P. 64 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 6301-6400   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	smo0 6301	The null set is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 20-Nov-2011.)
⊢ Smo ∅
Theorem	smofvon 6302	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 𝐵 ∧ 𝐴 ∈ dom 𝐵) → (𝐵‘𝐴) ∈ On)
Theorem	smoel 6303	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 𝐵 ∧ 𝐴 ∈ dom 𝐵 ∧ 𝐶 ∈ 𝐴) → (𝐵‘𝐶) ∈ (𝐵‘𝐴))
Theorem	smoiun 6304*	The value of a strictly monotone ordinal function contains its indexed union. (Contributed by Andrew Salmon, 22-Nov-2011.)
⊢ ((Smo 𝐵 ∧ 𝐴 ∈ dom 𝐵) → ∪ 𝑥 ∈ 𝐴 (𝐵‘𝑥) ⊆ (𝐵‘𝐴))
Theorem	smoiso 6305	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 𝐴 ∧ 𝐵 ⊆ On) → Smo 𝐹)
Theorem	smoel2 6306	A strictly monotone ordinal function preserves the epsilon relation. (Contributed by Mario Carneiro, 12-Mar-2013.)
⊢ (((𝐹 Fn 𝐴 ∧ Smo 𝐹) ∧ (𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐵)) → (𝐹‘𝐶) ∈ (𝐹‘𝐵))
2.6.20  "Strong" transfinite recursion
Syntax	crecs 6307	Notation for a function defined by strong transfinite recursion.
class recs(𝐹)
Definition	df-recs 6308*	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 6373 for more details on why this definition is desirable. Unlike df-irdg 6373 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 6338 and tfri2d 6339 for the primary contract of this definition. (Contributed by Stefan O'Rear, 18-Jan-2015.)
⊢ recs(𝐹) = ∪ {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}
Theorem	recseq 6309	Equality theorem for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
⊢ (𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))
Theorem	nfrecs 6310	Bound-variable hypothesis builder for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
⊢ Ⅎ𝑥𝐹    ⇒   ⊢ Ⅎ𝑥recs(𝐹)
Theorem	tfrlem1 6311*	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.)
⊢ (𝜑 → 𝐴 ∈ On)    &   ⊢ (𝜑 → (Fun 𝐹 ∧ 𝐴 ⊆ dom 𝐹))    &   ⊢ (𝜑 → (Fun 𝐺 ∧ 𝐴 ⊆ dom 𝐺))    &   ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝐹‘𝑥) = (𝐵‘(𝐹 ↾ 𝑥)))    &   ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝐺‘𝑥) = (𝐵‘(𝐺 ↾ 𝑥)))    ⇒   ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝐹‘𝑥) = (𝐺‘𝑥))
Theorem	tfrlem3ag 6312*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    ⇒   ⊢ (𝐺 ∈ V → (𝐺 ∈ 𝐴 ↔ ∃𝑧 ∈ On (𝐺 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝐺‘𝑤) = (𝐹‘(𝐺 ↾ 𝑤)))))
Theorem	tfrlem3a 6313*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐺 ∈ V    ⇒   ⊢ (𝐺 ∈ 𝐴 ↔ ∃𝑧 ∈ On (𝐺 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝐺‘𝑤) = (𝐹‘(𝐺 ↾ 𝑤))))
Theorem	tfrlem3 6314*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    ⇒   ⊢ 𝐴 = {𝑔 ∣ ∃𝑧 ∈ On (𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐹‘(𝑔 ↾ 𝑤)))}
Theorem	tfrlem3-2d 6315*	Lemma for transfinite recursion which changes a bound variable (Contributed by Jim Kingdon, 2-Jul-2019.)
⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    ⇒   ⊢ (𝜑 → (Fun 𝐹 ∧ (𝐹‘𝑔) ∈ V))
Theorem	tfrlem4 6316*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    ⇒   ⊢ (𝑔 ∈ 𝐴 → Fun 𝑔)
Theorem	tfrlem5 6317*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    ⇒   ⊢ ((𝑔 ∈ 𝐴 ∧ ℎ ∈ 𝐴) → ((𝑥𝑔𝑢 ∧ 𝑥ℎ𝑣) → 𝑢 = 𝑣))
Theorem	recsfval 6318*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    ⇒   ⊢ recs(𝐹) = ∪ 𝐴
Theorem	tfrlem6 6319*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    ⇒   ⊢ Rel recs(𝐹)
Theorem	tfrlem7 6320*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    ⇒   ⊢ Fun recs(𝐹)
Theorem	tfrlem8 6321*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    ⇒   ⊢ Ord dom recs(𝐹)
Theorem	tfrlem9 6322*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    ⇒   ⊢ (𝐵 ∈ dom recs(𝐹) → (recs(𝐹)‘𝐵) = (𝐹‘(recs(𝐹) ↾ 𝐵)))
Theorem	tfrfun 6323	Transfinite recursion produces a function. (Contributed by Jim Kingdon, 20-Aug-2021.)
⊢ Fun recs(𝐹)
Theorem	tfr2a 6324	A weak version of transfinite recursion. (Contributed by Mario Carneiro, 24-Jun-2015.)
⊢ 𝐹 = recs(𝐺)    ⇒   ⊢ (𝐴 ∈ dom 𝐹 → (𝐹‘𝐴) = (𝐺‘(𝐹 ↾ 𝐴)))
Theorem	tfr0dm 6325	Transfinite recursion is defined at the empty set. (Contributed by Jim Kingdon, 8-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    ⇒   ⊢ ((𝐺‘∅) ∈ 𝑉 → ∅ ∈ dom 𝐹)
Theorem	tfr0 6326	Transfinite recursion at the empty set. (Contributed by Jim Kingdon, 8-May-2020.)
⊢ 𝐹 = recs(𝐺)    ⇒   ⊢ ((𝐺‘∅) ∈ 𝑉 → (𝐹‘∅) = (𝐺‘∅))
Theorem	tfrlemisucfn 6327*	We can extend an acceptable function by one element to produce a function. Lemma for tfrlemi1 6335. (Contributed by Jim Kingdon, 2-Jul-2019.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    &   ⊢ (𝜑 → 𝑧 ∈ On)    &   ⊢ (𝜑 → 𝑔 Fn 𝑧)    &   ⊢ (𝜑 → 𝑔 ∈ 𝐴)    ⇒   ⊢ (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐹‘𝑔)⟩}) Fn suc 𝑧)
Theorem	tfrlemisucaccv 6328*	We can extend an acceptable function by one element to produce an acceptable function. Lemma for tfrlemi1 6335. (Contributed by Jim Kingdon, 4-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    &   ⊢ (𝜑 → 𝑧 ∈ On)    &   ⊢ (𝜑 → 𝑔 Fn 𝑧)    &   ⊢ (𝜑 → 𝑔 ∈ 𝐴)    ⇒   ⊢ (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐹‘𝑔)⟩}) ∈ 𝐴)
Theorem	tfrlemibacc 6329*	Each element of 𝐵 is an acceptable function. Lemma for tfrlemi1 6335. (Contributed by Jim Kingdon, 14-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐹‘𝑔)⟩}))}    &   ⊢ (𝜑 → 𝑥 ∈ On)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐹‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → 𝐵 ⊆ 𝐴)
Theorem	tfrlemibxssdm 6330*	The union of 𝐵 is defined on all ordinals. Lemma for tfrlemi1 6335. (Contributed by Jim Kingdon, 18-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐹‘𝑔)⟩}))}    &   ⊢ (𝜑 → 𝑥 ∈ On)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐹‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → 𝑥 ⊆ dom ∪ 𝐵)
Theorem	tfrlemibfn 6331*	The union of 𝐵 is a function defined on 𝑥. Lemma for tfrlemi1 6335. (Contributed by Jim Kingdon, 18-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐹‘𝑔)⟩}))}    &   ⊢ (𝜑 → 𝑥 ∈ On)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐹‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∪ 𝐵 Fn 𝑥)
