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Theorem qseq1 6585
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 2674 . . 3 (𝐴 = 𝐵 → (∃𝑥𝐴 𝑦 = [𝑥]𝐶 ↔ ∃𝑥𝐵 𝑦 = [𝑥]𝐶))
21abbidv 2295 . 2 (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝐶} = {𝑦 ∣ ∃𝑥𝐵 𝑦 = [𝑥]𝐶})
3 df-qs 6543 . 2 (𝐴 / 𝐶) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝐶}
4 df-qs 6543 . 2 (𝐵 / 𝐶) = {𝑦 ∣ ∃𝑥𝐵 𝑦 = [𝑥]𝐶}
52, 3, 43eqtr4g 2235 1 (𝐴 = 𝐵 → (𝐴 / 𝐶) = (𝐵 / 𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1353  {cab 2163  wrex 2456  [cec 6535   / cqs 6536
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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-rex 2461  df-qs 6543
This theorem is referenced by: (None)
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