HomeHome Intuitionistic Logic Explorer
Theorem List (p. 65 of 164)
< Previous  Next >
Bad symbols? Try the
GIF version.

Mirrors  >  Metamath Home Page  >  ILE Home Page  >  Theorem List Contents  >  Recent Proofs       This page: Page List

Theorem List for Intuitionistic Logic Explorer - 6401-6500   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Syntaxwsmo 6401 Introduce the strictly monotone ordinal function. A strictly monotone function is one that is constantly increasing across the ordinals.
wff Smo 𝐴
 
Definitiondf-smo 6402* Definition of a strictly monotone ordinal function. Definition 7.46 in [TakeutiZaring] p. 50. (Contributed by Andrew Salmon, 15-Nov-2011.)
(Smo 𝐴 ↔ (𝐴:dom 𝐴⟶On ∧ Ord dom 𝐴 ∧ ∀𝑥 ∈ dom 𝐴𝑦 ∈ dom 𝐴(𝑥𝑦 → (𝐴𝑥) ∈ (𝐴𝑦))))
 
Theoremdfsmo2 6403* Alternate definition of a strictly monotone ordinal function. (Contributed by Mario Carneiro, 4-Mar-2013.)
(Smo 𝐹 ↔ (𝐹:dom 𝐹⟶On ∧ Ord dom 𝐹 ∧ ∀𝑥 ∈ dom 𝐹𝑦𝑥 (𝐹𝑦) ∈ (𝐹𝑥)))
 
Theoremissmo 6404* Conditions for which 𝐴 is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 15-Nov-2011.)
𝐴:𝐵⟶On    &   Ord 𝐵    &   ((𝑥𝐵𝑦𝐵) → (𝑥𝑦 → (𝐴𝑥) ∈ (𝐴𝑦)))    &   dom 𝐴 = 𝐵       Smo 𝐴
 
Theoremissmo2 6405* Alternate definition of a strictly monotone ordinal function. (Contributed by Mario Carneiro, 12-Mar-2013.)
(𝐹:𝐴𝐵 → ((𝐵 ⊆ On ∧ Ord 𝐴 ∧ ∀𝑥𝐴𝑦𝑥 (𝐹𝑦) ∈ (𝐹𝑥)) → Smo 𝐹))
 
Theoremsmoeq 6406 Equality theorem for strictly monotone functions. (Contributed by Andrew Salmon, 16-Nov-2011.)
(𝐴 = 𝐵 → (Smo 𝐴 ↔ Smo 𝐵))
 
Theoremsmodm 6407 The domain of a strictly monotone function is an ordinal. (Contributed by Andrew Salmon, 16-Nov-2011.)
(Smo 𝐴 → Ord dom 𝐴)
 
Theoremsmores 6408 A strictly monotone function restricted to an ordinal remains strictly monotone. (Contributed by Andrew Salmon, 16-Nov-2011.) (Proof shortened by Mario Carneiro, 5-Dec-2016.)
((Smo 𝐴𝐵 ∈ dom 𝐴) → Smo (𝐴𝐵))
 
Theoremsmores3 6409 A strictly monotone function restricted to an ordinal remains strictly monotone. (Contributed by Andrew Salmon, 19-Nov-2011.)
((Smo (𝐴𝐵) ∧ 𝐶 ∈ (dom 𝐴𝐵) ∧ Ord 𝐵) → Smo (𝐴𝐶))
 
Theoremsmores2 6410 A strictly monotone ordinal function restricted to an ordinal is still monotone. (Contributed by Mario Carneiro, 15-Mar-2013.)
((Smo 𝐹 ∧ Ord 𝐴) → Smo (𝐹𝐴))
 
Theoremsmodm2 6411 The domain of a strictly monotone ordinal function is an ordinal. (Contributed by Mario Carneiro, 12-Mar-2013.)
((𝐹 Fn 𝐴 ∧ Smo 𝐹) → Ord 𝐴)
 
Theoremsmofvon2dm 6412 The function values of a strictly monotone ordinal function are ordinals. (Contributed by Mario Carneiro, 12-Mar-2013.)
((Smo 𝐹𝐵 ∈ dom 𝐹) → (𝐹𝐵) ∈ On)
 
Theoremiordsmo 6413 The identity relation restricted to the ordinals is a strictly monotone function. (Contributed by Andrew Salmon, 16-Nov-2011.)
Ord 𝐴       Smo ( I ↾ 𝐴)
 
Theoremsmo0 6414 The null set is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 20-Nov-2011.)
Smo ∅
 
Theoremsmofvon 6415 If 𝐵 is a strictly monotone ordinal function, and 𝐴 is in the domain of 𝐵, then the value of the function at 𝐴 is an ordinal. (Contributed by Andrew Salmon, 20-Nov-2011.)
((Smo 𝐵𝐴 ∈ dom 𝐵) → (𝐵𝐴) ∈ On)
 
Theoremsmoel 6416 If 𝑥 is less than 𝑦 then a strictly monotone function's value will be strictly less at 𝑥 than at 𝑦. (Contributed by Andrew Salmon, 22-Nov-2011.)
((Smo 𝐵𝐴 ∈ dom 𝐵𝐶𝐴) → (𝐵𝐶) ∈ (𝐵𝐴))
 
Theoremsmoiun 6417* The value of a strictly monotone ordinal function contains its indexed union. (Contributed by Andrew Salmon, 22-Nov-2011.)
((Smo 𝐵𝐴 ∈ dom 𝐵) → 𝑥𝐴 (𝐵𝑥) ⊆ (𝐵𝐴))
 