Theorem	tfrlemibex 6332*	The set 𝐵 exists. Lemma for tfrlemi1 6335. (Contributed by Jim Kingdon, 17-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐹‘𝑔)⟩}))}    &   ⊢ (𝜑 → 𝑥 ∈ On)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐹‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → 𝐵 ∈ V)
Theorem	tfrlemiubacc 6333*	The union of 𝐵 satisfies the recursion rule (lemma for tfrlemi1 6335). (Contributed by Jim Kingdon, 22-Apr-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐹‘𝑔)⟩}))}    &   ⊢ (𝜑 → 𝑥 ∈ On)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐹‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∀𝑢 ∈ 𝑥 (∪ 𝐵‘𝑢) = (𝐹‘(∪ 𝐵 ↾ 𝑢)))
Theorem	tfrlemiex 6334*	Lemma for tfrlemi1 6335. (Contributed by Jim Kingdon, 18-Mar-2019.) (Proof shortened by Mario Carneiro, 24-May-2019.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐹‘𝑔)⟩}))}    &   ⊢ (𝜑 → 𝑥 ∈ On)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝑥 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐹‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∃𝑓(𝑓 Fn 𝑥 ∧ ∀𝑢 ∈ 𝑥 (𝑓‘𝑢) = (𝐹‘(𝑓 ↾ 𝑢))))
Theorem	tfrlemi1 6335*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    ⇒   ⊢ ((𝜑 ∧ 𝐶 ∈ On) → ∃𝑔(𝑔 Fn 𝐶 ∧ ∀𝑢 ∈ 𝐶 (𝑔‘𝑢) = (𝐹‘(𝑔 ↾ 𝑢))))
Theorem	tfrlemi14d 6336*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    ⇒   ⊢ (𝜑 → dom recs(𝐹) = On)
Theorem	tfrexlem 6337*	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 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐹 ∧ (𝐹‘𝑥) ∈ V))    ⇒   ⊢ ((𝜑 ∧ 𝐶 ∈ 𝑉) → (recs(𝐹)‘𝐶) ∈ V)
Theorem	tfri1d 6338*	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 (𝐺‘𝑥) ∈ V. Alternately, ∀𝑥 ∈ On∀𝑓(𝑓 Fn 𝑥 → 𝑓 ∈ 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 ∧ (𝐺‘𝑥) ∈ V))    ⇒   ⊢ (𝜑 → 𝐹 Fn On)
Theorem	tfri2d 6339*	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 6368). 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 ∧ (𝐺‘𝑥) ∈ V))    ⇒   ⊢ ((𝜑 ∧ 𝐴 ∈ On) → (𝐹‘𝐴) = (𝐺‘(𝐹 ↾ 𝐴)))
Theorem	tfr1onlem3ag 6340*	Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3ag 6312 but for tfr1on 6353 related lemmas). (Contributed by Jim Kingdon, 13-Mar-2022.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    ⇒   ⊢ (𝐻 ∈ 𝑉 → (𝐻 ∈ 𝐴 ↔ ∃𝑧 ∈ 𝑋 (𝐻 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝐻‘𝑤) = (𝐺‘(𝐻 ↾ 𝑤)))))
Theorem	tfr1onlem3 6341*	Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3 6314 but for tfr1on 6353 related lemmas). (Contributed by Jim Kingdon, 14-Mar-2022.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    ⇒   ⊢ 𝐴 = {𝑔 ∣ ∃𝑧 ∈ 𝑋 (𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤)))}
Theorem	tfr1onlemssrecs 6342*	Lemma for tfr1on 6353. The union of functions acceptable for tfr1on 6353 is a subset of recs. (Contributed by Jim Kingdon, 15-Mar-2022.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → Ord 𝑋)    ⇒   ⊢ (𝜑 → ∪ 𝐴 ⊆ recs(𝐺))
Theorem	tfr1onlemsucfn 6343*	We can extend an acceptable function by one element to produce a function. Lemma for tfr1on 6353. (Contributed by Jim Kingdon, 12-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓 Fn 𝑥) → (𝐺‘𝑓) ∈ V)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → 𝑧 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑔 Fn 𝑧)    &   ⊢ (𝜑 → 𝑔 ∈ 𝐴)    ⇒   ⊢ (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}) Fn suc 𝑧)
