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Theorem List for Intuitionistic Logic Explorer - 6101-6200   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremtpos0 6101 Transposition of the empty set. (Contributed by NM, 10-Sep-2015.)
tpos ∅ = ∅

Theoremtposco 6102 Transposition of a composition. (Contributed by Mario Carneiro, 4-Oct-2015.)
tpos (𝐹𝐺) = (𝐹 ∘ tpos 𝐺)

Theoremtpossym 6103* Two ways to say a function is symmetric. (Contributed by Mario Carneiro, 4-Oct-2015.)
(𝐹 Fn (𝐴 × 𝐴) → (tpos 𝐹 = 𝐹 ↔ ∀𝑥𝐴𝑦𝐴 (𝑥𝐹𝑦) = (𝑦𝐹𝑥)))

Theoremtposeqi 6104 Equality theorem for transposition. (Contributed by Mario Carneiro, 10-Sep-2015.)
𝐹 = 𝐺       tpos 𝐹 = tpos 𝐺

Theoremtposex 6105 A transposition is a set. (Contributed by Mario Carneiro, 10-Sep-2015.)
𝐹 ∈ V       tpos 𝐹 ∈ V

Theoremnftpos 6106 Hypothesis builder for transposition. (Contributed by Mario Carneiro, 10-Sep-2015.)
𝑥𝐹       𝑥tpos 𝐹

Theoremtposoprab 6107* Transposition of a class of ordered triples. (Contributed by Mario Carneiro, 10-Sep-2015.)
𝐹 = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ 𝜑}       tpos 𝐹 = {⟨⟨𝑦, 𝑥⟩, 𝑧⟩ ∣ 𝜑}

Theoremtposmpo 6108* Transposition of a two-argument mapping. (Contributed by Mario Carneiro, 10-Sep-2015.)
𝐹 = (𝑥𝐴, 𝑦𝐵𝐶)       tpos 𝐹 = (𝑦𝐵, 𝑥𝐴𝐶)

2.6.17  Undefined values

Theorempwuninel2 6109 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 6110 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 6111* 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 6112 Introduce the strictly monotone ordinal function. A strictly monotone function is one that is constantly increasing across the ordinals.
wff Smo 𝐴

Definitiondf-smo 6113* 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 6114* 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 6115* 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 6116* 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 6117 Equality theorem for strictly monotone functions. (Contributed by Andrew Salmon, 16-Nov-2011.)
(𝐴 = 𝐵 → (Smo 𝐴 ↔ Smo 𝐵))

Theoremsmodm 6118 The domain of a strictly monotone function is an ordinal. (Contributed by Andrew Salmon, 16-Nov-2011.)
(Smo 𝐴 → Ord dom 𝐴)

Theoremsmores 6119 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 6120 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 6121 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 6122 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 6123 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 6124 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 6125 The null set is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 20-Nov-2011.)
Smo ∅

Theoremsmofvon 6126 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 6127 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 6128* 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 6129 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 6130 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 6131 Notation for a function defined by strong transfinite recursion.
class recs(𝐹)

Definitiondf-recs 6132* 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 6197 for more details on why this definition is desirable. Unlike df-irdg 6197 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 6162 and tfri2d 6163 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 6133 Equality theorem for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
(𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))

Theoremnfrecs 6134 Bound-variable hypothesis builder for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
𝑥𝐹       𝑥recs(𝐹)

Theoremtfrlem1 6135* 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 6136* 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 6137* 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 6138* 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 6139* 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 6140* 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 6141* 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 6142* 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 6143* 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 6144* 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 6145* 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 6146* 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 6147 Transfinite recursion produces a function. (Contributed by Jim Kingdon, 20-Aug-2021.)
Fun recs(𝐹)

Theoremtfr2a 6148 A weak version of transfinite recursion. (Contributed by Mario Carneiro, 24-Jun-2015.)
𝐹 = recs(𝐺)       (𝐴 ∈ dom 𝐹 → (𝐹𝐴) = (𝐺‘(𝐹𝐴)))

Theoremtfr0dm 6149 Transfinite recursion is defined at the empty set. (Contributed by Jim Kingdon, 8-Mar-2022.)
𝐹 = recs(𝐺)       ((𝐺‘∅) ∈ 𝑉 → ∅ ∈ dom 𝐹)

Theoremtfr0 6150 Transfinite recursion at the empty set. (Contributed by Jim Kingdon, 8-May-2020.)
𝐹 = recs(𝐺)       ((𝐺‘∅) ∈ 𝑉 → (𝐹‘∅) = (𝐺‘∅))

Theoremtfrlemisucfn 6151* We can extend an acceptable function by one element to produce a function. Lemma for tfrlemi1 6159. (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 6152* We can extend an acceptable function by one element to produce an acceptable function. Lemma for tfrlemi1 6159. (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 6153* Each element of 𝐵 is an acceptable function. Lemma for tfrlemi1 6159. (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 6154* The union of 𝐵 is defined on all ordinals. Lemma for tfrlemi1 6159. (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 6155* The union of 𝐵 is a function defined on 𝑥. Lemma for tfrlemi1 6159. (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 6156* The set 𝐵 exists. Lemma for tfrlemi1 6159. (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 6157* The union of 𝐵 satisfies the recursion rule (lemma for tfrlemi1 6159). (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 6158* Lemma for tfrlemi1 6159. (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 6159* 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 6160* 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 6161* 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 6162* 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 6163* 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 6192). 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 6164* Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3ag 6136 but for tfr1on 6177 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 6165* Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3 6138 but for tfr1on 6177 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 6166* Lemma for tfr1on 6177. The union of functions acceptable for tfr1on 6177 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 6167* We can extend an acceptable function by one element to produce a function. Lemma for tfr1on 6177. (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 6168* Lemma for tfr1on 6177. 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 6169* Lemma for tfr1on 6177. 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 6170* Lemma for tfr1on 6177. 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 6171* Lemma for tfr1on 6177. 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 6172* Lemma for tfr1on 6177. 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 6173* Lemma for tfr1on 6177. 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 6174* Lemma for tfr1on 6177. (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 6175* 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 6176* Lemma for tfr1on 6177. 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 6177* 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 6178* Alternate proof of tfri1d 6162 in terms of tfr1on 6177.

Although this does show that the tfr1on 6177 proof is general enough to also prove tfri1d 6162, the tfri1d 6162 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 6179* Lemma for tfrcl 6191. The union of functions acceptable for tfrcl 6191 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 6180* We can extend an acceptable function by one element to produce a function. Lemma for tfrcl 6191. (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 6181* Lemma for tfrcl 6191. 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 6182* Lemma for tfrcl 6191. 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 6183* Lemma for tfrcl 6191. 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 6184* Lemma for tfrcl 6191. 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 6185* Lemma for tfrcl 6191. 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 6186* Lemma for tfrcl 6191. 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 6187* Lemma for tfrcl 6191. (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 6188* 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 6189* Lemma for tfr1on 6177. 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 6190* 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 6191* Closure for transfinite recursion. As with tfr1on 6177, 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 6192* 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 6193* 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 6192). 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 6194* 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 6192). 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 6195* 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 6196 Extend class notation with the recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class rec(𝐹, 𝐼)

Definitiondf-irdg 6197* 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 6132 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 6218 and for suitable characteristic functions df-frec 6218 yields the same result as rec restricted to ω, as seen at frecrdg 6235.

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 6198 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 6199 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 6200 The recursive definition generator is a function. (Contributed by Mario Carneiro, 16-Nov-2014.)
Fun rec(𝐹, 𝐴)

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