbidvvOLD	⊢ (𝜑 → (∀𝑥 ∈ 𝐴 𝜓 ↔ ∀𝑥 ∈ 𝐵 𝜒))

Distinct variable groups:   𝜑,𝑥   𝑥,𝐴   𝑥,𝐵

Allowed substitution hints:   𝜓(𝑥)   𝜒(𝑥)

Proof of Theorem raleqbidvvOLD

Step	Hyp	Ref	Expression
1		raleqbidvv.2	. . . . . 6 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
2	1	alrimiv 1927	. . . . 5 ⊢ (𝜑 → ∀𝑥(𝜓 ↔ 𝜒))
3		raleqbidvv.1	. . . . . 6 ⊢ (𝜑 → 𝐴 = 𝐵)
4		dfcleq 2730	. . . . . 6 ⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
5	3, 4	sylib 218	. . . . 5 ⊢ (𝜑 → ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
6		19.26 1870	. . . . 5 ⊢ (∀𝑥((𝜓 ↔ 𝜒) ∧ (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)) ↔ (∀𝑥(𝜓 ↔ 𝜒) ∧ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
7	2, 5, 6	sylanbrc 583	. . . 4 ⊢ (𝜑 → ∀𝑥((𝜓 ↔ 𝜒) ∧ (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)))
8		imbi12 346	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵) → ((𝜓 ↔ 𝜒) → ((𝑥 ∈ 𝐴 → 𝜓) ↔ (𝑥 ∈ 𝐵 → 𝜒))))
9	8	impcom 407	. . . 4 ⊢ (((𝜓 ↔ 𝜒) ∧ (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)) → ((𝑥 ∈ 𝐴 → 𝜓) ↔ (𝑥 ∈ 𝐵 → 𝜒)))
10	7, 9	sylg 1823	. . 3 ⊢ (𝜑 → ∀𝑥((𝑥 ∈ 𝐴 → 𝜓) ↔ (𝑥 ∈ 𝐵 → 𝜒)))
11		albi 1818	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐴 → 𝜓) ↔ (𝑥 ∈ 𝐵 → 𝜒)) → (∀𝑥(𝑥 ∈ 𝐴 → 𝜓) ↔ ∀𝑥(𝑥 ∈ 𝐵 → 𝜒)))
12	10, 11	syl 17	. 2 ⊢ (𝜑 → (∀𝑥(𝑥 ∈ 𝐴 → 𝜓) ↔ ∀𝑥(𝑥 ∈ 𝐵 → 𝜒)))
13		df-ral 3062	. 2 ⊢ (∀𝑥 ∈ 𝐴 𝜓 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜓))
14		df-ral 3062	. 2 ⊢ (∀𝑥 ∈ 𝐵 𝜒 ↔ ∀𝑥(𝑥 ∈ 𝐵 → 𝜒))
15	12, 13, 14	3bitr4g 314	1 ⊢ (𝜑 → (∀𝑥 ∈ 𝐴 𝜓 ↔ ∀𝑥 ∈ 𝐵 𝜒))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1538   = wceq 1540   ∈ wcel 2108  ∀wral 3061

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-9 2118  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-cleq 2729  df-ral 3062

This theorem is referenced by: (None)