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Theorem eqelbid 32936
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 (𝜑𝐵𝐴)
eqelbid.2 (𝜑𝐶𝐴)
Assertion
Ref Expression
eqelbid (𝜑 → (∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶) ↔ 𝐵 = 𝐶))
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝑥,𝐶   𝜑,𝑥

Proof of Theorem eqelbid
StepHypRef Expression
1 eqeq1 2766 . . . . 5 (𝑥 = 𝐵 → (𝑥 = 𝐵𝐵 = 𝐵))
2 eqeq1 2766 . . . . 5 (𝑥 = 𝐵 → (𝑥 = 𝐶𝐵 = 𝐶))
31, 2bibi12d 348 . . . 4 (𝑥 = 𝐵 → ((𝑥 = 𝐵𝑥 = 𝐶) ↔ (𝐵 = 𝐵𝐵 = 𝐶)))
4 eqid 2762 . . . . . 6 𝐵 = 𝐵
54tbt 372 . . . . 5 (𝐵 = 𝐶 ↔ (𝐵 = 𝐶𝐵 = 𝐵))
6 bicom 225 . . . . 5 ((𝐵 = 𝐶𝐵 = 𝐵) ↔ (𝐵 = 𝐵𝐵 = 𝐶))
75, 6bitri 278 . . . 4 (𝐵 = 𝐶 ↔ (𝐵 = 𝐵𝐵 = 𝐶))
83, 7bitr4di 292 . . 3 (𝑥 = 𝐵 → ((𝑥 = 𝐵𝑥 = 𝐶) ↔ 𝐵 = 𝐶))
9 simpr 490 . . 3 ((𝜑 ∧ ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶)) → ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶))
10 eqelbid.1 . . . 4 (𝜑𝐵𝐴)
1110adantr 486 . . 3 ((𝜑 ∧ ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶)) → 𝐵𝐴)
128, 9, 11rspcdva 3580 . 2 ((𝜑 ∧ ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶)) → 𝐵 = 𝐶)
13 simplr 781 . . . 4 (((𝜑𝐵 = 𝐶) ∧ 𝑥𝐴) → 𝐵 = 𝐶)
1413eqeq2d 2773 . . 3 (((𝜑𝐵 = 𝐶) ∧ 𝑥𝐴) → (𝑥 = 𝐵𝑥 = 𝐶))
1514ralrimiva 3156 . 2 ((𝜑𝐵 = 𝐶) → ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶))
1612, 15impbida 813 1 (𝜑 → (∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶) ↔ 𝐵 = 𝐶))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wb 209  wa 401   = wceq 1570  wcel 2145  wral 3078
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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3079
This theorem is used by:  ply1moneq  33985
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