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Theorem List for Metamath Proof Explorer - 8201-8300   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremmpocurryvald 8201* The value of a curried operation given in maps-to notation is a function over the second argument of the original operation. (Contributed by AV, 27-Oct-2019.)
𝐹 = (𝑥𝑋, 𝑦𝑌𝐶)    &   (𝜑 → ∀𝑥𝑋𝑦𝑌 𝐶𝑉)    &   (𝜑𝑌 ≠ ∅)    &   (𝜑𝑌𝑊)    &   (𝜑𝐴𝑋)       (𝜑 → (curry 𝐹𝐴) = (𝑦𝑌𝐴 / 𝑥𝐶))
 
Theoremfvmpocurryd 8202* The value of the value of a curried operation given in maps-to notation is the operation value of the original operation. (Contributed by AV, 27-Oct-2019.)
𝐹 = (𝑥𝑋, 𝑦𝑌𝐶)    &   (𝜑 → ∀𝑥𝑋𝑦𝑌 𝐶𝑉)    &   (𝜑𝑌𝑊)    &   (𝜑𝐴𝑋)    &   (𝜑𝐵𝑌)       (𝜑 → ((curry 𝐹𝐴)‘𝐵) = (𝐴𝐹𝐵))
 
2.4.16  Undefined values
 
Syntaxcund 8203 Extend class notation with undefined value function.
class Undef
 
Definitiondf-undef 8204 Define the undefined value function, whose value at set 𝑠 is guaranteed not to be a member of 𝑠 (see pwuninel 8206). (Contributed by NM, 15-Sep-2011.)
Undef = (𝑠 ∈ V ↦ 𝒫 𝑠)
 
Theorempwuninel2 8205 Direct proof of pwuninel 8206 avoiding functions and thus several ZF axioms. (Contributed by Stefan O'Rear, 22-Feb-2015.)
( 𝐴𝑉 → ¬ 𝒫 𝐴𝐴)
 
Theorempwuninel 8206 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. See also pwuninel2 8205. (Contributed by NM, 27-Jun-2008.) (Proof shortened by Mario Carneiro, 23-Dec-2016.)
¬ 𝒫 𝐴𝐴
 
Theoremundefval 8207 Value of the undefined value function. Normally we will not reference the explicit value but will use undefnel 8209 instead. (Contributed by NM, 15-Sep-2011.) (Revised by Mario Carneiro, 24-Dec-2016.)
(𝑆𝑉 → (Undef‘𝑆) = 𝒫 𝑆)
 
Theoremundefnel2 8208 The undefined value generated from a set is not a member of the set. (Contributed by NM, 15-Sep-2011.)
(𝑆𝑉 → ¬ (Undef‘𝑆) ∈ 𝑆)
 
Theoremundefnel 8209 The undefined value generated from a set is not a member of the set. (Contributed by NM, 15-Sep-2011.)
(𝑆𝑉 → (Undef‘𝑆) ∉ 𝑆)
 
Theoremundefne0 8210 The undefined value generated from a set is not empty. (Contributed by NM, 3-Sep-2018.)
(𝑆𝑉 → (Undef‘𝑆) ≠ ∅)
 
2.4.17  Well-founded recursion
 
Syntaxcfrecs 8211 Declare the syntax for the well-founded recursion generator. See df-frecs 8212.
class frecs(𝑅, 𝐴, 𝐹)
 
Definitiondf-frecs 8212* This is the definition for the well-founded recursion generator. Similar to df-wrecs 8243 and df-recs 8317, it is a direct definition form of normally recursive relationships. Unlike the former two definitions, it only requires a well-founded set-like relationship for its properties, not a well-ordered relationship. This proof requires either a partial order or the axiom of infinity. We develop the theorems twice, once with a partial order and once without. The second development occurs later in the database, after ax-inf 9574 has been introduced. (Contributed by Scott Fenton, 23-Dec-2021.)
frecs(𝑅, 𝐴, 𝐹) = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐹(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}
 
Theoremfrecseq123 8213 Equality theorem for the well-founded recursion generator. (Contributed by Scott Fenton, 23-Dec-2021.)
((𝑅 = 𝑆𝐴 = 𝐵𝐹 = 𝐺) → frecs(𝑅, 𝐴, 𝐹) = frecs(𝑆, 𝐵, 𝐺))
 
Theoremnffrecs 8214 Bound-variable hypothesis builder for the well-founded recursion generator. (Contributed by Scott Fenton, 23-Dec-2021.)
𝑥𝑅    &   𝑥𝐴    &   𝑥𝐹       𝑥frecs(𝑅, 𝐴, 𝐹)
 
Theoremcsbfrecsg 8215 Move class substitution in and out of the well-founded recursive function generator. (Contributed by Scott Fenton, 18-Nov-2024.)
(𝐴𝑉𝐴 / 𝑥frecs(𝑅, 𝐷, 𝐹) = frecs(𝐴 / 𝑥𝑅, 𝐴 / 𝑥𝐷, 𝐴 / 𝑥𝐹))
 
