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Theorem qseq1 6816
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 2741 . . 3 (𝐴 = 𝐵 → (∃𝑥𝐴 𝑦 = [𝑥]𝐶 ↔ ∃𝑥𝐵 𝑦 = [𝑥]𝐶))
21abbidv 2352 . 2 (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝐶} = {𝑦 ∣ ∃𝑥𝐵 𝑦 = [𝑥]𝐶})
3 df-qs 6772 . 2 (𝐴 / 𝐶) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝐶}
4 df-qs 6772 . 2 (𝐵 / 𝐶) = {𝑦 ∣ ∃𝑥𝐵 𝑦 = [𝑥]𝐶}
52, 3, 43eqtr4g 2290 1 (𝐴 = 𝐵 → (𝐴 / 𝐶) = (𝐵 / 𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1398  {cab 2218  wrex 2521  [cec 6764   / cqs 6765
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2214
This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-rex 2526  df-qs 6772
This theorem is referenced by: (None)
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