Theoremsmoiso 6418 If 𝐹 is an isomorphism from an ordinal 𝐴 onto 𝐵, which is a subset of the ordinals, then 𝐹 is a strictly monotonic function. Exercise 3 in [TakeutiZaring] p. 50. (Contributed by Andrew Salmon, 24-Nov-2011.)
((𝐹 Isom E , E (𝐴, 𝐵) ∧ Ord 𝐴𝐵 ⊆ On) → Smo 𝐹)
 
Theoremsmoel2 6419 A strictly monotone ordinal function preserves the epsilon relation. (Contributed by Mario Carneiro, 12-Mar-2013.)
(((𝐹 Fn 𝐴 ∧ Smo 𝐹) ∧ (𝐵𝐴𝐶𝐵)) → (𝐹𝐶) ∈ (𝐹𝐵))
 
2.6.20  "Strong" transfinite recursion
 
Syntaxcrecs 6420 Notation for a function defined by strong transfinite recursion.
class recs(𝐹)
 
Definitiondf-recs 6421* Define a function recs(𝐹) on On, the class of ordinal numbers, by transfinite recursion given a rule 𝐹 which sets the next value given all values so far. See df-irdg 6486 for more details on why this definition is desirable. Unlike df-irdg 6486 which restricts the update rule to use only the previous value, this version allows the update rule to use all previous values, which is why it is described as "strong", although it is actually more primitive. See tfri1d 6451 and tfri2d 6452 for the primary contract of this definition.

(Contributed by Stefan O'Rear, 18-Jan-2015.)

recs(𝐹) = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}
 
Theoremrecseq 6422 Equality theorem for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
(𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))
 
Theoremnfrecs 6423 Bound-variable hypothesis builder for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
𝑥𝐹       𝑥recs(𝐹)
 
Theoremtfrlem1 6424* A technical lemma for transfinite recursion. Compare Lemma 1 of [TakeutiZaring] p. 47. (Contributed by NM, 23-Mar-1995.) (Revised by Mario Carneiro, 24-May-2019.)
(𝜑𝐴 ∈ On)    &   (𝜑 → (Fun 𝐹𝐴 ⊆ dom 𝐹))    &   (𝜑 → (Fun 𝐺𝐴 ⊆ dom 𝐺))    &   (𝜑 → ∀𝑥𝐴 (𝐹𝑥) = (𝐵‘(𝐹𝑥)))    &   (𝜑 → ∀𝑥𝐴 (𝐺𝑥) = (𝐵‘(𝐺𝑥)))       (𝜑 → ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥))
 
Theoremtfrlem3ag 6425* Lemma for transfinite recursion. This lemma just changes some bound variables in 𝐴 for later use. (Contributed by Jim Kingdon, 5-Jul-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}       (𝐺 ∈ V → (𝐺𝐴 ↔ ∃𝑧 ∈ On (𝐺 Fn 𝑧 ∧ ∀𝑤𝑧 (𝐺𝑤) = (𝐹‘(𝐺𝑤)))))
 
Theoremtfrlem3a 6426* Lemma for transfinite recursion. Let 𝐴 be the class of "acceptable" functions. The final thing we're interested in is the union of all these acceptable functions. This lemma just changes some bound variables in 𝐴 for later use. (Contributed by NM, 9-Apr-1995.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   𝐺 ∈ V       (𝐺𝐴 ↔ ∃𝑧 ∈ On (𝐺 Fn 𝑧 ∧ ∀𝑤𝑧 (𝐺𝑤) = (𝐹‘(𝐺𝑤))))
 
Theoremtfrlem3 6427* Lemma for transfinite recursion. Let 𝐴 be the class of "acceptable" functions. The final thing we're interested in is the union of all these acceptable functions. This lemma just changes some bound variables in 𝐴 for later use. (Contributed by NM, 9-Apr-1995.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}       𝐴 = {𝑔 ∣ ∃𝑧 ∈ On (𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐹‘(𝑔𝑤)))}
 
Theoremtfrlem3-2d 6428* Lemma for transfinite recursion which changes a bound variable (Contributed by Jim Kingdon, 2-Jul-2019.)
(𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))       (𝜑 → (Fun 𝐹 ∧ (𝐹𝑔) ∈ V))
 
Theoremtfrlem4 6429* Lemma for transfinite recursion. 𝐴 is the class of all "acceptable" functions, and 𝐹 is their union. First we show that an acceptable function is in fact a function. (Contributed by NM, 9-Apr-1995.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}       (𝑔𝐴 → Fun 𝑔)
 
Theoremtfrlem5 6430* Lemma for transfinite recursion. The values of two acceptable functions are the same within their domains. (Contributed by NM, 9-Apr-1995.) (Revised by Mario Carneiro, 24-May-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}       ((𝑔𝐴𝐴) → ((𝑥𝑔𝑢𝑥𝑣) → 𝑢 = 𝑣))
 
Theoremrecsfval 6431* Lemma for transfinite recursion. The definition recs is the union of all acceptable functions. (Contributed by Mario Carneiro, 9-May-2015.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}       recs(𝐹) = 𝐴
 
Theoremtfrlem6 6432* Lemma for transfinite recursion. The union of all acceptable functions is a relation. (Contributed by NM, 8-Aug-1994.) (Revised by Mario Carneiro, 9-May-2015.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}       Rel recs(𝐹)
 
Theoremtfrlem7 6433* Lemma for transfinite recursion. The union of all acceptable functions is a function. (Contributed by NM, 9-Aug-1994.) (Revised by Mario Carneiro, 24-May-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}       Fun recs(𝐹)
 