Theoremfpr3g 8216* Functions defined by well-founded recursion over a partial order are identical up to relation, domain, and characteristic function. This version of frr3g 9692 does not require infinity. (Contributed by Scott Fenton, 24-Aug-2022.)
(((𝑅 Fr 𝐴𝑅 Po 𝐴𝑅 Se 𝐴) ∧ (𝐹 Fn 𝐴 ∧ ∀𝑦𝐴 (𝐹𝑦) = (𝑦𝐻(𝐹 ↾ Pred(𝑅, 𝐴, 𝑦)))) ∧ (𝐺 Fn 𝐴 ∧ ∀𝑦𝐴 (𝐺𝑦) = (𝑦𝐻(𝐺 ↾ Pred(𝑅, 𝐴, 𝑦))))) → 𝐹 = 𝐺)
 
Theoremfrrlem1 8217* Lemma for well-founded recursion. The final item we are interested in is the union of acceptable functions 𝐵. This lemma just changes bound variables for later use. (Contributed by Paul Chapman, 21-Apr-2012.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}       𝐵 = {𝑔 ∣ ∃𝑧(𝑔 Fn 𝑧 ∧ (𝑧𝐴 ∧ ∀𝑤𝑧 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑧) ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝑤𝐺(𝑔 ↾ Pred(𝑅, 𝐴, 𝑤))))}
 
Theoremfrrlem2 8218* Lemma for well-founded recursion. An acceptable function is a function. (Contributed by Paul Chapman, 21-Apr-2012.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}       (𝑔𝐵 → Fun 𝑔)
 
Theoremfrrlem3 8219* Lemma for well-founded recursion. An acceptable function's domain is a subset of 𝐴. (Contributed by Paul Chapman, 21-Apr-2012.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}       (𝑔𝐵 → dom 𝑔𝐴)
 
Theoremfrrlem4 8220* Lemma for well-founded recursion. Properties of the restriction of an acceptable function to the domain of another acceptable function. (Contributed by Paul Chapman, 21-Apr-2012.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}       ((𝑔𝐵𝐵) → ((𝑔 ↾ (dom 𝑔 ∩ dom )) Fn (dom 𝑔 ∩ dom ) ∧ ∀𝑎 ∈ (dom 𝑔 ∩ dom )((𝑔 ↾ (dom 𝑔 ∩ dom ))‘𝑎) = (𝑎𝐺((𝑔 ↾ (dom 𝑔 ∩ dom )) ↾ Pred(𝑅, (dom 𝑔 ∩ dom ), 𝑎)))))
 
Theoremfrrlem5 8221* Lemma for well-founded recursion. State the well-founded recursion generator in terms of the acceptable functions. (Contributed by Scott Fenton, 27-Aug-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐹 = frecs(𝑅, 𝐴, 𝐺)       𝐹 = 𝐵
 
Theoremfrrlem6 8222* Lemma for well-founded recursion. The well-founded recursion generator is a relation. (Contributed by Scott Fenton, 27-Aug-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐹 = frecs(𝑅, 𝐴, 𝐺)       Rel 𝐹
 
Theoremfrrlem7 8223* Lemma for well-founded recursion. The well-founded recursion generator's domain is a subclass of 𝐴. (Contributed by Scott Fenton, 27-Aug-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐹 = frecs(𝑅, 𝐴, 𝐺)       dom 𝐹𝐴
 
Theoremfrrlem8 8224* Lemma for well-founded recursion. dom 𝐹 is closed under predecessor classes. (Contributed by Scott Fenton, 6-Dec-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐹 = frecs(𝑅, 𝐴, 𝐺)       (𝑧 ∈ dom 𝐹 → Pred(𝑅, 𝐴, 𝑧) ⊆ dom 𝐹)
 
Theoremfrrlem9 8225* Lemma for well-founded recursion. Show that the well-founded recursive generator produces a function. Hypothesis three will be eliminated using different induction rules depending on if we use partial orders or the axiom of infinity. (Contributed by Scott Fenton, 27-Aug-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ((𝜑 ∧ (𝑔𝐵𝐵)) → ((𝑥𝑔𝑢𝑥𝑣) → 𝑢 = 𝑣))       (𝜑 → Fun 𝐹)
 
Theoremfrrlem10 8226* Lemma for well-founded recursion. Under the compatibility hypothesis, compute the value of 𝐹 within its domain. (Contributed by Scott Fenton, 6-Dec-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ((𝜑 ∧ (𝑔𝐵𝐵)) → ((𝑥𝑔𝑢𝑥𝑣) → 𝑢 = 𝑣))       ((𝜑𝑦 ∈ dom 𝐹) → (𝐹𝑦) = (𝑦𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑦))))
 
