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Theorem eqelbid 32150
Description: A variable elimination law for equality within a given set 𝐴. See equvel 2447. (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 2728 . . . . 5 (𝑥 = 𝐵 → (𝑥 = 𝐵𝐵 = 𝐵))
2 eqeq1 2728 . . . . 5 (𝑥 = 𝐵 → (𝑥 = 𝐶𝐵 = 𝐶))
31, 2bibi12d 345 . . . 4 (𝑥 = 𝐵 → ((𝑥 = 𝐵𝑥 = 𝐶) ↔ (𝐵 = 𝐵𝐵 = 𝐶)))
4 eqid 2724 . . . . . 6 𝐵 = 𝐵
54tbt 369 . . . . 5 (𝐵 = 𝐶 ↔ (𝐵 = 𝐶𝐵 = 𝐵))
6 bicom 221 . . . . 5 ((𝐵 = 𝐶𝐵 = 𝐵) ↔ (𝐵 = 𝐵𝐵 = 𝐶))
75, 6bitri 275 . . . 4 (𝐵 = 𝐶 ↔ (𝐵 = 𝐵𝐵 = 𝐶))
83, 7bitr4di 289 . . 3 (𝑥 = 𝐵 → ((𝑥 = 𝐵𝑥 = 𝐶) ↔ 𝐵 = 𝐶))
9 simpr 484 . . 3 ((𝜑 ∧ ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶)) → ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶))
10 eqelbid.1 . . . 4 (𝜑𝐵𝐴)
1110adantr 480 . . 3 ((𝜑 ∧ ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶)) → 𝐵𝐴)
128, 9, 11rspcdva 3605 . 2 ((𝜑 ∧ ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶)) → 𝐵 = 𝐶)
13 simplr 766 . . . 4 (((𝜑𝐵 = 𝐶) ∧ 𝑥𝐴) → 𝐵 = 𝐶)
1413eqeq2d 2735 . . 3 (((𝜑𝐵 = 𝐶) ∧ 𝑥𝐴) → (𝑥 = 𝐵𝑥 = 𝐶))
1514ralrimiva 3138 . 2 ((𝜑𝐵 = 𝐶) → ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶))
1612, 15impbida 798 1 (𝜑 → (∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶) ↔ 𝐵 = 𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wa 395   = wceq 1533  wcel 2098  wral 3053
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2695
This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1536  df-ex 1774  df-sb 2060  df-clab 2702  df-cleq 2716  df-clel 2802  df-ral 3054
This theorem is referenced by:  ply1moneq  33096
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