Theoremtfrlem8 6434* Lemma for transfinite recursion. The domain of recs is ordinal. (Contributed by NM, 14-Aug-1994.) (Proof shortened by Alan Sare, 11-Mar-2008.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}       Ord dom recs(𝐹)
 
Theoremtfrlem9 6435* Lemma for transfinite recursion. Here we compute the value of recs (the union of all acceptable functions). (Contributed by NM, 17-Aug-1994.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}       (𝐵 ∈ dom recs(𝐹) → (recs(𝐹)‘𝐵) = (𝐹‘(recs(𝐹) ↾ 𝐵)))
 
Theoremtfrfun 6436 Transfinite recursion produces a function. (Contributed by Jim Kingdon, 20-Aug-2021.)
Fun recs(𝐹)
 
Theoremtfr2a 6437 A weak version of transfinite recursion. (Contributed by Mario Carneiro, 24-Jun-2015.)
𝐹 = recs(𝐺)       (𝐴 ∈ dom 𝐹 → (𝐹𝐴) = (𝐺‘(𝐹𝐴)))
 
Theoremtfr0dm 6438 Transfinite recursion is defined at the empty set. (Contributed by Jim Kingdon, 8-Mar-2022.)
𝐹 = recs(𝐺)       ((𝐺‘∅) ∈ 𝑉 → ∅ ∈ dom 𝐹)
 
Theoremtfr0 6439 Transfinite recursion at the empty set. (Contributed by Jim Kingdon, 8-May-2020.)
𝐹 = recs(𝐺)       ((𝐺‘∅) ∈ 𝑉 → (𝐹‘∅) = (𝐺‘∅))
 
Theoremtfrlemisucfn 6440* We can extend an acceptable function by one element to produce a function. Lemma for tfrlemi1 6448. (Contributed by Jim Kingdon, 2-Jul-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))    &   (𝜑𝑧 ∈ On)    &   (𝜑𝑔 Fn 𝑧)    &   (𝜑𝑔𝐴)       (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐹𝑔)⟩}) Fn suc 𝑧)
 
Theoremtfrlemisucaccv 6441* We can extend an acceptable function by one element to produce an acceptable function. Lemma for tfrlemi1 6448. (Contributed by Jim Kingdon, 4-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))    &   (𝜑𝑧 ∈ On)    &   (𝜑𝑔 Fn 𝑧)    &   (𝜑𝑔𝐴)       (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐹𝑔)⟩}) ∈ 𝐴)
 
Theoremtfrlemibacc 6442* Each element of 𝐵 is an acceptable function. Lemma for tfrlemi1 6448. (Contributed by Jim Kingdon, 14-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))    &   𝐵 = { ∣ ∃𝑧𝑥𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐹𝑔)⟩}))}    &   (𝜑𝑥 ∈ On)    &   (𝜑 → ∀𝑧𝑥𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐹‘(𝑔𝑤))))       (𝜑𝐵𝐴)
 
Theoremtfrlemibxssdm 6443* The union of 𝐵 is defined on all ordinals. Lemma for tfrlemi1 6448. (Contributed by Jim Kingdon, 18-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))    &   𝐵 = { ∣ ∃𝑧𝑥𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐹𝑔)⟩}))}    &   (𝜑𝑥 ∈ On)    &   (𝜑 → ∀𝑧𝑥𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐹‘(𝑔𝑤))))       (𝜑𝑥 ⊆ dom 𝐵)
 
Theoremtfrlemibfn 6444* The union of 𝐵 is a function defined on 𝑥. Lemma for tfrlemi1 6448. (Contributed by Jim Kingdon, 18-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))    &   𝐵 = { ∣ ∃𝑧𝑥𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐹𝑔)⟩}))}    &   (𝜑𝑥 ∈ On)    &   (𝜑 → ∀𝑧𝑥𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐹‘(𝑔𝑤))))       (𝜑 𝐵 Fn 𝑥)
 
Theoremtfrlemibex 6445* The set 𝐵 exists. Lemma for tfrlemi1 6448. (Contributed by Jim Kingdon, 17-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))    &   𝐵 = { ∣ ∃𝑧𝑥𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐹𝑔)⟩}))}    &   (𝜑𝑥 ∈ On)    &   (𝜑 → ∀𝑧𝑥𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐹‘(𝑔𝑤))))       (𝜑𝐵 ∈ V)
 
Theoremtfrlemiubacc 6446* The union of 𝐵 satisfies the recursion rule (lemma for tfrlemi1 6448). (Contributed by Jim Kingdon, 22-Apr-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))    &   𝐵 = { ∣ ∃𝑧𝑥𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐹𝑔)⟩}))}    &   (𝜑𝑥 ∈ On)    &   (𝜑 → ∀𝑧𝑥𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐹‘(𝑔𝑤))))       (𝜑 → ∀𝑢𝑥 ( 𝐵𝑢) = (𝐹‘( 𝐵𝑢)))
 
Theoremtfrlemiex 6447* Lemma for tfrlemi1 6448. (Contributed by Jim Kingdon, 18-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))    &   𝐵 = { ∣ ∃𝑧𝑥𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐹𝑔)⟩}))}    &   (𝜑𝑥 ∈ On)    &   (𝜑 → ∀𝑧𝑥𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐹‘(𝑔𝑤))))       (𝜑 → ∃𝑓(𝑓 Fn 𝑥 ∧ ∀𝑢𝑥 (𝑓𝑢) = (𝐹‘(𝑓𝑢))))
 
Theoremtfrlemi1 6448* We can define an acceptable function on any ordinal.