Theoremfrrlem11 8227* Lemma for well-founded recursion. For the next several theorems we will be aiming to prove that dom 𝐹 = 𝐴. To do this, we set up a function 𝐶 that supposedly contains an element of 𝐴 that is not in dom 𝐹 and we show that the element must be in dom 𝐹. Our choice of what to restrict 𝐹 to depends on if we assume partial orders or the axiom of infinity. To begin with, we establish the functionality of 𝐶. (Contributed by Scott Fenton, 7-Dec-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ((𝜑 ∧ (𝑔𝐵𝐵)) → ((𝑥𝑔𝑢𝑥𝑣) → 𝑢 = 𝑣))    &   𝐶 = ((𝐹𝑆) ∪ {⟨𝑧, (𝑧𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})       ((𝜑𝑧 ∈ (𝐴 ∖ dom 𝐹)) → 𝐶 Fn ((𝑆 ∩ dom 𝐹) ∪ {𝑧}))
 
Theoremfrrlem12 8228* Lemma for well-founded recursion. Next, we calculate the value of 𝐶. (Contributed by Scott Fenton, 7-Dec-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ((𝜑 ∧ (𝑔𝐵𝐵)) → ((𝑥𝑔𝑢𝑥𝑣) → 𝑢 = 𝑣))    &   𝐶 = ((𝐹𝑆) ∪ {⟨𝑧, (𝑧𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})    &   (𝜑𝑅 Fr 𝐴)    &   ((𝜑𝑧𝐴) → Pred(𝑅, 𝐴, 𝑧) ⊆ 𝑆)    &   ((𝜑𝑧𝐴) → ∀𝑤𝑆 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑆)       ((𝜑𝑧 ∈ (𝐴 ∖ dom 𝐹) ∧ 𝑤 ∈ ((𝑆 ∩ dom 𝐹) ∪ {𝑧})) → (𝐶𝑤) = (𝑤𝐺(𝐶 ↾ Pred(𝑅, 𝐴, 𝑤))))
 
Theoremfrrlem13 8229* Lemma for well-founded recursion. Assuming that 𝑆 is a subset of 𝐴 and that 𝑧 is 𝑅-minimal, then 𝐶 is an acceptable function. (Contributed by Scott Fenton, 7-Dec-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ((𝜑 ∧ (𝑔𝐵𝐵)) → ((𝑥𝑔𝑢𝑥𝑣) → 𝑢 = 𝑣))    &   𝐶 = ((𝐹𝑆) ∪ {⟨𝑧, (𝑧𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})    &   (𝜑𝑅 Fr 𝐴)    &   ((𝜑𝑧𝐴) → Pred(𝑅, 𝐴, 𝑧) ⊆ 𝑆)    &   ((𝜑𝑧𝐴) → ∀𝑤𝑆 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑆)    &   ((𝜑𝑧𝐴) → 𝑆 ∈ V)    &   ((𝜑𝑧𝐴) → 𝑆𝐴)       ((𝜑 ∧ (𝑧 ∈ (𝐴 ∖ dom 𝐹) ∧ Pred(𝑅, (𝐴 ∖ dom 𝐹), 𝑧) = ∅)) → 𝐶𝐵)
 
Theoremfrrlem14 8230* Lemma for well-founded recursion. Finally, we tie all these threads together and show that dom 𝐹 = 𝐴 when given the right 𝑆. Specifically, we prove that there can be no 𝑅-minimal element of (𝐴 ∖ dom 𝐹). (Contributed by Scott Fenton, 7-Dec-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ((𝜑 ∧ (𝑔𝐵𝐵)) → ((𝑥𝑔𝑢𝑥𝑣) → 𝑢 = 𝑣))    &   𝐶 = ((𝐹𝑆) ∪ {⟨𝑧, (𝑧𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})    &   (𝜑𝑅 Fr 𝐴)    &   ((𝜑𝑧𝐴) → Pred(𝑅, 𝐴, 𝑧) ⊆ 𝑆)    &   ((𝜑𝑧𝐴) → ∀𝑤𝑆 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑆)    &   ((𝜑𝑧𝐴) → 𝑆 ∈ V)    &   ((𝜑𝑧𝐴) → 𝑆𝐴)    &   ((𝜑 ∧ (𝐴 ∖ dom 𝐹) ≠ ∅) → ∃𝑧 ∈ (𝐴 ∖ dom 𝐹)Pred(𝑅, (𝐴 ∖ dom 𝐹), 𝑧) = ∅)       (𝜑 → dom 𝐹 = 𝐴)
 
Theoremfprlem1 8231* Lemma for well-founded recursion with a partial order. Two acceptable functions are compatible. (Contributed by Scott Fenton, 11-Sep-2023.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐹 = frecs(𝑅, 𝐴, 𝐺)       (((𝑅 Fr 𝐴𝑅 Po 𝐴𝑅 Se 𝐴) ∧ (𝑔𝐵𝐵)) → ((𝑥𝑔𝑢𝑥𝑣) → 𝑢 = 𝑣))
 
