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Theorem recsfval 6559
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 6549 . 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 3929 . 2 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}
41, 3eqtr4i 2258 1 recs(𝐹) = 𝐴
Colors of variables: wff set class
Syntax hints:  wa 104   = wceq 1398  {cab 2220  wral 2522  wrex 2523   cuni 3919  Oncon0 4489  cres 4756   Fn wfn 5352  cfv 5357  recscrecs 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 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 2216
This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-rex 2528  df-uni 3920  df-recs 6549
This theorem is referenced by:  tfrlem6  6560  tfrlem7  6561  tfrlem8  6562  tfrlem9  6563  tfrlemibfn  6572  tfrlemiubacc  6574  tfrlemi14d  6577  tfrexlem  6578
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