As with many of the transfinite recursion theorems, we have a hypothesis that states that 𝐹 is a function and that it is defined for all ordinals. (Contributed by Jim Kingdon, 4-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)

𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))       ((𝜑𝐶 ∈ On) → ∃𝑔(𝑔 Fn 𝐶 ∧ ∀𝑢𝐶 (𝑔𝑢) = (𝐹‘(𝑔𝑢))))
 
Theoremtfrlemi14d 6449* The domain of recs is all ordinals (lemma for transfinite recursion). (Contributed by Jim Kingdon, 9-Jul-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))       (𝜑 → dom recs(𝐹) = On)
 
Theoremtfrexlem 6450* The transfinite recursion function is set-like if the input is. (Contributed by Mario Carneiro, 3-Jul-2019.)
𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}    &   (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹𝑥) ∈ V))       ((𝜑𝐶𝑉) → (recs(𝐹)‘𝐶) ∈ V)
 
Theoremtfri1d 6451* Principle of Transfinite Recursion, part 1 of 3. Theorem 7.41(1) of [TakeutiZaring] p. 47, with an additional condition.

The condition is that 𝐺 is defined "everywhere", which is stated here as (𝐺𝑥) ∈ V. Alternately, 𝑥 ∈ On∀𝑓(𝑓 Fn 𝑥𝑓 ∈ dom 𝐺) would suffice.

Given a function 𝐺 satisfying that condition, we define a class 𝐴 of all "acceptable" functions. The final function we're interested in is the union 𝐹 = recs(𝐺) of them. 𝐹 is then said to be defined by transfinite recursion. The purpose of the 3 parts of this theorem is to demonstrate properties of 𝐹. In this first part we show that 𝐹 is a function whose domain is all ordinal numbers. (Contributed by Jim Kingdon, 4-May-2019.) (Revised by Mario Carneiro, 24-May-2019.)

𝐹 = recs(𝐺)    &   (𝜑 → ∀𝑥(Fun 𝐺 ∧ (𝐺𝑥) ∈ V))       (𝜑𝐹 Fn On)
 
Theoremtfri2d 6452* Principle of Transfinite Recursion, part 2 of 3. Theorem 7.41(2) of [TakeutiZaring] p. 47, with an additional condition on the recursion rule 𝐺 ( as described at tfri1 6481). Here we show that the function 𝐹 has the property that for any function 𝐺 satisfying that condition, the "next" value of 𝐹 is 𝐺 recursively applied to all "previous" values of 𝐹. (Contributed by Jim Kingdon, 4-May-2019.)
𝐹 = recs(𝐺)    &   (𝜑 → ∀𝑥(Fun 𝐺 ∧ (𝐺𝑥) ∈ V))       ((𝜑𝐴 ∈ On) → (𝐹𝐴) = (𝐺‘(𝐹𝐴)))
 
Theoremtfr1onlem3ag 6453* Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3ag 6425 but for tfr1on 6466 related lemmas). (Contributed by Jim Kingdon, 13-Mar-2022.)
𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}       (𝐻𝑉 → (𝐻𝐴 ↔ ∃𝑧𝑋 (𝐻 Fn 𝑧 ∧ ∀𝑤𝑧 (𝐻𝑤) = (𝐺‘(𝐻𝑤)))))
 
Theoremtfr1onlem3 6454* Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3 6427 but for tfr1on 6466 related lemmas). (Contributed by Jim Kingdon, 14-Mar-2022.)
𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}       𝐴 = {𝑔 ∣ ∃𝑧𝑋 (𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤)))}
 
Theoremtfr1onlemssrecs 6455* Lemma for tfr1on 6466. The union of functions acceptable for tfr1on 6466 is a subset of recs. (Contributed by Jim Kingdon, 15-Mar-2022.)
𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   (𝜑 → Ord 𝑋)       (𝜑 𝐴 ⊆ recs(𝐺))
 
Theoremtfr1onlemsucfn 6456* We can extend an acceptable function by one element to produce a function. Lemma for tfr1on 6466. (Contributed by Jim Kingdon, 12-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓 Fn 𝑥) → (𝐺𝑓) ∈ V)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   (𝜑𝑧𝑋)    &   (𝜑𝑔 Fn 𝑧)    &   (𝜑𝑔𝐴)       (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}) Fn suc 𝑧)
 
Theoremtfr1onlemsucaccv 6457* Lemma for tfr1on 6466. We can extend an acceptable function by one element to produce an acceptable function. (Contributed by Jim Kingdon, 12-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓 Fn 𝑥) → (𝐺𝑓) ∈ V)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   (𝜑𝑌𝑋)    &   (𝜑𝑧𝑌)    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝑔 Fn 𝑧)    &   (𝜑𝑔𝐴)       (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}) ∈ 𝐴)
 
Theoremtfr1onlembacc 6458* Lemma for tfr1on 6466. Each element of 𝐵 is an acceptable function. (Contributed by Jim Kingdon, 14-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓 Fn 𝑥) → (𝐺𝑓) ∈ V)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑𝐵𝐴)
 
Theoremtfr1onlembxssdm 6459* Lemma for tfr1on 6466. The union of 𝐵 is defined on all elements of 𝑋. (Contributed by Jim Kingdon, 14-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓 Fn 𝑥) → (𝐺𝑓) ∈ V)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑𝐷 ⊆ dom 𝐵)
 
