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Theorem qseq1 6485
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 2630 . . 3 (𝐴 = 𝐵 → (∃𝑥𝐴 𝑦 = [𝑥]𝐶 ↔ ∃𝑥𝐵 𝑦 = [𝑥]𝐶))
21abbidv 2258 . 2 (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝐶} = {𝑦 ∣ ∃𝑥𝐵 𝑦 = [𝑥]𝐶})
3 df-qs 6443 . 2 (𝐴 / 𝐶) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝐶}
4 df-qs 6443 . 2 (𝐵 / 𝐶) = {𝑦 ∣ ∃𝑥𝐵 𝑦 = [𝑥]𝐶}
52, 3, 43eqtr4g 2198 1 (𝐴 = 𝐵 → (𝐴 / 𝐶) = (𝐵 / 𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1332  {cab 2126  wrex 2418  [cec 6435   / cqs 6436
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122
This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-rex 2423  df-qs 6443
This theorem is referenced by: (None)
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