Theorem	tfr1onlemsucaccv 6344*	Lemma for tfr1on 6353. 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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓 Fn 𝑥) → (𝐺‘𝑓) ∈ V)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → 𝑌 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑧 ∈ 𝑌)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑔 Fn 𝑧)    &   ⊢ (𝜑 → 𝑔 ∈ 𝐴)    ⇒   ⊢ (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}) ∈ 𝐴)
Theorem	tfr1onlembacc 6345*	Lemma for tfr1on 6353. Each element of 𝐵 is an acceptable function. (Contributed by Jim Kingdon, 14-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓 Fn 𝑥) → (𝐺‘𝑓) ∈ V)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → 𝐵 ⊆ 𝐴)
Theorem	tfr1onlembxssdm 6346*	Lemma for tfr1on 6353. 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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓 Fn 𝑥) → (𝐺‘𝑓) ∈ V)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → 𝐷 ⊆ dom ∪ 𝐵)
Theorem	tfr1onlembfn 6347*	Lemma for tfr1on 6353. 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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓 Fn 𝑥) → (𝐺‘𝑓) ∈ V)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∪ 𝐵 Fn 𝐷)
Theorem	tfr1onlembex 6348*	Lemma for tfr1on 6353. The set 𝐵 exists. (Contributed by Jim Kingdon, 14-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓 Fn 𝑥) → (𝐺‘𝑓) ∈ V)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → 𝐵 ∈ V)
Theorem	tfr1onlemubacc 6349*	Lemma for tfr1on 6353. The union of 𝐵 satisfies the recursion rule. (Contributed by Jim Kingdon, 15-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓 Fn 𝑥) → (𝐺‘𝑓) ∈ V)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∀𝑢 ∈ 𝐷 (∪ 𝐵‘𝑢) = (𝐺‘(∪ 𝐵 ↾ 𝑢)))
Theorem	tfr1onlemex 6350*	Lemma for tfr1on 6353. (Contributed by Jim Kingdon, 16-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓 Fn 𝑥) → (𝐺‘𝑓) ∈ V)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔 Fn 𝑧 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∃𝑓(𝑓 Fn 𝐷 ∧ ∀𝑢 ∈ 𝐷 (𝑓‘𝑢) = (𝐺‘(𝑓 ↾ 𝑢))))
Theorem	tfr1onlemaccex 6351*	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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓 Fn 𝑥) → (𝐺‘𝑓) ∈ V)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    ⇒   ⊢ ((𝜑 ∧ 𝐶 ∈ 𝑋) → ∃𝑔(𝑔 Fn 𝐶 ∧ ∀𝑢 ∈ 𝐶 (𝑔‘𝑢) = (𝐺‘(𝑔 ↾ 𝑢))))
Theorem	tfr1onlemres 6352*	Lemma for tfr1on 6353. 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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓 Fn 𝑥) → (𝐺‘𝑓) ∈ V)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓 Fn 𝑥 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑌 ∈ 𝑋)    ⇒   ⊢ (𝜑 → 𝑌 ⊆ dom 𝐹)
Theorem	tfr1on 6353*	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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓 Fn 𝑥) → (𝐺‘𝑓) ∈ V)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑌 ∈ 𝑋)    ⇒   ⊢ (𝜑 → 𝑌 ⊆ dom 𝐹)
Theorem	tfri1dALT 6354*	Alternate proof of tfri1d 6338 in terms of tfr1on 6353. Although this does show that the tfr1on 6353 proof is general enough to also prove tfri1d 6338, the tfri1d 6338 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 ∧ (𝐺‘𝑥) ∈ V))    ⇒   ⊢ (𝜑 → 𝐹 Fn On)
Theorem	tfrcllemssrecs 6355*	Lemma for tfrcl 6367. The union of functions acceptable for tfrcl 6367 is a subset of recs. (Contributed by Jim Kingdon, 25-Mar-2022.)
⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → Ord 𝑋)    ⇒   ⊢ (𝜑 → ∪ 𝐴 ⊆ recs(𝐺))
Theorem	tfrcllemsucfn 6356*	We can extend an acceptable function by one element to produce a function. Lemma for tfrcl 6367. (Contributed by Jim Kingdon, 24-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → 𝑧 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑔:𝑧⟶𝑆)    &   ⊢ (𝜑 → 𝑔 ∈ 𝐴)    ⇒   ⊢ (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}):suc 𝑧⟶𝑆)
Theorem	tfrcllemsucaccv 6357*	Lemma for tfrcl 6367. 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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ (𝜑 → 𝑌 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑧 ∈ 𝑌)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑔:𝑧⟶𝑆)    &   ⊢ (𝜑 → 𝑔 ∈ 𝐴)    ⇒   ⊢ (𝜑 → (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}) ∈ 𝐴)
Theorem	tfrcllembacc 6358*	Lemma for tfrcl 6367. Each element of 𝐵 is an acceptable function. (Contributed by Jim Kingdon, 25-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → 𝐵 ⊆ 𝐴)
Theorem	tfrcllembxssdm 6359*	Lemma for tfrcl 6367. 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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → 𝐷 ⊆ dom ∪ 𝐵)
Theorem	tfrcllembfn 6360*	Lemma for tfrcl 6367. 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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∪ 𝐵:𝐷⟶𝑆)
Theorem	tfrcllembex 6361*	Lemma for tfrcl 6367. The set 𝐵 exists. (Contributed by Jim Kingdon, 25-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → 𝐵 ∈ V)
Theorem	tfrcllemubacc 6362*	Lemma for tfrcl 6367. The union of 𝐵 satisfies the recursion rule. (Contributed by Jim Kingdon, 25-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∀𝑢 ∈ 𝐷 (∪ 𝐵‘𝑢) = (𝐺‘(∪ 𝐵 ↾ 𝑢)))
Theorem	tfrcllemex 6363*	Lemma for tfrcl 6367. (Contributed by Jim Kingdon, 26-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∃𝑓(𝑓:𝐷⟶𝑆 ∧ ∀𝑢 ∈ 𝐷 (𝑓‘𝑢) = (𝐺‘(𝑓 ↾ 𝑢))))
Theorem	tfrcllemaccex 6364*	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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    ⇒   ⊢ ((𝜑 ∧ 𝐶 ∈ 𝑋) → ∃𝑔(𝑔:𝐶⟶𝑆 ∧ ∀𝑢 ∈ 𝐶 (𝑔‘𝑢) = (𝐺‘(𝑔 ↾ 𝑢))))
Theorem	tfrcllemres 6365*	Lemma for tfr1on 6353. 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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑌 ∈ 𝑋)    ⇒   ⊢ (𝜑 → 𝑌 ⊆ dom 𝐹)
Theorem	tfrcldm 6366*	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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑌 ∈ ∪ 𝑋)    ⇒   ⊢ (𝜑 → 𝑌 ∈ dom 𝐹)
Theorem	tfrcl 6367*	Closure for transfinite recursion. As with tfr1on 6353, 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)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑌 ∈ ∪ 𝑋)    ⇒   ⊢ (𝜑 → (𝐹‘𝑌) ∈ 𝑆)
Theorem	tfri1 6368*	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 (𝐺‘𝑥) ∈ V. Alternately, ∀𝑥 ∈ On∀𝑓(𝑓 Fn 𝑥 → 𝑓 ∈ 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 ∧ (𝐺‘𝑥) ∈ V)    ⇒   ⊢ 𝐹 Fn On
Theorem	tfri2 6369*	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 6368). 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 ∧ (𝐺‘𝑥) ∈ V)    ⇒   ⊢ (𝐴 ∈ On → (𝐹‘𝐴) = (𝐺‘(𝐹 ↾ 𝐴)))
Theorem	tfri3 6370*	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 6368). 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 ∧ (𝐺‘𝑥) ∈ V)    ⇒   ⊢ ((𝐵 Fn On ∧ ∀𝑥 ∈ On (𝐵‘𝑥) = (𝐺‘(𝐵 ↾ 𝑥))) → 𝐵 = 𝐹)
Theorem	tfrex 6371*	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 ∧ (𝐺‘𝑥) ∈ V))    ⇒   ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑉) → (𝐹‘𝐴) ∈ V)
2.6.21  Recursive definition generator
Syntax	crdg 6372	Extend class notation with the recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class rec(𝐹, 𝐼)
Definition	df-irdg 6373*	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 6308 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 6394 and for suitable characteristic functions df-frec 6394 yields the same result as rec restricted to ω, as seen at frecrdg 6411. 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‘(𝑔‘𝑥)))))
Theorem	rdgeq1 6374	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))
Theorem	rdgeq2 6375	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))
Theorem	rdgfun 6376	The recursive definition generator is a function. (Contributed by Mario Carneiro, 16-Nov-2014.)