Theoremtfr1onlembfn 6460* Lemma for tfr1on 6466. The union of 𝐵 is a function defined on 𝑥. (Contributed by Jim Kingdon, 15-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓 Fn 𝑥) → (𝐺𝑓) ∈ V)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑 𝐵 Fn 𝐷)
 
Theoremtfr1onlembex 6461* Lemma for tfr1on 6466. The set 𝐵 exists. (Contributed by Jim Kingdon, 14-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓 Fn 𝑥) → (𝐺𝑓) ∈ V)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑𝐵 ∈ V)
 
Theoremtfr1onlemubacc 6462* Lemma for tfr1on 6466. The union of 𝐵 satisfies the recursion rule. (Contributed by Jim Kingdon, 15-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓 Fn 𝑥) → (𝐺𝑓) ∈ V)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑 → ∀𝑢𝐷 ( 𝐵𝑢) = (𝐺‘( 𝐵𝑢)))
 
Theoremtfr1onlemex 6463* Lemma for tfr1on 6466. (Contributed by Jim Kingdon, 16-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓 Fn 𝑥) → (𝐺𝑓) ∈ V)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔 Fn 𝑧𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑 → ∃𝑓(𝑓 Fn 𝐷 ∧ ∀𝑢𝐷 (𝑓𝑢) = (𝐺‘(𝑓𝑢))))
 
Theoremtfr1onlemaccex 6464* We can define an acceptable function on any element of 𝑋.

As with many of the transfinite recursion theorems, we have hypotheses that state that 𝐹 is a function and that it is defined up to 𝑋. (Contributed by Jim Kingdon, 16-Mar-2022.)

𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓 Fn 𝑥) → (𝐺𝑓) ∈ V)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)       ((𝜑𝐶𝑋) → ∃𝑔(𝑔 Fn 𝐶 ∧ ∀𝑢𝐶 (𝑔𝑢) = (𝐺‘(𝑔𝑢))))
 
Theoremtfr1onlemres 6465* Lemma for tfr1on 6466. Recursion is defined on an ordinal if the characteristic function is defined up to a suitable point. (Contributed by Jim Kingdon, 18-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓 Fn 𝑥) → (𝐺𝑓) ∈ V)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝑌𝑋)       (𝜑𝑌 ⊆ dom 𝐹)
 
Theoremtfr1on 6466* Recursion is defined on an ordinal if the characteristic function is defined up to a suitable point. (Contributed by Jim Kingdon, 12-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓 Fn 𝑥) → (𝐺𝑓) ∈ V)    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝑌𝑋)       (𝜑𝑌 ⊆ dom 𝐹)
 
Theoremtfri1dALT 6467* Alternate proof of tfri1d 6451 in terms of tfr1on 6466.

Although this does show that the tfr1on 6466 proof is general enough to also prove tfri1d 6451, the tfri1d 6451 proof is simpler in places because it does not need to deal with 𝑋 being any ordinal. For that reason, we have both proofs. (Proof modification is discouraged.) (New usage is discouraged.) (Contributed by Jim Kingdon, 20-Mar-2022.)

𝐹 = recs(𝐺)    &   (𝜑 → ∀𝑥(Fun 𝐺 ∧ (𝐺𝑥) ∈ V))       (𝜑𝐹 Fn On)
 
Theoremtfrcllemssrecs 6468* Lemma for tfrcl 6480. The union of functions acceptable for tfrcl 6480 is a subset of recs. (Contributed by Jim Kingdon, 25-Mar-2022.)
𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓:𝑥𝑆 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   (𝜑 → Ord 𝑋)       (𝜑 𝐴 ⊆ recs(𝐺))
 
Theoremtfrcllemsucfn 6469* We can extend an acceptable function by one element to produce a function. Lemma for tfrcl 6480. (Contributed by Jim Kingdon, 24-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓:𝑥𝑆 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   (𝜑𝑧𝑋)    &   (𝜑𝑔:𝑧𝑆)    &   (𝜑𝑔𝐴)       (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}):suc 𝑧𝑆)
 
Theoremtfrcllemsucaccv 6470* Lemma for tfrcl 6480. We can extend an acceptable function by one element to produce an acceptable function. (Contributed by Jim Kingdon, 24-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓:𝑥𝑆 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   (𝜑𝑌𝑋)    &   (𝜑𝑧𝑌)    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝑔:𝑧𝑆)    &   (𝜑𝑔𝐴)       (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}) ∈ 𝐴)
 
Theoremtfrcllembacc 6471* Lemma for tfrcl 6480. Each element of 𝐵 is an acceptable function. (Contributed by Jim Kingdon, 25-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓:𝑥𝑆 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔:𝑧𝑆𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔:𝑧𝑆 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑𝐵𝐴)
 
Theoremtfrcllembxssdm 6472* Lemma for tfrcl 6480. The union of 𝐵 is defined on all elements of 𝑋. (Contributed by Jim Kingdon, 25-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓:𝑥𝑆 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔:𝑧𝑆𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔:𝑧𝑆 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑𝐷 ⊆ dom 𝐵)
 
Theoremtfrcllembfn 6473* Lemma for tfrcl 6480. The union of 𝐵 is a function defined on 𝑥. (Contributed by Jim Kingdon, 25-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓:𝑥𝑆 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔:𝑧𝑆𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔:𝑧𝑆 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑 𝐵:𝐷𝑆)
 
Theoremtfrcllembex 6474* Lemma for tfrcl 6480. The set 𝐵 exists. (Contributed by Jim Kingdon, 25-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓:𝑥𝑆 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔:𝑧𝑆𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔:𝑧𝑆 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑𝐵 ∈ V)
 
