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Theorem qsinxp 8769
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 8768 . . . . 5 (((𝑅𝐴) ⊆ 𝐴𝑥𝐴) → [𝑥]𝑅 = [𝑥](𝑅 ∩ (𝐴 × 𝐴)))
21eqeq2d 2741 . . . 4 (((𝑅𝐴) ⊆ 𝐴𝑥𝐴) → (𝑦 = [𝑥]𝑅𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))))
32rexbidva 3156 . . 3 ((𝑅𝐴) ⊆ 𝐴 → (∃𝑥𝐴 𝑦 = [𝑥]𝑅 ↔ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))))
43abbidv 2796 . 2 ((𝑅𝐴) ⊆ 𝐴 → {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝑅} = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))})
5 df-qs 8680 . 2 (𝐴 / 𝑅) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝑅}
6 df-qs 8680 . 2 (𝐴 / (𝑅 ∩ (𝐴 × 𝐴))) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥](𝑅 ∩ (𝐴 × 𝐴))}
74, 5, 63eqtr4g 2790 1 ((𝑅𝐴) ⊆ 𝐴 → (𝐴 / 𝑅) = (𝐴 / (𝑅 ∩ (𝐴 × 𝐴))))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1540  wcel 2109  {cab 2708  wrex 3054  cin 3916  wss 3917   × cxp 5639  cima 5644  [cec 8672   / cqs 8673
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 2008  ax-8 2111  ax-9 2119  ax-11 2158  ax-ext 2702  ax-sep 5254  ax-nul 5264  ax-pr 5390
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-ral 3046  df-rex 3055  df-rab 3409  df-v 3452  df-dif 3920  df-un 3922  df-in 3924  df-ss 3934  df-nul 4300  df-if 4492  df-sn 4593  df-pr 4595  df-op 4599  df-br 5111  df-opab 5173  df-xp 5647  df-rel 5648  df-cnv 5649  df-dm 5651  df-rn 5652  df-res 5653  df-ima 5654  df-ec 8676  df-qs 8680
This theorem is referenced by:  pi1buni  24947  pi1bas3  24950
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