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Theorem List for Intuitionistic Logic Explorer - 6001-6100   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
2.6.17  Undefined values
 
Theorempwuninel2 6001 The power set of the union of a set does not belong to the set. This theorem provides a way of constructing a new set that doesn't belong to a given set. (Contributed by Stefan O'Rear, 22-Feb-2015.)
( 𝐴𝑉 → ¬ 𝒫 𝐴𝐴)
 
Theorem2pwuninelg 6002 The power set of the power set of the union of a set does not belong to the set. This theorem provides a way of constructing a new set that doesn't belong to a given set. (Contributed by Jim Kingdon, 14-Jan-2020.)
(𝐴𝑉 → ¬ 𝒫 𝒫 𝐴𝐴)
 
2.6.18  Functions on ordinals; strictly monotone ordinal functions
 
Theoremiunon 6003* The indexed union of a set of ordinal numbers 𝐵(𝑥) is an ordinal number. (Contributed by NM, 13-Oct-2003.) (Revised by Mario Carneiro, 5-Dec-2016.)
((𝐴𝑉 ∧ ∀𝑥𝐴 𝐵 ∈ On) → 𝑥𝐴 𝐵 ∈ On)
 
Syntaxwsmo 6004 Introduce the strictly monotone ordinal function. A strictly monotone function is one that is constantly increasing across the ordinals.
wff Smo 𝐴
 
Definitiondf-smo 6005* 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 6006* 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 6007* 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 6008* 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 6009 Equality theorem for strictly monotone functions. (Contributed by Andrew Salmon, 16-Nov-2011.)
(𝐴 = 𝐵 → (Smo 𝐴 ↔ Smo 𝐵))
 
Theoremsmodm 6010 The domain of a strictly monotone function is an ordinal. (Contributed by Andrew Salmon, 16-Nov-2011.)
(Smo 𝐴 → Ord dom 𝐴)
 
Theoremsmores 6011 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 6012 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 6013 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 6014 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 6015 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 6016 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 6017 The null set is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 20-Nov-2011.)
Smo ∅
 
Theoremsmofvon 6018 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 6019 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 6020* 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 6021 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 6022 A strictly monotone ordinal function preserves the epsilon relation. (Contributed by Mario Carneiro, 12-Mar-2013.)
(((𝐹 Fn 𝐴 ∧ Smo 𝐹) ∧ (𝐵𝐴𝐶𝐵)) → (𝐹𝐶) ∈ (𝐹𝐵))
 
2.6.19  "Strong" transfinite recursion
 
Syntaxcrecs 6023 Notation for a function defined by strong transfinite recursion.
class recs(𝐹)
 
Definitiondf-recs 6024* 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 6089 for more details on why this definition is desirable. Unlike df-irdg 6089 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 6054 and tfri2d 6055 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 6025 Equality theorem for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
(𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))
 
Theoremnfrecs 6026 Bound-variable hypothesis builder for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
𝑥𝐹       𝑥recs(𝐹)
 
Theoremtfrlem1 6027* 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 6028* 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 6029* 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 6030* 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 6031* 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 6032* 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 6033* 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 6034* 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 6035* 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 6036* 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 6037* 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 6038* 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 6039 Transfinite recursion produces a function. (Contributed by Jim Kingdon, 20-Aug-2021.)
Fun recs(𝐹)
 
Theoremtfr2a 6040 A weak version of transfinite recursion. (Contributed by Mario Carneiro, 24-Jun-2015.)
𝐹 = recs(𝐺)       (𝐴 ∈ dom 𝐹 → (𝐹𝐴) = (𝐺‘(𝐹𝐴)))
 
Theoremtfr0dm 6041 Transfinite recursion is defined at the empty set. (Contributed by Jim Kingdon, 8-Mar-2022.)
𝐹 = recs(𝐺)       ((𝐺‘∅) ∈ 𝑉 → ∅ ∈ dom 𝐹)
 
