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	tposf2 6301	The domain and codomain of a transposition. (Contributed by NM, 10-Sep-2015.)
⊢ (Rel 𝐴 → (𝐹:𝐴⟶𝐵 → tpos 𝐹:◡𝐴⟶𝐵))
Theorem	tposf12 6302	Condition for an injective transposition. (Contributed by NM, 10-Sep-2015.)
⊢ (Rel 𝐴 → (𝐹:𝐴–1-1→𝐵 → tpos 𝐹:◡𝐴–1-1→𝐵))
Theorem	tposf1o2 6303	Condition of a bijective transposition. (Contributed by NM, 10-Sep-2015.)
⊢ (Rel 𝐴 → (𝐹:𝐴–1-1-onto→𝐵 → tpos 𝐹:◡𝐴–1-1-onto→𝐵))
Theorem	tposfo 6304	The domain and codomain/range of a transposition. (Contributed by NM, 10-Sep-2015.)
⊢ (𝐹:(𝐴 × 𝐵)–onto→𝐶 → tpos 𝐹:(𝐵 × 𝐴)–onto→𝐶)
Theorem	tposf 6305	The domain and codomain of a transposition. (Contributed by NM, 10-Sep-2015.)
⊢ (𝐹:(𝐴 × 𝐵)⟶𝐶 → tpos 𝐹:(𝐵 × 𝐴)⟶𝐶)
Theorem	tposfn 6306	Functionality of a transposition. (Contributed by Mario Carneiro, 4-Oct-2015.)
⊢ (𝐹 Fn (𝐴 × 𝐵) → tpos 𝐹 Fn (𝐵 × 𝐴))
Theorem	tpos0 6307	Transposition of the empty set. (Contributed by NM, 10-Sep-2015.)
⊢ tpos ∅ = ∅
Theorem	tposco 6308	Transposition of a composition. (Contributed by Mario Carneiro, 4-Oct-2015.)
⊢ tpos (𝐹 ∘ 𝐺) = (𝐹 ∘ tpos 𝐺)
Theorem	tpossym 6309*	Two ways to say a function is symmetric. (Contributed by Mario Carneiro, 4-Oct-2015.)
⊢ (𝐹 Fn (𝐴 × 𝐴) → (tpos 𝐹 = 𝐹 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝐹𝑦) = (𝑦𝐹𝑥)))
Theorem	tposeqi 6310	Equality theorem for transposition. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ 𝐹 = 𝐺    ⇒   ⊢ tpos 𝐹 = tpos 𝐺
Theorem	tposex 6311	A transposition is a set. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ 𝐹 ∈ V    ⇒   ⊢ tpos 𝐹 ∈ V
Theorem	nftpos 6312	Hypothesis builder for transposition. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ Ⅎ𝑥𝐹    ⇒   ⊢ Ⅎ𝑥tpos 𝐹
Theorem	tposoprab 6313*	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}
Theorem	tposmpo 6314*	Transposition of a two-argument mapping. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ 𝐹 = (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐵 ↦ 𝐶)    ⇒   ⊢ tpos 𝐹 = (𝑦 ∈ 𝐵, 𝑥 ∈ 𝐴 ↦ 𝐶)
2.6.18  Undefined values
Theorem	pwuninel2 6315	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.)
⊢ (∪ 𝐴 ∈ 𝑉 → ¬ 𝒫 ∪ 𝐴 ∈ 𝐴)
Theorem	2pwuninelg 6316	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.)
⊢ (𝐴 ∈ 𝑉 → ¬ 𝒫 𝒫 ∪ 𝐴 ∈ 𝐴)
2.6.19  Functions on ordinals; strictly monotone ordinal functions
Theorem	iunon 6317*	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.)
