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Theorem qsinxp 8832
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 8831 . . . . 5 (((𝑅𝐴) ⊆ 𝐴𝑥𝐴) → [𝑥]𝑅 = [𝑥](𝑅 ∩ (𝐴 × 𝐴)))
21eqeq2d 2746 . . . 4 (((𝑅𝐴) ⊆ 𝐴𝑥𝐴) → (𝑦 = [𝑥]𝑅𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))))
32rexbidva 3175 . . 3 ((𝑅𝐴) ⊆ 𝐴 → (∃𝑥𝐴 𝑦 = [𝑥]𝑅 ↔ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))))
43abbidv 2806 . 2 ((𝑅𝐴) ⊆ 𝐴 → {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝑅} = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))})
5 df-qs 8750 . 2 (𝐴 / 𝑅) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝑅}
6 df-qs 8750 . 2 (𝐴 / (𝑅 ∩ (𝐴 × 𝐴))) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))}
74, 5, 63eqtr4g 2800 1 ((𝑅𝐴) ⊆ 𝐴 → (𝐴 / 𝑅) = (𝐴 / (𝑅 ∩ (𝐴 × 𝐴))))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1537  wcel 2106  {cab 2712  wrex 3068  cin 3962  wss 3963   × cxp 5687  cima 5692  [cec 8742   / cqs 8743
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-11 2155  ax-ext 2706  ax-sep 5302  ax-nul 5312  ax-pr 5438
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ral 3060  df-rex 3069  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-in 3970  df-ss 3980  df-nul 4340  df-if 4532  df-sn 4632  df-pr 4634  df-op 4638  df-br 5149  df-opab 5211  df-xp 5695  df-rel 5696  df-cnv 5697  df-dm 5699  df-rn 5700  df-res 5701  df-ima 5702  df-ec 8746  df-qs 8750
This theorem is referenced by:  pi1buni  25087  pi1bas3  25090
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