⊢ Fun rec(𝐹, 𝐴)
Theorem	rdgtfr 6377*	The recursion rule for the recursive definition generator is defined everywhere. (Contributed by Jim Kingdon, 14-May-2020.)
⊢ ((∀𝑧(𝐹‘𝑧) ∈ V ∧ 𝐴 ∈ 𝑉) → (Fun (𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) ∧ ((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))‘𝑓) ∈ V))
Theorem	rdgruledefgg 6378*	The recursion rule for the recursive definition generator is defined everywhere. (Contributed by Jim Kingdon, 4-Jul-2019.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉) → (Fun (𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) ∧ ((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))‘𝑓) ∈ V))
Theorem	rdgruledefg 6379*	The recursion rule for the recursive definition generator is defined everywhere. (Contributed by Jim Kingdon, 4-Jul-2019.)
⊢ 𝐹 Fn V    ⇒   ⊢ (𝐴 ∈ 𝑉 → (Fun (𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) ∧ ((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))‘𝑓) ∈ V))
Theorem	rdgexggg 6380	The recursive definition generator produces a set on a set input. (Contributed by Jim Kingdon, 4-Jul-2019.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (rec(𝐹, 𝐴)‘𝐵) ∈ V)
Theorem	rdgexgg 6381	The recursive definition generator produces a set on a set input. (Contributed by Jim Kingdon, 4-Jul-2019.)
⊢ 𝐹 Fn V    ⇒   ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (rec(𝐹, 𝐴)‘𝐵) ∈ V)
Theorem	rdgifnon 6382	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 6389; in cases like df-oadd 6423 either presumably could work). (Contributed by Jim Kingdon, 13-Jul-2019.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉) → rec(𝐹, 𝐴) Fn On)
Theorem	rdgifnon2 6383*	The recursive definition generator is a function on ordinal numbers. (Contributed by Jim Kingdon, 14-May-2020.)
⊢ ((∀𝑧(𝐹‘𝑧) ∈ V ∧ 𝐴 ∈ 𝑉) → rec(𝐹, 𝐴) Fn On)
Theorem	rdgivallem 6384*	Value of the recursive definition generator. Lemma for rdgival 6385 which simplifies the value further. (Contributed by Jim Kingdon, 13-Jul-2019.) (New usage is discouraged.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ On) → (rec(𝐹, 𝐴)‘𝐵) = (𝐴 ∪ ∪ 𝑥 ∈ 𝐵 (𝐹‘((rec(𝐹, 𝐴) ↾ 𝐵)‘𝑥))))
Theorem	rdgival 6385*	Value of the recursive definition generator. (Contributed by Jim Kingdon, 26-Jul-2019.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ On) → (rec(𝐹, 𝐴)‘𝐵) = (𝐴 ∪ ∪ 𝑥 ∈ 𝐵 (𝐹‘(rec(𝐹, 𝐴)‘𝑥))))
Theorem	rdgss 6386	Subset and recursive definition generator. (Contributed by Jim Kingdon, 15-Jul-2019.)