Theoremtfrcllemubacc 6475* Lemma for tfrcl 6480. The union of 𝐵 satisfies the recursion rule. (Contributed by Jim Kingdon, 25-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓:𝑥𝑆 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔:𝑧𝑆𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔:𝑧𝑆 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑 → ∀𝑢𝐷 ( 𝐵𝑢) = (𝐺‘( 𝐵𝑢)))
 
Theoremtfrcllemex 6476* Lemma for tfrcl 6480. (Contributed by Jim Kingdon, 26-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓:𝑥𝑆 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   𝐵 = { ∣ ∃𝑧𝐷𝑔(𝑔:𝑧𝑆𝑔𝐴 = (𝑔 ∪ {⟨𝑧, (𝐺𝑔)⟩}))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝐷𝑋)    &   (𝜑 → ∀𝑧𝐷𝑔(𝑔:𝑧𝑆 ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐺‘(𝑔𝑤))))       (𝜑 → ∃𝑓(𝑓:𝐷𝑆 ∧ ∀𝑢𝐷 (𝑓𝑢) = (𝐺‘(𝑓𝑢))))
 
Theoremtfrcllemaccex 6477* We can define an acceptable function on any element of 𝑋.

As with many of the transfinite recursion theorems, we have hypotheses that state that 𝐹 is a function and that it is defined up to 𝑋. (Contributed by Jim Kingdon, 26-Mar-2022.)

𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓:𝑥𝑆 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)       ((𝜑𝐶𝑋) → ∃𝑔(𝑔:𝐶𝑆 ∧ ∀𝑢𝐶 (𝑔𝑢) = (𝐺‘(𝑔𝑢))))
 
Theoremtfrcllemres 6478* Lemma for tfr1on 6466. Recursion is defined on an ordinal if the characteristic function is defined up to a suitable point. (Contributed by Jim Kingdon, 18-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   𝐴 = {𝑓 ∣ ∃𝑥𝑋 (𝑓:𝑥𝑆 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓𝑦)))}    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝑌𝑋)       (𝜑𝑌 ⊆ dom 𝐹)
 
Theoremtfrcldm 6479* Recursion is defined on an ordinal if the characteristic function satisfies a closure hypothesis up to a suitable point. (Contributed by Jim Kingdon, 26-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝑌 𝑋)       (𝜑𝑌 ∈ dom 𝐹)
 
Theoremtfrcl 6480* Closure for transfinite recursion. As with tfr1on 6466, the characteristic function must be defined up to a suitable point, not necessarily on all ordinals. (Contributed by Jim Kingdon, 25-Mar-2022.)
𝐹 = recs(𝐺)    &   (𝜑 → Fun 𝐺)    &   (𝜑 → Ord 𝑋)    &   ((𝜑𝑥𝑋𝑓:𝑥𝑆) → (𝐺𝑓) ∈ 𝑆)    &   ((𝜑𝑥 𝑋) → suc 𝑥𝑋)    &   (𝜑𝑌 𝑋)       (𝜑 → (𝐹𝑌) ∈ 𝑆)
 
Theoremtfri1 6481* Principle of Transfinite Recursion, part 1 of 3. Theorem 7.41(1) of [TakeutiZaring] p. 47, with an additional condition.

The condition is that 𝐺 is defined "everywhere", which is stated here as (𝐺𝑥) ∈ V. Alternately, 𝑥 ∈ On∀𝑓(𝑓 Fn 𝑥𝑓 ∈ dom 𝐺) would suffice.

Given a function 𝐺 satisfying that condition, we define a class 𝐴 of all "acceptable" functions. The final function we're interested in is the union 𝐹 = recs(𝐺) of them. 𝐹 is then said to be defined by transfinite recursion. The purpose of the 3 parts of this theorem is to demonstrate properties of 𝐹. In this first part we show that 𝐹 is a function whose domain is all ordinal numbers. (Contributed by Jim Kingdon, 4-May-2019.) (Revised by Mario Carneiro, 24-May-2019.)

𝐹 = recs(𝐺)    &   (Fun 𝐺 ∧ (𝐺𝑥) ∈ V)       𝐹 Fn On
 
Theoremtfri2 6482* Principle of Transfinite Recursion, part 2 of 3. Theorem 7.41(2) of [TakeutiZaring] p. 47, with an additional condition on the recursion rule 𝐺 ( as described at tfri1 6481). Here we show that the function 𝐹 has the property that for any function 𝐺 satisfying that condition, the "next" value of 𝐹 is 𝐺 recursively applied to all "previous" values of 𝐹. (Contributed by Jim Kingdon, 4-May-2019.)
𝐹 = recs(𝐺)    &   (Fun 𝐺 ∧ (𝐺𝑥) ∈ V)       (𝐴 ∈ On → (𝐹𝐴) = (𝐺‘(𝐹𝐴)))
 
Theoremtfri3 6483* Principle of Transfinite Recursion, part 3 of 3. Theorem 7.41(3) of [TakeutiZaring] p. 47, with an additional condition on the recursion rule 𝐺 ( as described at tfri1 6481). Finally, we show that 𝐹 is unique. We do this by showing that any class 𝐵 with the same properties of 𝐹 that we showed in parts 1 and 2 is identical to 𝐹. (Contributed by Jim Kingdon, 4-May-2019.)
𝐹 = recs(𝐺)    &   (Fun 𝐺 ∧ (𝐺𝑥) ∈ V)       ((𝐵 Fn On ∧ ∀𝑥 ∈ On (𝐵𝑥) = (𝐺‘(𝐵𝑥))) → 𝐵 = 𝐹)
 
