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Theorem recseq 6301
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 5510 . . . . . . . 8 (𝐹 = 𝐺 → (𝐹‘(𝑎𝑐)) = (𝐺‘(𝑎𝑐)))
21eqeq2d 2189 . . . . . . 7 (𝐹 = 𝐺 → ((𝑎𝑐) = (𝐹‘(𝑎𝑐)) ↔ (𝑎𝑐) = (𝐺‘(𝑎𝑐))))
32ralbidv 2477 . . . . . 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 2478 . . . 4 (𝐹 = 𝐺 → (∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐))) ↔ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))))
65abbidv 2295 . . 3 (𝐹 = 𝐺 → {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))} = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))})
76unieqd 3818 . 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 6300 . 2 recs(𝐹) = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐹‘(𝑎𝑐)))}
9 df-recs 6300 . 2 recs(𝐺) = {𝑎 ∣ ∃𝑏 ∈ On (𝑎 Fn 𝑏 ∧ ∀𝑐𝑏 (𝑎𝑐) = (𝐺‘(𝑎𝑐)))}
107, 8, 93eqtr4g 2235 1 (𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104   = wceq 1353  {cab 2163  wral 2455  wrex 2456   cuni 3807  Oncon0 4360  cres 4625   Fn wfn 5207  cfv 5212  recscrecs 6299
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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-rex 2461  df-uni 3808  df-br 4001  df-iota 5174  df-fv 5220  df-recs 6300
This theorem is referenced by:  rdgeq1  6366  rdgeq2  6367  freceq1  6387  freceq2  6388  frecsuclem  6401
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