⊢ ((𝐴 ∈ 𝑉 ∧ ∀𝑥 ∈ 𝐴 𝐵 ∈ On) → ∪ 𝑥 ∈ 𝐴 𝐵 ∈ On)
Syntax	wsmo 6318	Introduce the strictly monotone ordinal function. A strictly monotone function is one that is constantly increasing across the ordinals.
wff Smo 𝐴
Definition	df-smo 6319*	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 𝐴∀𝑦 ∈ dom 𝐴(𝑥 ∈ 𝑦 → (𝐴‘𝑥) ∈ (𝐴‘𝑦))))
Theorem	dfsmo2 6320*	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 𝐹∀𝑦 ∈ 𝑥 (𝐹‘𝑦) ∈ (𝐹‘𝑥)))
Theorem	issmo 6321*	Conditions for which 𝐴 is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 15-Nov-2011.)
⊢ 𝐴:𝐵⟶On    &   ⊢ Ord 𝐵    &   ⊢ ((𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵) → (𝑥 ∈ 𝑦 → (𝐴‘𝑥) ∈ (𝐴‘𝑦)))    &   ⊢ dom 𝐴 = 𝐵    ⇒   ⊢ Smo 𝐴
Theorem	issmo2 6322*	Alternate definition of a strictly monotone ordinal function. (Contributed by Mario Carneiro, 12-Mar-2013.)
⊢ (𝐹:𝐴⟶𝐵 → ((𝐵 ⊆ On ∧ Ord 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝑥 (𝐹‘𝑦) ∈ (𝐹‘𝑥)) → Smo 𝐹))
Theorem	smoeq 6323	Equality theorem for strictly monotone functions. (Contributed by Andrew Salmon, 16-Nov-2011.)
⊢ (𝐴 = 𝐵 → (Smo 𝐴 ↔ Smo 𝐵))
Theorem	smodm 6324	The domain of a strictly monotone function is an ordinal. (Contributed by Andrew Salmon, 16-Nov-2011.)
⊢ (Smo 𝐴 → Ord dom 𝐴)
Theorem	smores 6325	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 𝐴 ∧ 𝐵 ∈ dom 𝐴) → Smo (𝐴 ↾ 𝐵))
Theorem	smores3 6326	A strictly monotone function restricted to an ordinal remains strictly monotone. (Contributed by Andrew Salmon, 19-Nov-2011.)
⊢ ((Smo (𝐴 ↾ 𝐵) ∧ 𝐶 ∈ (dom 𝐴 ∩ 𝐵) ∧ Ord 𝐵) → Smo (𝐴 ↾ 𝐶))
Theorem	smores2 6327	A strictly monotone ordinal function restricted to an ordinal is still monotone. (Contributed by Mario Carneiro, 15-Mar-2013.)
⊢ ((Smo 𝐹 ∧ Ord 𝐴) → Smo (𝐹 ↾ 𝐴))
Theorem	smodm2 6328	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)
Theorem	smofvon2dm 6329	The function values of a strictly monotone ordinal function are ordinals. (Contributed by Mario Carneiro, 12-Mar-2013.)
⊢ ((Smo 𝐹 ∧ 𝐵 ∈ dom 𝐹) → (𝐹‘𝐵) ∈ On)
Theorem	iordsmo 6330	The identity relation restricted to the ordinals is a strictly monotone function. (Contributed by Andrew Salmon, 16-Nov-2011.)
⊢ Ord 𝐴    ⇒   ⊢ Smo ( I ↾ 𝐴)
Theorem	smo0 6331	The null set is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 20-Nov-2011.)
⊢ Smo ∅
Theorem	smofvon 6332	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 6333	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 6334*	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 6335	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 6336	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 6337	Notation for a function defined by strong transfinite recursion.
class recs(𝐹)
Definition	df-recs 6338*	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 6403 for more details on why this definition is desirable. Unlike df-irdg 6403 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 6368 and tfri2d 6369 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 6339	Equality theorem for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
⊢ (𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))
Theorem	nfrecs 6340	Bound-variable hypothesis builder for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
⊢ Ⅎ𝑥𝐹    ⇒   ⊢ Ⅎ𝑥recs(𝐹)
Theorem	tfrlem1 6341*	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 6342*	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 6343*	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 6344*	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 6345*	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 6346*	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 6347*	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 6348*	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 6349*	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 6350*	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 6351*	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 6352*	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 6353	Transfinite recursion produces a function. (Contributed by Jim Kingdon, 20-Aug-2021.)
