P. 66 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 6501-6600   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	rntpos 6501	The range of tpos 𝐹 when dom 𝐹 is a relation. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ (Rel dom 𝐹 → ran tpos 𝐹 = ran 𝐹)
Theorem	tposexg 6502	The transposition of a set is a set. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ (𝐹 ∈ 𝑉 → tpos 𝐹 ∈ V)
Theorem	ovtposg 6503	The transposition swaps the arguments in a two-argument function. When 𝐹 is a matrix, which is to say a function from ( 1 ... m ) × ( 1 ... n ) to the reals or some ring, tpos 𝐹 is the transposition of 𝐹, which is where the name comes from. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴tpos 𝐹𝐵) = (𝐵𝐹𝐴))
Theorem	tposfun 6504	The transposition of a function is a function. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ (Fun 𝐹 → Fun tpos 𝐹)
Theorem	dftpos2 6505*	Alternate definition of tpos when 𝐹 has relational domain. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ (Rel dom 𝐹 → tpos 𝐹 = (𝐹 ∘ (𝑥 ∈ ◡dom 𝐹 ↦ ∪ ◡{𝑥})))
Theorem	dftpos3 6506*	Alternate definition of tpos when 𝐹 has relational domain. Compare df-cnv 4762. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ (Rel dom 𝐹 → tpos 𝐹 = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ⟨𝑦, 𝑥⟩𝐹𝑧})
Theorem	dftpos4 6507*	Alternate definition of tpos. (Contributed by Mario Carneiro, 4-Oct-2015.)
⊢ tpos 𝐹 = (𝐹 ∘ (𝑥 ∈ ((V × V) ∪ {∅}) ↦ ∪ ◡{𝑥}))
Theorem	tpostpos 6508	Value of the double transposition for a general class 𝐹. (Contributed by Mario Carneiro, 16-Sep-2015.)
⊢ tpos tpos 𝐹 = (𝐹 ∩ (((V × V) ∪ {∅}) × V))
Theorem	tpostpos2 6509	Value of the double transposition for a relation on triples. (Contributed by Mario Carneiro, 16-Sep-2015.)
⊢ ((Rel 𝐹 ∧ Rel dom 𝐹) → tpos tpos 𝐹 = 𝐹)
Theorem	tposfn2 6510	The domain of a transposition. (Contributed by NM, 10-Sep-2015.)
⊢ (Rel 𝐴 → (𝐹 Fn 𝐴 → tpos 𝐹 Fn ◡𝐴))
Theorem	tposfo2 6511	Condition for a surjective transposition. (Contributed by NM, 10-Sep-2015.)
⊢ (Rel 𝐴 → (𝐹:𝐴–onto→𝐵 → tpos 𝐹:◡𝐴–onto→𝐵))
Theorem	tposf2 6512	The domain and codomain of a transposition. (Contributed by NM, 10-Sep-2015.)
⊢ (Rel 𝐴 → (𝐹:𝐴⟶𝐵 → tpos 𝐹:◡𝐴⟶𝐵))
Theorem	tposf12 6513	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 6514	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 6515	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 6516	The domain and codomain of a transposition. (Contributed by NM, 10-Sep-2015.)
⊢ (𝐹:(𝐴 × 𝐵)⟶𝐶 → tpos 𝐹:(𝐵 × 𝐴)⟶𝐶)
Theorem	tposfn 6517	Functionality of a transposition. (Contributed by Mario Carneiro, 4-Oct-2015.)
⊢ (𝐹 Fn (𝐴 × 𝐵) → tpos 𝐹 Fn (𝐵 × 𝐴))
Theorem	tpos0 6518	Transposition of the empty set. (Contributed by NM, 10-Sep-2015.)
⊢ tpos ∅ = ∅
Theorem	tposco 6519	Transposition of a composition. (Contributed by Mario Carneiro, 4-Oct-2015.)
⊢ tpos (𝐹 ∘ 𝐺) = (𝐹 ∘ tpos 𝐺)
Theorem	tpossym 6520*	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 6521	Equality theorem for transposition. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ 𝐹 = 𝐺    ⇒   ⊢ tpos 𝐹 = tpos 𝐺
Theorem	tposex 6522	A transposition is a set. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ 𝐹 ∈ V    ⇒   ⊢ tpos 𝐹 ∈ V
Theorem	nftpos 6523	Hypothesis builder for transposition. (Contributed by Mario Carneiro, 10-Sep-2015.)
⊢ Ⅎ𝑥𝐹    ⇒   ⊢ Ⅎ𝑥tpos 𝐹
Theorem	tposoprab 6524*	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 6525*	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)
2.6.19  Undefined values
Theorem	pwuninel2 6526	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 6527	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)
2.6.20  Functions on ordinals; strictly monotone ordinal functions
Theorem	iunon 6528*	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 6529	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 6530*	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 6531*	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 6532*	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 6533*	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 6534	Equality theorem for strictly monotone functions. (Contributed by Andrew Salmon, 16-Nov-2011.)
