Theorem eqelbid 32832

Description: A variable elimination law for equality within a given set 𝐴. See equvel 2487. (Contributed by Thierry Arnoux, 20-Feb-2025.)

Hypotheses

Ref	Expression
eqelbid.1	⊢ (𝜑 → 𝐵 ∈ 𝐴)
eqelbid.2	⊢ (𝜑 → 𝐶 ∈ 𝐴)

Assertion

Ref	Expression
eqelbid	⊢ (𝜑 → (∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶) ↔ 𝐵 = 𝐶))

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

Proof of Theorem eqelbid

Step	Hyp	Ref	Expression
1		eqeq1 2766	. . . . 5 ⊢ (𝑥 = 𝐵 → (𝑥 = 𝐵 ↔ 𝐵 = 𝐵))
2		eqeq1 2766	. . . . 5 ⊢ (𝑥 = 𝐵 → (𝑥 = 𝐶 ↔ 𝐵 = 𝐶))
3	1, 2	bibi12d 348	. . . 4 ⊢ (𝑥 = 𝐵 → ((𝑥 = 𝐵 ↔ 𝑥 = 𝐶) ↔ (𝐵 = 𝐵 ↔ 𝐵 = 𝐶)))
4		eqid 2762	. . . . . 6 ⊢ 𝐵 = 𝐵
5	4	tbt 372	. . . . 5 ⊢ (𝐵 = 𝐶 ↔ (𝐵 = 𝐶 ↔ 𝐵 = 𝐵))
6		bicom 225	. . . . 5 ⊢ ((𝐵 = 𝐶 ↔ 𝐵 = 𝐵) ↔ (𝐵 = 𝐵 ↔ 𝐵 = 𝐶))
7	5, 6	bitri 278	. . . 4 ⊢ (𝐵 = 𝐶 ↔ (𝐵 = 𝐵 ↔ 𝐵 = 𝐶))
8	3, 7	bitr4di 292	. . 3 ⊢ (𝑥 = 𝐵 → ((𝑥 = 𝐵 ↔ 𝑥 = 𝐶) ↔ 𝐵 = 𝐶))
9		simpr 489	. . 3 ⊢ ((𝜑 ∧ ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶)) → ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶))
10		eqelbid.1	. . . 4 ⊢ (𝜑 → 𝐵 ∈ 𝐴)
11	10	adantr 485	. . 3 ⊢ ((𝜑 ∧ ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶)) → 𝐵 ∈ 𝐴)
12	8, 9, 11	rspcdva 3581	. 2 ⊢ ((𝜑 ∧ ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶)) → 𝐵 = 𝐶)
13		simplr 780	. . . 4 ⊢ (((𝜑 ∧ 𝐵 = 𝐶) ∧ 𝑥 ∈ 𝐴) → 𝐵 = 𝐶)
14	13	eqeq2d 2773	. . 3 ⊢ (((𝜑 ∧ 𝐵 = 𝐶) ∧ 𝑥 ∈ 𝐴) → (𝑥 = 𝐵 ↔ 𝑥 = 𝐶))
15	14	ralrimiva 3156	. 2 ⊢ ((𝜑 ∧ 𝐵 = 𝐶) → ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶))
16	12, 15	impbida 812	1 ⊢ (𝜑 → (∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶) ↔ 𝐵 = 𝐶))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1569   ∈ wcel 2142  ∀wral 3078

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3079

This theorem is used by:  ply1moneq  33887