⊢ Fun recs(𝐹)
Theorem	tfr2a 6354	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 6355	Transfinite recursion is defined at the empty set. (Contributed by Jim Kingdon, 8-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    ⇒   ⊢ ((𝐺‘∅) ∈ 𝑉 → ∅ ∈ dom 𝐹)
Theorem	tfr0 6356	Transfinite recursion at the empty set. (Contributed by Jim Kingdon, 8-May-2020.)
⊢ 𝐹 = recs(𝐺)    ⇒   ⊢ ((𝐺‘∅) ∈ 𝑉 → (𝐹‘∅) = (𝐺‘∅))
Theorem	tfrlemisucfn 6357*	We can extend an acceptable function by one element to produce a function. Lemma for tfrlemi1 6365. (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 6358*	We can extend an acceptable function by one element to produce an acceptable function. Lemma for tfrlemi1 6365. (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 6359*	Each element of 𝐵 is an acceptable function. Lemma for tfrlemi1 6365. (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 6360*	The union of 𝐵 is defined on all ordinals. Lemma for tfrlemi1 6365. (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 6361*	The union of 𝐵 is a function defined on 𝑥. Lemma for tfrlemi1 6365. (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 6362*	The set 𝐵 exists. Lemma for tfrlemi1 6365. (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 6363*	The union of 𝐵 satisfies the recursion rule (lemma for tfrlemi1 6365). (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 6364*	Lemma for tfrlemi1 6365. (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 6365*	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 6366*	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 6367*	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 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)    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐺 ∧ (𝐺‘𝑥) ∈ V))    ⇒   ⊢ (𝜑 → 𝐹 Fn On)
Theorem	tfri2d 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 6398). 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 6370*	Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3ag 6342 but for tfr1on 6383 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 6371*	Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3 6344 but for tfr1on 6383 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 6372*	Lemma for tfr1on 6383. The union of functions acceptable for tfr1on 6383 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 6373*	We can extend an acceptable function by one element to produce a function. Lemma for tfr1on 6383. (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 6374*	Lemma for tfr1on 6383. 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 6375*	Lemma for tfr1on 6383. 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 6376*	Lemma for tfr1on 6383. 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 6377*	Lemma for tfr1on 6383. 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 6378*	Lemma for tfr1on 6383. 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 6379*	Lemma for tfr1on 6383. 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 6380*	Lemma for tfr1on 6383. (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 6381*	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 6382*	Lemma for tfr1on 6383. 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 6383*	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 6384*	Alternate proof of tfri1d 6368 in terms of tfr1on 6383. Although this does show that the tfr1on 6383 proof is general enough to also prove tfri1d 6368, the tfri1d 6368 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 6385*	Lemma for tfrcl 6397. The union of functions acceptable for tfrcl 6397 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 6386*	We can extend an acceptable function by one element to produce a function. Lemma for tfrcl 6397. (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 6387*	Lemma for tfrcl 6397. 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 6388*	Lemma for tfrcl 6397. 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 6389*	Lemma for tfrcl 6397. 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 6390*	Lemma for tfrcl 6397. 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 6391*	Lemma for tfrcl 6397. 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 6392*	Lemma for tfrcl 6397. 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 6393*	Lemma for tfrcl 6397. (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 6394*	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 6395*	Lemma for tfr1on 6383. 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 6396*	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 6397*	Closure for transfinite recursion. As with tfr1on 6383, 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 6398*	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 6399*	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 6398). 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 6400*	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 6398). 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)