⊢ (𝜑 → 𝐹 Fn V)    &   ⊢ (𝜑 → 𝐼 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐴 ∈ On)    &   ⊢ (𝜑 → 𝐵 ∈ On)    &   ⊢ (𝜑 → 𝐴 ⊆ 𝐵)    ⇒   ⊢ (𝜑 → (rec(𝐹, 𝐼)‘𝐴) ⊆ (rec(𝐹, 𝐼)‘𝐵))
Theorem	rdgisuc1 6387*	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 ∪ 𝑥 ∈ 𝐵(𝐹‘(rec(𝐹, 𝐴)‘𝑥)). If we add conditions on the characteristic function, we can show tighter results such as rdgisucinc 6388. (Contributed by Jim Kingdon, 9-Jun-2019.)
⊢ (𝜑 → 𝐹 Fn V)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐵 ∈ On)    ⇒   ⊢ (𝜑 → (rec(𝐹, 𝐴)‘suc 𝐵) = (𝐴 ∪ (∪ 𝑥 ∈ 𝐵 (𝐹‘(rec(𝐹, 𝐴)‘𝑥)) ∪ (𝐹‘(rec(𝐹, 𝐴)‘𝐵)))))
Theorem	rdgisucinc 6388*	Value of the recursive definition generator at a successor. This can be thought of as a generalization of oasuc 6467 and omsuc 6475. (Contributed by Jim Kingdon, 29-Aug-2019.)
⊢ (𝜑 → 𝐹 Fn V)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐵 ∈ On)    &   ⊢ (𝜑 → ∀𝑥 𝑥 ⊆ (𝐹‘𝑥))    ⇒   ⊢ (𝜑 → (rec(𝐹, 𝐴)‘suc 𝐵) = (𝐹‘(rec(𝐹, 𝐴)‘𝐵)))
Theorem	rdgon 6389*	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.)
⊢ (𝜑 → 𝐴 ∈ On)    &   ⊢ (𝜑 → ∀𝑥 ∈ On (𝐹‘𝑥) ∈ On)    ⇒   ⊢ ((𝜑 ∧ 𝐵 ∈ On) → (rec(𝐹, 𝐴)‘𝐵) ∈ On)
Theorem	rdg0 6390	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
Theorem	rdg0g 6391	The initial value of the recursive definition generator. (Contributed by NM, 25-Apr-1995.)
⊢ (𝐴 ∈ 𝐶 → (rec(𝐹, 𝐴)‘∅) = 𝐴)
Theorem	rdgexg 6392	The recursive definition generator produces a set on a set input. (Contributed by Mario Carneiro, 3-Jul-2019.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐹 Fn V    ⇒   ⊢ (𝐵 ∈ 𝑉 → (rec(𝐹, 𝐴)‘𝐵) ∈ V)
2.6.22  Finite recursion
Syntax	cfrec 6393	Extend class notation with the finite recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class frec(𝐹, 𝐼)
Definition	df-frec 6394*	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 6308 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 6400 and frecsuc 6410. 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 4605. 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 6411, this definition and df-irdg 6373 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 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐼))})) ↾ ω)
Theorem	freceq1 6395	Equality theorem for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)
⊢ (𝐹 = 𝐺 → frec(𝐹, 𝐴) = frec(𝐺, 𝐴))
Theorem	freceq2 6396	Equality theorem for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)
⊢ (𝐴 = 𝐵 → frec(𝐹, 𝐴) = frec(𝐹, 𝐵))
Theorem	frecex 6397	Finite recursion produces a set. (Contributed by Jim Kingdon, 20-Aug-2021.)
⊢ frec(𝐹, 𝐴) ∈ V
Theorem	frecfun 6398	Finite recursion produces a function. See also frecfnom 6404 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)
Theorem	nffrec 6399	Bound-variable hypothesis builder for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)
⊢ Ⅎ𝑥𝐹    &   ⊢ Ⅎ𝑥𝐴    ⇒   ⊢ Ⅎ𝑥frec(𝐹, 𝐴)
Theorem	frec0g 6400	The initial value resulting from finite recursive definition generation. (Contributed by Jim Kingdon, 7-May-2020.)
⊢ (𝐴 ∈ 𝑉 → (frec(𝐹, 𝐴)‘∅) = 𝐴)