Theoremtfrex 6484* The transfinite recursion function is set-like if the input is. (Contributed by Mario Carneiro, 3-Jul-2019.)
𝐹 = recs(𝐺)    &   (𝜑 → ∀𝑥(Fun 𝐺 ∧ (𝐺𝑥) ∈ V))       ((𝜑𝐴𝑉) → (𝐹𝐴) ∈ V)
 
2.6.21  Recursive definition generator
 
Syntaxcrdg 6485 Extend class notation with the recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class rec(𝐹, 𝐼)
 
Definitiondf-irdg 6486* Define a recursive definition generator on On (the class of ordinal numbers) with characteristic function 𝐹 and initial value 𝐼. This rather amazing operation allows us to define, with compact direct definitions, functions that are usually defined in textbooks only with indirect self-referencing recursive definitions. A recursive definition requires advanced metalogic to justify - in particular, eliminating a recursive definition is very difficult and often not even shown in textbooks. On the other hand, the elimination of a direct definition is a matter of simple mechanical substitution. The price paid is the daunting complexity of our rec operation (especially when df-recs 6421 that it is built on is also eliminated). But once we get past this hurdle, definitions that would otherwise be recursive become relatively simple. In classical logic it would be easier to divide this definition into cases based on whether the domain of 𝑔 is zero, a successor, or a limit ordinal. Cases do not (in general) work that way in intuitionistic logic, so instead we choose a definition which takes the union of all the results of the characteristic function for ordinals in the domain of 𝑔. This means that this definition has the expected properties for increasing and continuous ordinal functions, which include ordinal addition and multiplication.

For finite recursion we also define df-frec 6507 and for suitable characteristic functions df-frec 6507 yields the same result as rec restricted to ω, as seen at frecrdg 6524.

Note: We introduce rec with the philosophical goal of being able to eliminate all definitions with direct mechanical substitution and to verify easily the soundness of definitions. Metamath itself has no built-in technical limitation that prevents multiple-part recursive definitions in the traditional textbook style. (Contributed by Jim Kingdon, 19-May-2019.)

rec(𝐹, 𝐼) = recs((𝑔 ∈ V ↦ (𝐼 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))))
 
Theoremrdgeq1 6487 Equality theorem for the recursive definition generator. (Contributed by NM, 9-Apr-1995.) (Revised by Mario Carneiro, 9-May-2015.)
(𝐹 = 𝐺 → rec(𝐹, 𝐴) = rec(𝐺, 𝐴))
 
Theoremrdgeq2 6488 Equality theorem for the recursive definition generator. (Contributed by NM, 9-Apr-1995.) (Revised by Mario Carneiro, 9-May-2015.)
(𝐴 = 𝐵 → rec(𝐹, 𝐴) = rec(𝐹, 𝐵))
 
Theoremrdgfun 6489 The recursive definition generator is a function. (Contributed by Mario Carneiro, 16-Nov-2014.)
Fun rec(𝐹, 𝐴)
 
Theoremrdgtfr 6490* The recursion rule for the recursive definition generator is defined everywhere. (Contributed by Jim Kingdon, 14-May-2020.)
((∀𝑧(𝐹𝑧) ∈ V ∧ 𝐴𝑉) → (Fun (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))) ∧ ((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥))))‘𝑓) ∈ V))
 
Theoremrdgruledefgg 6491* The recursion rule for the recursive definition generator is defined everywhere. (Contributed by Jim Kingdon, 4-Jul-2019.)
((𝐹 Fn V ∧ 𝐴𝑉) → (Fun (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))) ∧ ((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥))))‘𝑓) ∈ V))
 
Theoremrdgruledefg 6492* The recursion rule for the recursive definition generator is defined everywhere. (Contributed by Jim Kingdon, 4-Jul-2019.)
𝐹 Fn V       (𝐴𝑉 → (Fun (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))) ∧ ((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥))))‘𝑓) ∈ V))
 
Theoremrdgexggg 6493 The recursive definition generator produces a set on a set input. (Contributed by Jim Kingdon, 4-Jul-2019.)
((𝐹 Fn V ∧ 𝐴𝑉𝐵𝑊) → (rec(𝐹, 𝐴)‘𝐵) ∈ V)
 
Theoremrdgexgg 6494 The recursive definition generator produces a set on a set input. (Contributed by Jim Kingdon, 4-Jul-2019.)
𝐹 Fn V       ((𝐴𝑉𝐵𝑊) → (rec(𝐹, 𝐴)‘𝐵) ∈ V)
 
Theoremrdgifnon 6495 The recursive definition generator is a function on ordinal numbers. The 𝐹 Fn V condition states that the characteristic function is defined for all sets (being defined for all ordinals might be enough if being used in a manner similar to rdgon 6502; in cases like df-oadd 6536 either presumably could work). (Contributed by Jim Kingdon, 13-Jul-2019.)
((𝐹 Fn V ∧ 𝐴𝑉) → rec(𝐹, 𝐴) Fn On)
 
Theoremrdgifnon2 6496* The recursive definition generator is a function on ordinal numbers. (Contributed by Jim Kingdon, 14-May-2020.)
((∀𝑧(𝐹𝑧) ∈ V ∧ 𝐴𝑉) → rec(𝐹, 𝐴) Fn On)
 
