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Theorem qseq1 6597
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 2684 . . 3 (𝐴 = 𝐵 → (∃𝑥𝐴 𝑦 = [𝑥]𝐶 ↔ ∃𝑥𝐵 𝑦 = [𝑥]𝐶))
21abbidv 2305 . 2 (𝐴 = 𝐵 → {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝐶} = {𝑦 ∣ ∃𝑥𝐵 𝑦 = [𝑥]𝐶})
3 df-qs 6555 . 2 (𝐴 / 𝐶) = {𝑦 ∣ ∃𝑥𝐴 𝑦 = [𝑥]𝐶}
4 df-qs 6555 . 2 (𝐵 / 𝐶) = {𝑦 ∣ ∃𝑥𝐵 𝑦 = [𝑥]𝐶}
52, 3, 43eqtr4g 2245 1 (𝐴 = 𝐵 → (𝐴 / 𝐶) = (𝐵 / 𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1363  {cab 2173  wrex 2466  [cec 6547   / cqs 6548
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 710  ax-5 1457  ax-7 1458  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-10 1515  ax-11 1516  ax-i12 1517  ax-bndl 1519  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544  ax-i5r 1545  ax-ext 2169
This theorem depends on definitions:  df-bi 117  df-tru 1366  df-nf 1471  df-sb 1773  df-clab 2174  df-cleq 2180  df-clel 2183  df-nfc 2318  df-rex 2471  df-qs 6555
This theorem is referenced by: (None)
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