Theoremfprlem2 8232* Lemma for well-founded recursion with a partial order. Establish a subset relation. (Contributed by Scott Fenton, 11-Sep-2023.)
(((𝑅 Fr 𝐴𝑅 Po 𝐴𝑅 Se 𝐴) ∧ 𝑧𝐴) → ∀𝑤 ∈ Pred (𝑅, 𝐴, 𝑧)Pred(𝑅, 𝐴, 𝑤) ⊆ Pred(𝑅, 𝐴, 𝑧))
 
Theoremfpr2a 8233 Weak version of fpr2 8235 which is useful for proofs that avoid the axiom of replacement. (Contributed by Scott Fenton, 18-Nov-2024.)
𝐹 = frecs(𝑅, 𝐴, 𝐺)       (((𝑅 Fr 𝐴𝑅 Po 𝐴𝑅 Se 𝐴) ∧ 𝑋 ∈ dom 𝐹) → (𝐹𝑋) = (𝑋𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
 
Theoremfpr1 8234 Law of well-founded recursion over a partial order, part one. Establish the functionality and domain of the recursive function generator. Note that by requiring a partial order we can avoid using the axiom of infinity. (Contributed by Scott Fenton, 11-Sep-2023.)
𝐹 = frecs(𝑅, 𝐴, 𝐺)       ((𝑅 Fr 𝐴𝑅 Po 𝐴𝑅 Se 𝐴) → 𝐹 Fn 𝐴)
 
Theoremfpr2 8235 Law of well-founded recursion over a partial order, part two. Now we establish the value of 𝐹 within 𝐴. (Contributed by Scott Fenton, 11-Sep-2023.) (Proof shortened by Scott Fenton, 18-Nov-2024.)
𝐹 = frecs(𝑅, 𝐴, 𝐺)       (((𝑅 Fr 𝐴𝑅 Po 𝐴𝑅 Se 𝐴) ∧ 𝑋𝐴) → (𝐹𝑋) = (𝑋𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
 
Theoremfpr3 8236* Law of well-founded recursion over a partial order, part three. Finally, we show that 𝐹 is unique. We do this by showing that any function 𝐻 with the same properties we proved of 𝐹 in fpr1 8234 and fpr2 8235 is identical to 𝐹. (Contributed by Scott Fenton, 11-Sep-2023.)
𝐹 = frecs(𝑅, 𝐴, 𝐺)       (((𝑅 Fr 𝐴𝑅 Po 𝐴𝑅 Se 𝐴) ∧ (𝐻 Fn 𝐴 ∧ ∀𝑧𝐴 (𝐻𝑧) = (𝑧𝐺(𝐻 ↾ Pred(𝑅, 𝐴, 𝑧))))) → 𝐹 = 𝐻)
 
Theoremfrrrel 8237 Show without using the axiom of replacement that the well-founded recursion generator gives a relation. (Contributed by Scott Fenton, 18-Nov-2024.)
𝐹 = frecs(𝑅, 𝐴, 𝐺)       Rel 𝐹
 
Theoremfrrdmss 8238 Show without using the axiom of replacement that the domain of the well-founded recursion generator is a subclass of 𝐴. (Contributed by Scott Fenton, 18-Nov-2024.)
𝐹 = frecs(𝑅, 𝐴, 𝐺)       dom 𝐹𝐴
 
Theoremfrrdmcl 8239 Show without using the axiom of replacement that for a "function" defined by well-founded recursion, the predecessor class of an element of its domain is a subclass of its domain. (Contributed by Scott Fenton, 21-Apr-2011.) (Proof shortened by Scott Fenton, 17-Nov-2024.)
𝐹 = frecs(𝑅, 𝐴, 𝐺)       (𝑋 ∈ dom 𝐹 → Pred(𝑅, 𝐴, 𝑋) ⊆ dom 𝐹)
 
Theoremfprfung 8240 A "function" defined by well-founded recursion is indeed a function when the relation is a partial order. Avoids the axiom of replacement. (Contributed by Scott Fenton, 18-Nov-2024.)
𝐹 = frecs(𝑅, 𝐴, 𝐺)       ((𝑅 Fr 𝐴𝑅 Po 𝐴𝑅 Se 𝐴) → Fun 𝐹)
 
Theoremfprresex 8241 The restriction of a function defined by well-founded recursion to the predecessor of an element of its domain is a set. Avoids the axiom of replacement. (Contributed by Scott Fenton, 18-Nov-2024.)
𝐹 = frecs(𝑅, 𝐴, 𝐺)       (((𝑅 Fr 𝐴𝑅 Po 𝐴𝑅 Se 𝐴) ∧ 𝑋 ∈ dom 𝐹) → (𝐹 ↾ Pred(𝑅, 𝐴, 𝑋)) ∈ V)
 