Theoremrdgivallem 6497* Value of the recursive definition generator. Lemma for rdgival 6498 which simplifies the value further. (Contributed by Jim Kingdon, 13-Jul-2019.) (New usage is discouraged.)
((𝐹 Fn V ∧ 𝐴𝑉𝐵 ∈ On) → (rec(𝐹, 𝐴)‘𝐵) = (𝐴 𝑥𝐵 (𝐹‘((rec(𝐹, 𝐴) ↾ 𝐵)‘𝑥))))
 
Theoremrdgival 6498* Value of the recursive definition generator. (Contributed by Jim Kingdon, 26-Jul-2019.)
((𝐹 Fn V ∧ 𝐴𝑉𝐵 ∈ On) → (rec(𝐹, 𝐴)‘𝐵) = (𝐴 𝑥𝐵 (𝐹‘(rec(𝐹, 𝐴)‘𝑥))))
 
Theoremrdgss 6499 Subset and recursive definition generator. (Contributed by Jim Kingdon, 15-Jul-2019.)
(𝜑𝐹 Fn V)    &   (𝜑𝐼𝑉)    &   (𝜑𝐴 ∈ On)    &   (𝜑𝐵 ∈ On)    &   (𝜑𝐴𝐵)       (𝜑 → (rec(𝐹, 𝐼)‘𝐴) ⊆ (rec(𝐹, 𝐼)‘𝐵))
 
Theoremrdgisuc1 6500* One way of describing the value of the recursive definition generator at a successor. There is no condition on the characteristic function 𝐹 other than 𝐹 Fn V. Given that, the resulting expression encompasses both the expected successor term (𝐹‘(rec(𝐹, 𝐴)‘𝐵)) but also terms that correspond to the initial value 𝐴 and to limit ordinals 𝑥𝐵(𝐹‘(rec(𝐹, 𝐴)‘𝑥)).

If we add conditions on the characteristic function, we can show tighter results such as rdgisucinc 6501. (Contributed by Jim Kingdon, 9-Jun-2019.)

(𝜑𝐹 Fn V)    &   (𝜑𝐴𝑉)    &   (𝜑𝐵 ∈ On)       (𝜑 → (rec(𝐹, 𝐴)‘suc 𝐵) = (𝐴 ∪ ( 𝑥𝐵 (𝐹‘(rec(𝐹, 𝐴)‘𝑥)) ∪ (𝐹‘(rec(𝐹, 𝐴)‘𝐵)))))
    < Previous  Next >

Page List
Jump to page: Contents  1 1-100 2 101-200 3 201-300 4 301-400 5 401-500 6 501-600 7 601-700 8 701-800 9 801-900 10 901-1000 11 1001-1100 12 1101-1200 13 1201-1300 14 1301-1400 15 1401-1500 16 1501-1600 17 1601-1700 18 1701-1800 19 1801-1900 20 1901-2000 21 2001-2100 22 2101-2200 23 2201-2300 24 2301-2400 25 2401-2500 26 2501-2600 27 2601-2700 28 2701-2800 29 2801-2900 30 2901-3000 31 3001-3100 32 3101-3200 33 3201-3300 34 3301-3400 35 3401-3500 36 3501-3600 37 3601-3700 38 3701-3800 39 3801-3900 40 3901-4000 41 4001-4100 42 4101-4200 43 4201-4300 44 4301-4400 45 4401-4500 46 4501-4600 47 4601-4700 48 4701-4800 49 4801-4900 50 4901-5000 51 5001-5100 52 5101-5200 53 5201-5300 54 5301-5400 55 5401-5500 56 5501-5600 57 5601-5700 58 5701-5800 59 5801-5900 60 5901-6000 61 6001-6100 62 6101-6200 63 6201-6300 64 6301-6400 65 6401-6500 66 6501-6600 67 6601-6700 68 6701-6800 69 6801-6900 70 6901-7000 71 7001-7100 72 7101-7200 73 7201-7300 74 7301-7400 75 7401-7500 76 7501-7600 77 7601-7700 78 7701-7800 79 7801-7900 80 7901-8000 81 8001-8100 82 8101-8200 83 8201-8300 84 8301-8400 85 8401-8500 86 8501-8600 87 8601-8700 88 8701-8800 89 8801-8900 90 8901-9000 91 9001-9100 92 9101-9200 93 9201-9300 94 9301-9400 95 9401-9500 96 9501-9600 97 9601-9700 98 9701-9800 99 9801-9900 100 9901-10000 101 10001-10100 102 10101-10200 103 10201-10300 104 10301-10400 105 10401-10500 106 10501-10600 107 10601-10700 108 10701-10800 109 10801-10900 110 10901-11000 111 11001-11100 112 11101-11200 113 11201-11300 114 11301-11400 115 11401-11500 116 11501-11600 117 11601-11700 118 11701-11800 119 11801-11900 120 11901-12000 121 12001-12100 122 12101-12200 123 12201-12300 124 12301-12400 125 12401-12500 126 12501-12600 127 12601-12700 128 12701-12800 129 12801-12900 130 12901-13000 131 13001-13100 132 13101-13200 133 13201-13300 134 13301-13400 135 13401-13500 136 13501-13600 137 13601-13700 138 13701-13800 139 13801-13900 140 13901-14000 141 14001-14100 142 14101-14200 143 14201-14300 144 14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14800 149 14801-14900 150 14901-15000 151 15001-15100 152 15101-15200 153 15201-15300 154 15301-15400 155 15401-15500 156 15501-15600 157 15601-15700 158 15701-15800 159 15801-15900 160 15901-16000 161 16001-16100 162 16101-16200 163 16201-16300 164 16301-16363
  Copyright terms: Public domain < Previous  Next >