Theoremtfr0 6042 Transfinite recursion at the empty set. (Contributed by Jim Kingdon, 8-May-2020.)
𝐹 = recs(𝐺)       ((𝐺‘∅) ∈ 𝑉 → (𝐹‘∅) = (𝐺‘∅))
 
Theoremtfrlemisucfn 6043* We can extend an acceptable function by one element to produce a function. Lemma for tfrlemi1 6051. (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 6044* We can extend an acceptable function by one element to produce an acceptable function. Lemma for tfrlemi1 6051. (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 6045* Each element of 𝐵 is an acceptable function. Lemma for tfrlemi1 6051. (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 6046* The union of 𝐵 is defined on all ordinals. Lemma for tfrlemi1 6051. (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 6047* The union of 𝐵 is a function defined on 𝑥. Lemma for tfrlemi1 6051. (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 6048* The set 𝐵 exists. Lemma for tfrlemi1 6051. (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 6049* The union of 𝐵 satisfies the recursion rule (lemma for tfrlemi1 6051). (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 6050* Lemma for tfrlemi1 6051. (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 6051* 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 6052* 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 6053* 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 6054* 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 6055* 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 6084). 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 6056* Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3ag 6028 but for tfr1on 6069 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 6057* Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3 6030 but for tfr1on 6069 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 6058* Lemma for tfr1on 6069. The union of functions acceptable for tfr1on 6069 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 6059* We can extend an acceptable function by one element to produce a function. Lemma for tfr1on 6069. (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 6060* Lemma for tfr1on 6069. 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 6061* Lemma for tfr1on 6069. 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 6062* Lemma for tfr1on 6069. 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 6063* Lemma for tfr1on 6069. 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 6064* Lemma for tfr1on 6069. 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 6065* Lemma for tfr1on 6069. 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 6066* Lemma for tfr1on 6069. (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 6067* 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 6068* Lemma for tfr1on 6069. 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 6069* 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 6070* Alternate proof of tfri1d 6054 in terms of tfr1on 6069.

Although this does show that the tfr1on 6069 proof is general enough to also prove tfri1d 6054, the tfri1d 6054 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 6071* Lemma for tfrcl 6083. The union of functions acceptable for tfrcl 6083 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 6072* We can extend an acceptable function by one element to produce a function. Lemma for tfrcl 6083. (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 6073* Lemma for tfrcl 6083. 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 6074* Lemma for tfrcl 6083. 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 6075* Lemma for tfrcl 6083. 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 6076* Lemma for tfrcl 6083. 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 6077* Lemma for tfrcl 6083. 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 6078* Lemma for tfrcl 6083. 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 6079* Lemma for tfrcl 6083. (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 6080* 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 6081* Lemma for tfr1on 6069. 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 6082* 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 6083* Closure for transfinite recursion. As with tfr1on 6069, 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 6084* 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 6085* 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 6084). 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 6086* 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 6084). 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 6087* 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)
 
2.6.20  Recursive definition generator
 
Syntaxcrdg 6088 Extend class notation with the recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class rec(𝐹, 𝐼)
 
Definitiondf-irdg 6089* 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 6024 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 6110 and for suitable characteristic functions df-frec 6110 yields the same result as rec restricted to ω, as seen at frecrdg 6127.

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 6090 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 6091 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 6092 The recursive definition generator is a function. (Contributed by Mario Carneiro, 16-Nov-2014.)
Fun rec(𝐹, 𝐴)
 
Theoremrdgtfr 6093* 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 6094* 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 6095* 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 6096 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 6097 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 6098 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 6105; in cases like df-oadd 6139 either presumably could work). (Contributed by Jim Kingdon, 13-Jul-2019.)
((𝐹 Fn V ∧ 𝐴𝑉) → rec(𝐹, 𝐴) Fn On)
 
Theoremrdgifnon2 6099* 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 6100* Value of the recursive definition generator. Lemma for rdgival 6101 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))))
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