2.4.18  Well-ordered recursion
 
Syntaxcwrecs 8242 Declare syntax for the well-ordered recursive function generator.
class wrecs(𝑅, 𝐴, 𝐹)
 
Definitiondf-wrecs 8243 Define the well-ordered recursive function generator. This function takes the usual expressions from recursion theorems and forms a unified definition. Specifically, given a function 𝐹, a relation 𝑅, and a base set 𝐴, this definition generates a function 𝐺 = wrecs(𝑅, 𝐴, 𝐹) that has property that, at any point 𝑥𝐴, (𝐺𝑥) = (𝐹‘(𝐺 ↾ Pred(𝑅, 𝐴, 𝑥))). See wfr1 8281, wfr2 8282, and wfr3 8283. (Contributed by Scott Fenton, 7-Jun-2018.) (Revised by BJ, 27-Oct-2024.)
wrecs(𝑅, 𝐴, 𝐹) = frecs(𝑅, 𝐴, (𝐹 ∘ 2nd ))
 
TheoremdfwrecsOLD 8244* Obsolete definition of the well-ordered recursive function generator as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 7-Jun-2018.)
wrecs(𝑅, 𝐴, 𝐹) = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}
 
Theoremwrecseq123 8245 General equality theorem for the well-ordered recursive function generator. (Contributed by Scott Fenton, 7-Jun-2018.) (Proof shortened by Scott Fenton, 17-Nov-2024.)
((𝑅 = 𝑆𝐴 = 𝐵𝐹 = 𝐺) → wrecs(𝑅, 𝐴, 𝐹) = wrecs(𝑆, 𝐵, 𝐺))
 
Theoremwrecseq123OLD 8246 Obsolete proof of wrecseq123 8245 as of 17-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 7-Jun-2018.)
((𝑅 = 𝑆𝐴 = 𝐵𝐹 = 𝐺) → wrecs(𝑅, 𝐴, 𝐹) = wrecs(𝑆, 𝐵, 𝐺))
 
Theoremnfwrecs 8247 Bound-variable hypothesis builder for the well-ordered recursive function generator. (Contributed by Scott Fenton, 9-Jun-2018.) (Proof shortened by Scott Fenton, 17-Nov-2024.)
𝑥𝑅    &   𝑥𝐴    &   𝑥𝐹       𝑥wrecs(𝑅, 𝐴, 𝐹)
 
TheoremnfwrecsOLD 8248 Obsolete proof of nfwrecs 8247 as of 17-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 9-Jun-2018.)
𝑥𝑅    &   𝑥𝐴    &   𝑥𝐹       𝑥wrecs(𝑅, 𝐴, 𝐹)
 
Theoremwrecseq1 8249 Equality theorem for the well-ordered recursive function generator. (Contributed by Scott Fenton, 7-Jun-2018.)
(𝑅 = 𝑆 → wrecs(𝑅, 𝐴, 𝐹) = wrecs(𝑆, 𝐴, 𝐹))
 
Theoremwrecseq2 8250 Equality theorem for the well-ordered recursive function generator. (Contributed by Scott Fenton, 7-Jun-2018.)
(𝐴 = 𝐵 → wrecs(𝑅, 𝐴, 𝐹) = wrecs(𝑅, 𝐵, 𝐹))
 
Theoremwrecseq3 8251 Equality theorem for the well-ordered recursive function generator. (Contributed by Scott Fenton, 7-Jun-2018.)
(𝐹 = 𝐺 → wrecs(𝑅, 𝐴, 𝐹) = wrecs(𝑅, 𝐴, 𝐺))
 
Theoremcsbwrecsg 8252 Move class substitution in and out of the well-founded recursive function generator. (Contributed by ML, 25-Oct-2020.) (Revised by Scott Fenton, 18-Nov-2024.)
(𝐴𝑉𝐴 / 𝑥wrecs(𝑅, 𝐷, 𝐹) = wrecs(𝐴 / 𝑥𝑅, 𝐴 / 𝑥𝐷, 𝐴 / 𝑥𝐹))
 
Theoremwfr3g 8253* Functions defined by well-ordered recursion are identical up to relation, domain, and characteristic function. (Contributed by Scott Fenton, 11-Feb-2011.)
(((𝑅 We 𝐴𝑅 Se 𝐴) ∧ (𝐹 Fn 𝐴 ∧ ∀𝑦𝐴 (𝐹𝑦) = (𝐻‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑦)))) ∧ (𝐺 Fn 𝐴 ∧ ∀𝑦𝐴 (𝐺𝑦) = (𝐻‘(𝐺 ↾ Pred(𝑅, 𝐴, 𝑦))))) → 𝐹 = 𝐺)
 