⊢ (𝐴 = 𝐵 → (Smo 𝐴 ↔ Smo 𝐵))
Theorem	smodm 6535	The domain of a strictly monotone function is an ordinal. (Contributed by Andrew Salmon, 16-Nov-2011.)
⊢ (Smo 𝐴 → Ord dom 𝐴)
Theorem	smores 6536	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 6537	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 6538	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 6539	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 6540	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 6541	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 6542	The null set is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 20-Nov-2011.)
⊢ Smo ∅
Theorem	smofvon 6543	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 6544	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 6545*	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 6546	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 6547	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))
2.6.21  "Strong" transfinite recursion
Syntax	crecs 6548	Notation for a function defined by strong transfinite recursion.
class recs(𝐹)
Definition	df-recs 6549*	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 6614 for more details on why this definition is desirable. Unlike df-irdg 6614 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 6579 and tfri2d 6580 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 6550	Equality theorem for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
⊢ (𝐹 = 𝐺 → recs(𝐹) = recs(𝐺))
Theorem	nfrecs 6551	Bound-variable hypothesis builder for recs. (Contributed by Stefan O'Rear, 18-Jan-2015.)
⊢ Ⅎ𝑥𝐹    ⇒   ⊢ Ⅎ𝑥recs(𝐹)
Theorem	tfrlem1 6552*	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 6553*	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 6554*	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 6555*	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 6556*	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 6557*	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 6558*	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 6559*	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 6560*	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 6561*	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 6562*	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 6563*	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 6564	Transfinite recursion produces a function. (Contributed by Jim Kingdon, 20-Aug-2021.)
⊢ Fun recs(𝐹)
Theorem	tfr2a 6565	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 6566	Transfinite recursion is defined at the empty set. (Contributed by Jim Kingdon, 8-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    ⇒   ⊢ ((𝐺‘∅) ∈ 𝑉 → ∅ ∈ dom 𝐹)
Theorem	tfr0 6567	Transfinite recursion at the empty set. (Contributed by Jim Kingdon, 8-May-2020.)
⊢ 𝐹 = recs(𝐺)    ⇒   ⊢ ((𝐺‘∅) ∈ 𝑉 → (𝐹‘∅) = (𝐺‘∅))
Theorem	tfrlemisucfn 6568*	We can extend an acceptable function by one element to produce a function. Lemma for tfrlemi1 6576. (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 6569*	We can extend an acceptable function by one element to produce an acceptable function. Lemma for tfrlemi1 6576. (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 6570*	Each element of 𝐵 is an acceptable function. Lemma for tfrlemi1 6576. (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 6571*	The union of 𝐵 is defined on all ordinals. Lemma for tfrlemi1 6576. (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 6572*	The union of 𝐵 is a function defined on 𝑥. Lemma for tfrlemi1 6576. (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 6573*	The set 𝐵 exists. Lemma for tfrlemi1 6576. (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 6574*	The union of 𝐵 satisfies the recursion rule (lemma for tfrlemi1 6576). (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 6575*	Lemma for tfrlemi1 6576. (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 6576*	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 6577*	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 6578*	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 6579*	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 6580*	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 6609). 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 6581*	Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3ag 6553 but for tfr1on 6594 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 6582*	Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3 6555 but for tfr1on 6594 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 6583*	Lemma for tfr1on 6594. The union of functions acceptable for tfr1on 6594 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 6584*	We can extend an acceptable function by one element to produce a function. Lemma for tfr1on 6594. (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 6585*	Lemma for tfr1on 6594. 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 6586*	Lemma for tfr1on 6594. 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 6587*	Lemma for tfr1on 6594. 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 6588*	Lemma for tfr1on 6594. 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 6589*	Lemma for tfr1on 6594. 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 6590*	Lemma for tfr1on 6594. 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 6591*	Lemma for tfr1on 6594. (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 6592*	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 6593*	Lemma for tfr1on 6594. 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 6594*	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 6595*	Alternate proof of tfri1d 6579 in terms of tfr1on 6594. Although this does show that the tfr1on 6594 proof is general enough to also prove tfri1d 6579, the tfri1d 6579 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 6596*	Lemma for tfrcl 6608. The union of functions acceptable for tfrcl 6608 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 6597*	We can extend an acceptable function by one element to produce a function. Lemma for tfrcl 6608. (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 6598*	Lemma for tfrcl 6608. 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 6599*	Lemma for tfrcl 6608. 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 6600*	Lemma for tfrcl 6608. 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)