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Theorem recseq 6335
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 5536 . . . . . . . 8 (𝐹 = 𝐺 → (𝐹‘(𝑎𝑐)) = (𝐺‘(𝑎𝑐)))
21eqeq2d 2201 . . . . . . 7 (𝐹 = 𝐺 → ((𝑎𝑐) = (𝐹‘(𝑎𝑐)) ↔ (𝑎𝑐) = (𝐺‘(𝑎𝑐))))
32ralbidv 2490 . . . . . 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 2491 . . . 4 (𝐹 = 𝐺 → (∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐))) ↔ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))))
65abbidv 2307 . . 3 (𝐹 = 𝐺 → {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))} = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))})
76unieqd 3838 . 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 6334 . 2 recs(𝐹) = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))}
9 df-recs 6334 . 2 recs(𝐺) = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))}
107, 8, 93eqtr4g 2247 1 (𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104   = wceq 1364  {cab 2175  wral 2468  wrex 2469   cuni 3827  Oncon0 4384  cres 4649   Fn wfn 5233  cfv 5238  recscrecs 6333
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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2171
This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2176  df-cleq 2182  df-clel 2185  df-nfc 2321  df-ral 2473  df-rex 2474  df-uni 3828  df-br 4022  df-iota 5199  df-fv 5246  df-recs 6334
This theorem is referenced by:  rdgeq1  6400  rdgeq2  6401  freceq1  6421  freceq2  6422  frecsuclem  6435
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