Theoremwfrlem1OLD 8254* Lemma for well-ordered recursion. The final item we are interested in is the union of acceptable functions 𝐵. This lemma just changes bound variables for later use. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}       𝐵 = {𝑔 ∣ ∃𝑧(𝑔 Fn 𝑧 ∧ (𝑧𝐴 ∧ ∀𝑤𝑧 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑧) ∧ ∀𝑤𝑧 (𝑔𝑤) = (𝐹‘(𝑔 ↾ Pred(𝑅, 𝐴, 𝑤))))}
 
Theoremwfrlem2OLD 8255* Lemma for well-ordered recursion. An acceptable function is a function. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}       (𝑔𝐵 → Fun 𝑔)
 
Theoremwfrlem3OLD 8256* Lemma for well-ordered recursion. An acceptable function's domain is a subset of 𝐴. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}       (𝑔𝐵 → dom 𝑔𝐴)
 
Theoremwfrlem3OLDa 8257* Lemma for well-ordered recursion. Show membership in the class of acceptable functions. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 31-Jul-2020.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   𝐺 ∈ V       (𝐺𝐵 ↔ ∃𝑧(𝐺 Fn 𝑧 ∧ (𝑧𝐴 ∧ ∀𝑤𝑧 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑧) ∧ ∀𝑤𝑧 (𝐺𝑤) = (𝐹‘(𝐺 ↾ Pred(𝑅, 𝐴, 𝑤)))))
 
Theoremwfrlem4OLD 8258* Lemma for well-ordered recursion. Properties of the restriction of an acceptable function to the domain of another one. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.) (Revised by AV, 18-Jul-2022.)
𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}       ((𝑔𝐵𝐵) → ((𝑔 ↾ (dom 𝑔 ∩ dom )) Fn (dom 𝑔 ∩ dom ) ∧ ∀𝑎 ∈ (dom 𝑔 ∩ dom )((𝑔 ↾ (dom 𝑔 ∩ dom ))‘𝑎) = (𝐹‘((𝑔 ↾ (dom 𝑔 ∩ dom )) ↾ Pred(𝑅, (dom 𝑔 ∩ dom ), 𝑎)))))
 
Theoremwfrlem5OLD 8259* Lemma for well-ordered recursion. The values of two acceptable functions agree within their domains. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}       ((𝑔𝐵𝐵) → ((𝑥𝑔𝑢𝑥𝑣) → 𝑢 = 𝑣))
 
TheoremwfrrelOLD 8260 Obsolete proof of wfrrel 8275 as of 17-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 8-Jun-2018.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       Rel 𝐹
 
TheoremwfrdmssOLD 8261 Obsolete proof of wfrdmss 8276 as of 17-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       dom 𝐹𝐴
 
Theoremwfrlem8OLD 8262 Lemma for well-ordered recursion. Compute the prececessor class for an 𝑅 minimal element of (𝐴 ∖ dom 𝐹). Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       (Pred(𝑅, (𝐴 ∖ dom 𝐹), 𝑋) = ∅ ↔ Pred(𝑅, 𝐴, 𝑋) = Pred(𝑅, dom 𝐹, 𝑋))
 
TheoremwfrdmclOLD 8263 Obsolete proof of wfrdmcl 8277 as of 17-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       (𝑋 ∈ dom 𝐹 → Pred(𝑅, 𝐴, 𝑋) ⊆ dom 𝐹)
 
Theoremwfrlem10OLD 8264* Lemma for well-ordered recursion. When 𝑧 is an 𝑅 minimal element of (𝐴 ∖ dom 𝐹), then its predecessor class is equal to dom 𝐹. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
𝑅 We 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)       ((𝑧 ∈ (𝐴 ∖ dom 𝐹) ∧ Pred(𝑅, (𝐴 ∖ dom 𝐹), 𝑧) = ∅) → Pred(𝑅, 𝐴, 𝑧) = dom 𝐹)
 
TheoremwfrfunOLD 8265 Obsolete proof of wfrfun 8278 as of 17-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)       Fun 𝐹
 
Theoremwfrlem12OLD 8266* Lemma for well-ordered recursion. Here, we compute the value of the recursive definition generator. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)       (𝑦 ∈ dom 𝐹 → (𝐹𝑦) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑦))))
 
Theoremwfrlem13OLD 8267* Lemma for well-ordered recursion. From here through wfrlem16OLD 8270, we aim to prove that dom 𝐹 = 𝐴. We do this by supposing that there is an element 𝑧 of 𝐴 that is not in dom 𝐹. We then define 𝐶 by extending dom 𝐹 with the appropriate value at 𝑧. We then show that 𝑧 cannot be an 𝑅 minimal element of (𝐴 ∖ dom 𝐹), meaning that (𝐴 ∖ dom 𝐹) must be empty, so dom 𝐹 = 𝐴. Here, we show that 𝐶 is a function extending the domain of 𝐹 by one. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)    &   𝐶 = (𝐹 ∪ {⟨𝑧, (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})       (𝑧 ∈ (𝐴 ∖ dom 𝐹) → 𝐶 Fn (dom 𝐹 ∪ {𝑧}))
 
