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Theorem qseq1 6220
Description: Equality theorem for quotient set. (Contributed by NM, 23-Jul-1995.)
Assertion
Ref Expression
qseq1 (𝐴 = 𝐵 → (𝐴 / 𝐶) = (𝐵 / 𝐶))

Proof of Theorem qseq1
Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 rexeq 2551 . . 3 (𝐴 = 𝐵 → (∃𝑥𝐴 𝑦 = [𝑥]𝐶 ↔ ∃𝑥𝐵 𝑦 = [𝑥]𝐶))
21abbidv 2197 . 2 (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝐶} = {𝑦 ∣ ∃𝑥𝐵 𝑦 = [𝑥]𝐶})
3 df-qs 6178 . 2 (𝐴 / 𝐶) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝐶}
4 df-qs 6178 . 2 (𝐵 / 𝐶) = {𝑦 ∣ ∃𝑥𝐵 𝑦 = [𝑥]𝐶}
52, 3, 43eqtr4g 2139 1 (𝐴 = 𝐵 → (𝐴 / 𝐶) = (𝐵 / 𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1285  {cab 2068  wrex 2350  [cec 6170   / cqs 6171
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064
This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-nfc 2209  df-rex 2355  df-qs 6178
This theorem is referenced by: (None)
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