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Theorem recseq 5862
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 5120 . . . . . . . 8 (𝐹 = 𝐺 → (𝐹‘(𝑎𝑐)) = (𝐺‘(𝑎𝑐)))
21eqeq2d 2048 . . . . . . 7 (𝐹 = 𝐺 → ((𝑎𝑐) = (𝐹‘(𝑎𝑐)) ↔ (𝑎𝑐) = (𝐺‘(𝑎𝑐))))
32ralbidv 2320 . . . . . 6 (𝐹 = 𝐺 → (𝑐 𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)) ↔ 𝑐 𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐))))
43anbi2d 437 . . . . 5 (𝐹 = 𝐺 → ((𝑎 Fn 𝑏 𝑐 𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐))) ↔ (𝑎 Fn 𝑏 𝑐 𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))))
54rexbidv 2321 . . . 4 (𝐹 = 𝐺 → (𝑏 On (𝑎 Fn 𝑏 𝑐 𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐))) ↔ 𝑏 On (𝑎 Fn 𝑏 𝑐 𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))))
65abbidv 2152 . . 3 (𝐹 = 𝐺 → {𝑎𝑏 On (𝑎 Fn 𝑏 𝑐 𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))} = {𝑎𝑏 On (𝑎 Fn 𝑏 𝑐 𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))})
76unieqd 3582 . 2 (𝐹 = 𝐺 {𝑎𝑏 On (𝑎 Fn 𝑏 𝑐 𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))} = {𝑎𝑏 On (𝑎 Fn 𝑏 𝑐 𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))})
8 df-recs 5861 . 2 recs(𝐹) = {𝑎𝑏 On (𝑎 Fn 𝑏 𝑐 𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))}
9 df-recs 5861 . 2 recs(𝐺) = {𝑎𝑏 On (𝑎 Fn 𝑏 𝑐 𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))}
107, 8, 93eqtr4g 2094 1 (𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97   = wceq 1242  {cab 2023  wral 2300  wrex 2301   cuni 3571  Oncon0 4066  cres 4290   Fn wfn 4840  cfv 4845  recscrecs 5860
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ral 2305  df-rex 2306  df-uni 3572  df-br 3756  df-iota 4810  df-fv 4853  df-recs 5861
This theorem is referenced by:  rdgeq1  5898  rdgeq2  5899  freceq1  5919  freceq2  5920  frecsuclem1  5926  frecsuclem2  5928
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