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Theorem qsinxp 8800
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 8799 . . . . 5 (((𝑅𝐴) ⊆ 𝐴𝑥𝐴) → [𝑥]𝑅 = [𝑥](𝑅 ∩ (𝐴 × 𝐴)))
21eqeq2d 2777 . . . 4 (((𝑅𝐴) ⊆ 𝐴𝑥𝐴) → (𝑦 = [𝑥]𝑅𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))))
32rexbidva 3190 . . 3 ((𝑅𝐴) ⊆ 𝐴 → (∃𝑥𝐴 𝑦 = [𝑥]𝑅 ↔ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))))
43abbidv 2832 . 2 ((𝑅𝐴) ⊆ 𝐴 → {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝑅} = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))})
5 df-qs 8709 . 2 (𝐴 / 𝑅) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝑅}
6 df-qs 8709 . 2 (𝐴 / (𝑅 ∩ (𝐴 × 𝐴))) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))}
74, 5, 63eqtr4g 2826 1 ((𝑅𝐴) ⊆ 𝐴 → (𝐴 / 𝑅) = (𝐴 / (𝑅 ∩ (𝐴 × 𝐴))))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wa 401   = wceq 1570  wcel 2146  {cab 2744  wrex 3092  cin 3907  wss 3908   × cxp 5664  cima 5669  [cec 8701   / cqs 8702
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 2148  ax-9 2156  ax-11 2195  ax-ext 2738  ax-sep 5262  ax-pr 5409
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-br 5115  df-opab 5179  df-xp 5672  df-rel 5673  df-cnv 5674  df-dm 5676  df-rn 5677  df-res 5678  df-ima 5679  df-ec 8705  df-qs 8709
This theorem is used by:  pi1buni  25236  pi1bas3  25239
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