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Theorem eqelbid 32676
Description: A variable elimination law for equality within a given set 𝐴. See equvel 2489. (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 2768 . . . . 5 (𝑥 = 𝐵 → (𝑥 = 𝐵𝐵 = 𝐵))
2 eqeq1 2768 . . . . 5 (𝑥 = 𝐵 → (𝑥 = 𝐶𝐵 = 𝐶))
31, 2bibi12d 347 . . . 4 (𝑥 = 𝐵 → ((𝑥 = 𝐵𝑥 = 𝐶) ↔ (𝐵 = 𝐵𝐵 = 𝐶)))
4 eqid 2764 . . . . . 6 𝐵 = 𝐵
54tbt 371 . . . . 5 (𝐵 = 𝐶 ↔ (𝐵 = 𝐶𝐵 = 𝐵))
6 bicom 224 . . . . 5 ((𝐵 = 𝐶𝐵 = 𝐵) ↔ (𝐵 = 𝐵𝐵 = 𝐶))
75, 6bitri 277 . . . 4 (𝐵 = 𝐶 ↔ (𝐵 = 𝐵𝐵 = 𝐶))
83, 7bitr4di 291 . . 3 (𝑥 = 𝐵 → ((𝑥 = 𝐵𝑥 = 𝐶) ↔ 𝐵 = 𝐶))
9 simpr 488 . . 3 ((𝜑 ∧ ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶)) → ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶))
10 eqelbid.1 . . . 4 (𝜑𝐵𝐴)
1110adantr 484 . . 3 ((𝜑 ∧ ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶)) → 𝐵𝐴)
128, 9, 11rspcdva 3584 . 2 ((𝜑 ∧ ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶)) → 𝐵 = 𝐶)
13 simplr 778 . . . 4 (((𝜑𝐵 = 𝐶) ∧ 𝑥𝐴) → 𝐵 = 𝐶)
1413eqeq2d 2775 . . 3 (((𝜑𝐵 = 𝐶) ∧ 𝑥𝐴) → (𝑥 = 𝐵𝑥 = 𝐶))
1514ralrimiva 3156 . 2 ((𝜑𝐵 = 𝐶) → ∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶))
1612, 15impbida 810 1 (𝜑 → (∀𝑥𝐴 (𝑥 = 𝐵𝑥 = 𝐶) ↔ 𝐵 = 𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 208  wa 399   = wceq 1562  wcel 2144  wral 3078
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-8 2146  ax-9 2154  ax-ext 2736
This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1565  df-ex 1802  df-sb 2093  df-clab 2743  df-cleq 2756  df-clel 2839  df-ral 3079
This theorem is referenced by:  ply1moneq  33786
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