Theoremwfrlem14OLD 8268* Lemma for well-ordered recursion. Compute the value of 𝐶. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)    &   𝐶 = (𝐹 ∪ {⟨𝑧, (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})       (𝑧 ∈ (𝐴 ∖ dom 𝐹) → (𝑦 ∈ (dom 𝐹 ∪ {𝑧}) → (𝐶𝑦) = (𝐺‘(𝐶 ↾ Pred(𝑅, 𝐴, 𝑦)))))
 
Theoremwfrlem15OLD 8269* Lemma for well-ordered recursion. When 𝑧 is 𝑅 minimal, 𝐶 is an acceptable function. This step is where the Axiom of Replacement becomes required. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)    &   𝐶 = (𝐹 ∪ {⟨𝑧, (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})       ((𝑧 ∈ (𝐴 ∖ dom 𝐹) ∧ Pred(𝑅, (𝐴 ∖ dom 𝐹), 𝑧) = ∅) → 𝐶 ∈ {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥𝐴 ∧ ∀𝑦𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐺‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))})
 
Theoremwfrlem16OLD 8270* Lemma for well-ordered recursion. If 𝑧 is 𝑅 minimal in (𝐴 ∖ dom 𝐹), then 𝐶 is acceptable and thus a subset of 𝐹, but dom 𝐶 is bigger than dom 𝐹. Thus, 𝑧 cannot be minimal, so (𝐴 ∖ dom 𝐹) must be empty, and (due to wfrdmssOLD 8261), dom 𝐹 = 𝐴. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)    &   𝐶 = (𝐹 ∪ {⟨𝑧, (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})       dom 𝐹 = 𝐴
 
Theoremwfrlem17OLD 8271 Without using ax-rep 5242, show that all restrictions of wrecs are sets. Obsolete as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 31-Jul-2020.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)       (𝑋 ∈ dom 𝐹 → (𝐹 ↾ Pred(𝑅, 𝐴, 𝑋)) ∈ V)
 
Theoremwfr2aOLD 8272 Obsolete proof of wfr2a 8280 as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 30-Jul-2020.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)       (𝑋 ∈ dom 𝐹 → (𝐹𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
 
Theoremwfr1OLD 8273 Obsolete proof of wfr1 8281 as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 22-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)       𝐹 Fn 𝐴
 
Theoremwfr2OLD 8274 Obsolete proof of wfr2 8282 as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 18-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)       (𝑋𝐴 → (𝐹𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
 
Theoremwfrrel 8275 The well-ordered recursion generator generates a relation. Avoids the axiom of replacement. (Contributed by Scott Fenton, 8-Jun-2018.) (Proof shortened by Scott Fenton, 17-Nov-2024.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       Rel 𝐹
 
Theoremwfrdmss 8276 The domain of the well-ordered recursion generator is a subclass of 𝐴. Avoids the axiom of replacement. (Contributed by Scott Fenton, 21-Apr-2011.) (Proof shortened by Scott Fenton, 17-Nov-2024.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       dom 𝐹𝐴
 
Theoremwfrdmcl 8277 The predecessor class of an element of the well-ordered recursion generator's domain is a subset of its domain. Avoids the axiom of replacement. (Contributed by Scott Fenton, 21-Apr-2011.) (Proof shortened by Scott Fenton, 17-Nov-2024.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       (𝑋 ∈ dom 𝐹 → Pred(𝑅, 𝐴, 𝑋) ⊆ dom 𝐹)
 
Theoremwfrfun 8278 The "function" generated by the well-ordered recursion generator is indeed a function. Avoids the axiom of replacement. (Contributed by Scott Fenton, 21-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.) (Revised by Scott Fenton, 17-Nov-2024.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       ((𝑅 We 𝐴𝑅 Se 𝐴) → Fun 𝐹)
 
Theoremwfrresex 8279 Show without using the axiom of replacement that the restriction of the well-ordered recursion generator to a predecessor class is a set. (Contributed by Scott Fenton, 18-Nov-2024.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       (((𝑅 We 𝐴𝑅 Se 𝐴) ∧ 𝑋 ∈ dom 𝐹) → (𝐹 ↾ Pred(𝑅, 𝐴, 𝑋)) ∈ V)
 
Theoremwfr2a 8280 A weak version of wfr2 8282 which is useful for proofs that avoid the Axiom of Replacement. (Contributed by Scott Fenton, 30-Jul-2020.) (Proof shortened by Scott Fenton, 18-Nov-2024.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       (((𝑅 We 𝐴𝑅 Se 𝐴) ∧ 𝑋 ∈ dom 𝐹) → (𝐹𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
 
