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Theorem recseq 6422
Description: Equality theorem for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
Assertion
Ref Expression
recseq (𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))

Proof of Theorem recseq
Dummy variables 𝑎 𝑏 𝑐 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 fveq1 5602 . . . . . . . 8 (𝐹 = 𝐺 → (𝐹‘(𝑎𝑐)) = (𝐺‘(𝑎𝑐)))
21eqeq2d 2221 . . . . . . 7 (𝐹 = 𝐺 → ((𝑎𝑐) = (𝐹‘(𝑎𝑐)) ↔ (𝑎𝑐) = (𝐺‘(𝑎𝑐))))
32ralbidv 2510 . . . . . 6 (𝐹 = 𝐺 → (∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)) ↔ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐))))
43anbi2d 464 . . . . 5 (𝐹 = 𝐺 → ((𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐))) ↔ (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))))
54rexbidv 2511 . . . 4 (𝐹 = 𝐺 → (∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐))) ↔ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))))
65abbidv 2327 . . 3 (𝐹 = 𝐺 → {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))} = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))})
76unieqd 3878 . 2 (𝐹 = 𝐺 {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))} = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))})
8 df-recs 6421 . 2 recs(𝐹) = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))}
9 df-recs 6421 . 2 recs(𝐺) = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))}
107, 8, 93eqtr4g 2267 1 (𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104   = wceq 1375  {cab 2195  wral 2488  wrex 2489   cuni 3867  Oncon0 4431  cres 4698   Fn wfn 5289  cfv 5294  recscrecs 6420
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 713  ax-5 1473  ax-7 1474  ax-gen 1475  ax-ie1 1519  ax-ie2 1520  ax-8 1530  ax-10 1531  ax-11 1532  ax-i12 1533  ax-bndl 1535  ax-4 1536  ax-17 1552  ax-i9 1556  ax-ial 1560  ax-i5r 1561  ax-ext 2191
This theorem depends on definitions:  df-bi 117  df-tru 1378  df-nf 1487  df-sb 1789  df-clab 2196  df-cleq 2202  df-clel 2205  df-nfc 2341  df-ral 2493  df-rex 2494  df-uni 3868  df-br 4063  df-iota 5254  df-fv 5302  df-recs 6421
This theorem is referenced by:  rdgeq1  6487  rdgeq2  6488  freceq1  6508  freceq2  6509  frecsuclem  6522
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