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Theorem eqelbid 33140
Description: A variable elimination law for equality within a given set 𝐴. See equvel 2488. (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
StepHypRef Expression
1 eqeq1 2767 . . . . 5 (𝑥 = 𝐵 → (𝑥 = 𝐵 ↔ 𝐵 = 𝐵))
2 eqeq1 2767 . . . . 5 (𝑥 = 𝐵 → (𝑥 = 𝐶 ↔ 𝐵 = 𝐶))
31, 2bibi12d 348 . . . 4 (𝑥 = 𝐵 → ((𝑥 = 𝐵 ↔ 𝑥 = 𝐶) ↔ (𝐵 = 𝐵 ↔ 𝐵 = 𝐶)))
4 eqid 2763 . . . . . 6 𝐵 = 𝐵
54tbt 372 . . . . 5 (𝐵 = 𝐶 ↔ (𝐵 = 𝐶 ↔ 𝐵 = 𝐵))
6 bicom 225 . . . . 5 ((𝐵 = 𝐶 ↔ 𝐵 = 𝐵) ↔ (𝐵 = 𝐵 ↔ 𝐵 = 𝐶))
75, 6bitri 278 . . . 4 (𝐵 = 𝐶 ↔ (𝐵 = 𝐵 ↔ 𝐵 = 𝐶))
83, 7bitr4di 292 . . 3 (𝑥 = 𝐵 → ((𝑥 = 𝐵 ↔ 𝑥 = 𝐶) ↔ 𝐵 = 𝐶))
9 simpr 490 . . 3 ((𝜑 ∧ ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶)) → ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶))
10 eqelbid.1 . . . 4 (𝜑 → 𝐵 ∈ 𝐴)
1110adantr 486 . . 3 ((𝜑 ∧ ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶)) → 𝐵 ∈ 𝐴)
128, 9, 11rspcdva 3581 . 2 ((𝜑 ∧ ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶)) → 𝐵 = 𝐶)
13 simplr 781 . . . 4 (((𝜑 ∧ 𝐵 = 𝐶) ∧ 𝑥 ∈ 𝐴) → 𝐵 = 𝐶)
1413eqeq2d 2774 . . 3 (((𝜑 ∧ 𝐵 = 𝐶) ∧ 𝑥 ∈ 𝐴) → (𝑥 = 𝐵 ↔ 𝑥 = 𝐶))
1514ralrimiva 3158 . 2 ((𝜑 ∧ 𝐵 = 𝐶) → ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶))
1612, 15impbida 813 1 (𝜑 → (∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 ↔ 𝑥 = 𝐶) ↔ 𝐵 = 𝐶))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2147  ∀wral 3080
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081
This theorem is used by:  ply1moneq  34187
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