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Theorem recsfval 6220
Description: Lemma for transfinite recursion. The definition recs is the union of all acceptable functions. (Contributed by Mario Carneiro, 9-May-2015.)
Hypothesis
Ref Expression
tfrlem.1 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}
Assertion
Ref Expression
recsfval recs(𝐹) = 𝐴
Distinct variable group:   𝑥,𝑓,𝑦,𝐹
Allowed substitution hints:   𝐴(𝑥,𝑦,𝑓)

Proof of Theorem recsfval
StepHypRef Expression
1 df-recs 6210 . 2 recs(𝐹) = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}
2 tfrlem.1 . . 3 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}
32unieqi 3754 . 2 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}
41, 3eqtr4i 2164 1 recs(𝐹) = 𝐴
Colors of variables: wff set class
Syntax hints:  wa 103   = wceq 1332  {cab 2126  wral 2417  wrex 2418   cuni 3744  Oncon0 4293  cres 4549   Fn wfn 5126  cfv 5131  recscrecs 6209
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-uni 3745  df-recs 6210
This theorem is referenced by:  tfrlem6  6221  tfrlem7  6222  tfrlem8  6223  tfrlem9  6224  tfrlemibfn  6233  tfrlemiubacc  6235  tfrlemi14d  6238  tfrexlem  6239
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