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Theorem List for Intuitionistic Logic Explorer - 6201-6300   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremiordsmo 6201 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 6202 The null set is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 20-Nov-2011.)
Smo ∅

Theoremsmofvon 6203 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 6204 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 6205* 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 6206 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 6207 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 6208 Notation for a function defined by strong transfinite recursion.
class recs(𝐹)

Definitiondf-recs 6209* 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 6274 for more details on why this definition is desirable. Unlike df-irdg 6274 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 6239 and tfri2d 6240 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 6210 Equality theorem for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
(𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))

Theoremnfrecs 6211 Bound-variable hypothesis builder for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
𝑥𝐹       𝑥recs(𝐹)

Theoremtfrlem1 6212* 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 6213* 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 6214* 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 6215* 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 6216* 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 6217* 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 6218* 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 6219* 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 6220* 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 6221* 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 6222* 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 6223* 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 6224 Transfinite recursion produces a function. (Contributed by Jim Kingdon, 20-Aug-2021.)
Fun recs(𝐹)

Theoremtfr2a 6225 A weak version of transfinite recursion. (Contributed by Mario Carneiro, 24-Jun-2015.)
𝐹 = recs(𝐺)       (𝐴 ∈ dom 𝐹 → (𝐹𝐴) = (𝐺‘(𝐹𝐴)))

Theoremtfr0dm 6226 Transfinite recursion is defined at the empty set. (Contributed by Jim Kingdon, 8-Mar-2022.)
𝐹 = recs(𝐺)       ((𝐺‘∅) ∈ 𝑉 → ∅ ∈ dom 𝐹)

Theoremtfr0 6227 Transfinite recursion at the empty set. (Contributed by Jim Kingdon, 8-May-2020.)
𝐹 = recs(𝐺)       ((𝐺‘∅) ∈ 𝑉 → (𝐹‘∅) = (𝐺‘∅))

Theoremtfrlemisucfn 6228* We can extend an acceptable function by one element to produce a function. Lemma for tfrlemi1 6236. (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 6229* We can extend an acceptable function by one element to produce an acceptable function. Lemma for tfrlemi1 6236. (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 6230* Each element of 𝐵 is an acceptable function. Lemma for tfrlemi1 6236. (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 6231* The union of 𝐵 is defined on all ordinals. Lemma for tfrlemi1 6236. (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 6232* The union of 𝐵 is a function defined on 𝑥. Lemma for tfrlemi1 6236. (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 6233* The set 𝐵 exists. Lemma for tfrlemi1 6236. (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 6234* The union of 𝐵 satisfies the recursion rule (lemma for tfrlemi1 6236). (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 6235* Lemma for tfrlemi1 6236. (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 6236* 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 6237* 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 6238* 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 6239* 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 6240* 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 6269). 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 6241* Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3ag 6213 but for tfr1on 6254 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 6242* Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3 6215 but for tfr1on 6254 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 6243* Lemma for tfr1on 6254. The union of functions acceptable for tfr1on 6254 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 6244* We can extend an acceptable function by one element to produce a function. Lemma for tfr1on 6254. (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 6245* Lemma for tfr1on 6254. 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 6246* Lemma for tfr1on 6254. 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 6247* Lemma for tfr1on 6254. 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 6248* Lemma for tfr1on 6254. 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 6249* Lemma for tfr1on 6254. 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 6250* Lemma for tfr1on 6254. 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 6251* Lemma for tfr1on 6254. (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 6252* 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 6253* Lemma for tfr1on 6254. 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 6254* 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 6255* Alternate proof of tfri1d 6239 in terms of tfr1on 6254.

Although this does show that the tfr1on 6254 proof is general enough to also prove tfri1d 6239, the tfri1d 6239 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 6256* Lemma for tfrcl 6268. The union of functions acceptable for tfrcl 6268 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 6257* We can extend an acceptable function by one element to produce a function. Lemma for tfrcl 6268. (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 6258* Lemma for tfrcl 6268. 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 6259* Lemma for tfrcl 6268. 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 6260* Lemma for tfrcl 6268. 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 6261* Lemma for tfrcl 6268. 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 6262* Lemma for tfrcl 6268. 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 6263* Lemma for tfrcl 6268. 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 6264* Lemma for tfrcl 6268. (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 6265* 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 6266* Lemma for tfr1on 6254. 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 6267* 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 6268* Closure for transfinite recursion. As with tfr1on 6254, 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 6269* 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 6270* 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 6269). 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 6271* 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 6269). 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 6272* 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 6273 Extend class notation with the recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class rec(𝐹, 𝐼)

Definitiondf-irdg 6274* 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 6209 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 6295 and for suitable characteristic functions df-frec 6295 yields the same result as rec restricted to ω, as seen at frecrdg 6312.

