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Theorem qsinxp 8833
Description: Restrict the equivalence relation in a quotient set to the base set. (Contributed by Mario Carneiro, 23-Feb-2015.)
Assertion
Ref Expression
qsinxp ((𝑅𝐴) ⊆ 𝐴 → (𝐴 / 𝑅) = (𝐴 / (𝑅 ∩ (𝐴 × 𝐴))))

Proof of Theorem qsinxp
Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 ecinxp 8832 . . . . 5 (((𝑅𝐴) ⊆ 𝐴𝑥𝐴) → [𝑥]𝑅 = [𝑥](𝑅 ∩ (𝐴 × 𝐴)))
21eqeq2d 2748 . . . 4 (((𝑅𝐴) ⊆ 𝐴𝑥𝐴) → (𝑦 = [𝑥]𝑅𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))))
32rexbidva 3177 . . 3 ((𝑅𝐴) ⊆ 𝐴 → (∃𝑥𝐴 𝑦 = [𝑥]𝑅 ↔ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))))
43abbidv 2808 . 2 ((𝑅𝐴) ⊆ 𝐴 → {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝑅} = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))})
5 df-qs 8751 . 2 (𝐴 / 𝑅) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝑅}
6 df-qs 8751 . 2 (𝐴 / (𝑅 ∩ (𝐴 × 𝐴))) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))}
74, 5, 63eqtr4g 2802 1 ((𝑅𝐴) ⊆ 𝐴 → (𝐴 / 𝑅) = (𝐴 / (𝑅 ∩ (𝐴 × 𝐴))))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1540  wcel 2108  {cab 2714  wrex 3070  cin 3950  wss 3951   × cxp 5683  cima 5688  [cec 8743   / cqs 8744
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-11 2157  ax-ext 2708  ax-sep 5296  ax-nul 5306  ax-pr 5432
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-ral 3062  df-rex 3071  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-in 3958  df-ss 3968  df-nul 4334  df-if 4526  df-sn 4627  df-pr 4629  df-op 4633  df-br 5144  df-opab 5206  df-xp 5691  df-rel 5692  df-cnv 5693  df-dm 5695  df-rn 5696  df-res 5697  df-ima 5698  df-ec 8747  df-qs 8751
This theorem is referenced by:  pi1buni  25073  pi1bas3  25076
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