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Theorem recseq 6196
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 5413 . . . . . . . 8 (𝐹 = 𝐺 → (𝐹‘(𝑎𝑐)) = (𝐺‘(𝑎𝑐)))
21eqeq2d 2149 . . . . . . 7 (𝐹 = 𝐺 → ((𝑎𝑐) = (𝐹‘(𝑎𝑐)) ↔ (𝑎𝑐) = (𝐺‘(𝑎𝑐))))
32ralbidv 2435 . . . . . 6 (𝐹 = 𝐺 → (∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)) ↔ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐))))
43anbi2d 459 . . . . 5 (𝐹 = 𝐺 → ((𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐))) ↔ (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))))
54rexbidv 2436 . . . 4 (𝐹 = 𝐺 → (∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐))) ↔ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))))
65abbidv 2255 . . 3 (𝐹 = 𝐺 → {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))} = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))})
76unieqd 3742 . 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 6195 . 2 recs(𝐹) = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))}
9 df-recs 6195 . 2 recs(𝐺) = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))}
107, 8, 93eqtr4g 2195 1 (𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103   = wceq 1331  {cab 2123  wral 2414  wrex 2415   cuni 3731  Oncon0 4280  cres 4536   Fn wfn 5113  cfv 5118  recscrecs 6194
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 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119
This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-ral 2419  df-rex 2420  df-uni 3732  df-br 3925  df-iota 5083  df-fv 5126  df-recs 6195
This theorem is referenced by:  rdgeq1  6261  rdgeq2  6262  freceq1  6282  freceq2  6283  frecsuclem  6296
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