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 6275 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 6276 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 6277 The recursive definition generator is a function. (Contributed by Mario Carneiro, 16-Nov-2014.)
Fun rec(𝐹, 𝐴)

Theoremrdgtfr 6278* 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 6279* 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 6280* 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 6281 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 6282 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 6283 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 6290; in cases like df-oadd 6324 either presumably could work). (Contributed by Jim Kingdon, 13-Jul-2019.)
((𝐹 Fn V ∧ 𝐴𝑉) → rec(𝐹, 𝐴) Fn On)

Theoremrdgifnon2 6284* 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 6285* Value of the recursive definition generator. Lemma for rdgival 6286 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 6286* 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 6287 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 6288* 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 6289. (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)))))

Theoremrdgisucinc 6289* Value of the recursive definition generator at a successor.

This can be thought of as a generalization of oasuc 6367 and omsuc 6375. (Contributed by Jim Kingdon, 29-Aug-2019.)

(𝜑𝐹 Fn V)    &   (𝜑𝐴𝑉)    &   (𝜑𝐵 ∈ On)    &   (𝜑 → ∀𝑥 𝑥 ⊆ (𝐹𝑥))       (𝜑 → (rec(𝐹, 𝐴)‘suc 𝐵) = (𝐹‘(rec(𝐹, 𝐴)‘𝐵)))

Theoremrdgon 6290* Evaluating the recursive definition generator produces an ordinal. There is a hypothesis that the characteristic function produces ordinals on ordinal arguments. (Contributed by Jim Kingdon, 26-Jul-2019.) (Revised by Jim Kingdon, 13-Apr-2022.)
(𝜑𝐴 ∈ On)    &   (𝜑 → ∀𝑥 ∈ On (𝐹𝑥) ∈ On)       ((𝜑𝐵 ∈ On) → (rec(𝐹, 𝐴)‘𝐵) ∈ On)

Theoremrdg0 6291 The initial value of the recursive definition generator. (Contributed by NM, 23-Apr-1995.) (Revised by Mario Carneiro, 14-Nov-2014.)
𝐴 ∈ V       (rec(𝐹, 𝐴)‘∅) = 𝐴

Theoremrdg0g 6292 The initial value of the recursive definition generator. (Contributed by NM, 25-Apr-1995.)
(𝐴𝐶 → (rec(𝐹, 𝐴)‘∅) = 𝐴)

Theoremrdgexg 6293 The recursive definition generator produces a set on a set input. (Contributed by Mario Carneiro, 3-Jul-2019.)
𝐴 ∈ V    &   𝐹 Fn V       (𝐵𝑉 → (rec(𝐹, 𝐴)‘𝐵) ∈ V)

2.6.21  Finite recursion

Syntaxcfrec 6294 Extend class notation with the finite recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class frec(𝐹, 𝐼)

Definitiondf-frec 6295* Define a recursive definition generator on ω (the class of finite ordinals) 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 frec operation (especially when df-recs 6209 that it is built on is also eliminated). But once we get past this hurdle, definitions that would otherwise be recursive become relatively simple; see frec0g 6301 and frecsuc 6311.

Unlike with transfinite recursion, finite recurson can readily divide definitions and proofs into zero and successor cases, because even without excluded middle we have theorems such as nn0suc 4525. The analogous situation with transfinite recursion - being able to say that an ordinal is zero, successor, or limit - is enabled by excluded middle and thus is not available to us. For the characteristic functions which satisfy the conditions given at frecrdg 6312, this definition and df-irdg 6274 restricted to ω produce the same result.

Note: We introduce frec 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 Mario Carneiro and Jim Kingdon, 10-Aug-2019.)

frec(𝐹, 𝐼) = (recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚𝑥 ∈ (𝐹‘(𝑔𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥𝐼))})) ↾ ω)

Theoremfreceq1 6296 Equality theorem for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)
(𝐹 = 𝐺 → frec(𝐹, 𝐴) = frec(𝐺, 𝐴))

Theoremfreceq2 6297 Equality theorem for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)
(𝐴 = 𝐵 → frec(𝐹, 𝐴) = frec(𝐹, 𝐵))

Theoremfrecex 6298 Finite recursion produces a set. (Contributed by Jim Kingdon, 20-Aug-2021.)
frec(𝐹, 𝐴) ∈ V

Theoremfrecfun 6299 Finite recursion produces a function. See also frecfnom 6305 which also states that the domain of that function is ω but which puts conditions on 𝐴 and 𝐹. (Contributed by Jim Kingdon, 13-Feb-2022.)
Fun frec(𝐹, 𝐴)

Theoremnffrec 6300 Bound-variable hypothesis builder for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)
𝑥𝐹    &   𝑥𝐴       𝑥frec(𝐹, 𝐴)

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