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Theorem recseq 6285
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 5495 . . . . . . . 8 (𝐹 = 𝐺 → (𝐹‘(𝑎𝑐)) = (𝐺‘(𝑎𝑐)))
21eqeq2d 2182 . . . . . . 7 (𝐹 = 𝐺 → ((𝑎𝑐) = (𝐹‘(𝑎𝑐)) ↔ (𝑎𝑐) = (𝐺‘(𝑎𝑐))))
32ralbidv 2470 . . . . . 6 (𝐹 = 𝐺 → (∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)) ↔ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐))))
43anbi2d 461 . . . . 5 (𝐹 = 𝐺 → ((𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐))) ↔ (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))))
54rexbidv 2471 . . . 4 (𝐹 = 𝐺 → (∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐))) ↔ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))))
65abbidv 2288 . . 3 (𝐹 = 𝐺 → {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))} = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))})
76unieqd 3807 . 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 6284 . 2 recs(𝐹) = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))}
9 df-recs 6284 . 2 recs(𝐺) = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))}
107, 8, 93eqtr4g 2228 1 (𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103   = wceq 1348  {cab 2156  wral 2448  wrex 2449   cuni 3796  Oncon0 4348  cres 4613   Fn wfn 5193  cfv 5198  recscrecs 6283
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 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ral 2453  df-rex 2454  df-uni 3797  df-br 3990  df-iota 5160  df-fv 5206  df-recs 6284
This theorem is referenced by:  rdgeq1  6350  rdgeq2  6351  freceq1  6371  freceq2  6372  frecsuclem  6385
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