Theoremwfr1 8281 The Principle of Well-Ordered Recursion, part 1 of 3. We start with an arbitrary function 𝐺. Then, using a base class 𝐴 and a set-like well-ordering 𝑅 of 𝐴, we define a function 𝐹. This function is said to be defined by "well-ordered recursion". The purpose of these three theorems is to demonstrate the properties of 𝐹. We begin by showing that 𝐹 is a function over 𝐴. (Contributed by Scott Fenton, 22-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.) (Revised by Scott Fenton, 18-Nov-2024.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       ((𝑅 We 𝐴𝑅 Se 𝐴) → 𝐹 Fn 𝐴)
 
Theoremwfr2 8282 The Principle of Well-Ordered Recursion, part 2 of 3. Next, we show that the value of 𝐹 at any 𝑋𝐴 is 𝐺 applied to all "previous" values of 𝐹. (Contributed by Scott Fenton, 18-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       (((𝑅 We 𝐴𝑅 Se 𝐴) ∧ 𝑋𝐴) → (𝐹𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
 
Theoremwfr3 8283* The principle of Well-Ordered Recursion, part 3 of 3. Finally, we show that 𝐹 is unique. We do this by showing that any function 𝐻 with the same properties we proved of 𝐹 in wfr1 8281 and wfr2 8282 is identical to 𝐹. (Contributed by Scott Fenton, 18-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.) (Revised by Scott Fenton, 18-Nov-2024.)
𝐹 = wrecs(𝑅, 𝐴, 𝐺)       (((𝑅 We 𝐴𝑅 Se 𝐴) ∧ (𝐻 Fn 𝐴 ∧ ∀𝑧𝐴 (𝐻𝑧) = (𝐺‘(𝐻 ↾ Pred(𝑅, 𝐴, 𝑧))))) → 𝐹 = 𝐻)
 
Theoremwfr3OLD 8284* Obsolete form of wfr3 8283 as of 18-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 18-Apr-2011.) (Revised by Mario Carneiro, 26-Jun-2015.)
𝑅 We 𝐴    &   𝑅 Se 𝐴    &   𝐹 = wrecs(𝑅, 𝐴, 𝐺)       ((𝐻 Fn 𝐴 ∧ ∀𝑧𝐴 (𝐻𝑧) = (𝐺‘(𝐻 ↾ Pred(𝑅, 𝐴, 𝑧)))) → 𝐹 = 𝐻)
 
2.4.19  Functions on ordinals; strictly monotone ordinal functions
 
Theoremiunon 8285* The indexed union of a set of ordinal numbers 𝐵(𝑥) is an ordinal number. (Contributed by NM, 13-Oct-2003.) (Revised by Mario Carneiro, 5-Dec-2016.)
((𝐴𝑉 ∧ ∀𝑥𝐴 𝐵 ∈ On) → 𝑥𝐴 𝐵 ∈ On)
 
Theoremiinon 8286* The nonempty indexed intersection of a class of ordinal numbers 𝐵(𝑥) is an ordinal number. (Contributed by NM, 13-Oct-2003.) (Proof shortened by Mario Carneiro, 5-Dec-2016.)
((∀𝑥𝐴 𝐵 ∈ On ∧ 𝐴 ≠ ∅) → 𝑥𝐴 𝐵 ∈ On)
 
Theoremonfununi 8287* A property of functions on ordinal numbers. Generalization of Theorem Schema 8E of [Enderton] p. 218. (Contributed by Eric Schmidt, 26-May-2009.)
(Lim 𝑦 → (𝐹𝑦) = 𝑥𝑦 (𝐹𝑥))    &   ((𝑥 ∈ On ∧ 𝑦 ∈ On ∧ 𝑥𝑦) → (𝐹𝑥) ⊆ (𝐹𝑦))       ((𝑆𝑇𝑆 ⊆ On ∧ 𝑆 ≠ ∅) → (𝐹 𝑆) = 𝑥𝑆 (𝐹𝑥))
 
Theoremonovuni 8288* A variant of onfununi 8287 for operations. (Contributed by Eric Schmidt, 26-May-2009.) (Revised by Mario Carneiro, 11-Sep-2015.)
(Lim 𝑦 → (𝐴𝐹𝑦) = 𝑥𝑦 (𝐴𝐹𝑥))    &   ((𝑥 ∈ On ∧ 𝑦 ∈ On ∧ 𝑥𝑦) → (𝐴𝐹𝑥) ⊆ (𝐴𝐹𝑦))       ((𝑆𝑇𝑆 ⊆ On ∧ 𝑆 ≠ ∅) → (𝐴𝐹 𝑆) = 𝑥𝑆 (𝐴𝐹𝑥))
 
Theoremonoviun 8289* A variant of onovuni 8288 with indexed unions. (Contributed by Eric Schmidt, 26-May-2009.) (Proof shortened by Mario Carneiro, 5-Dec-2016.)
(Lim 𝑦 → (𝐴𝐹𝑦) = 𝑥𝑦 (𝐴𝐹𝑥))    &   ((𝑥 ∈ On ∧ 𝑦 ∈ On ∧ 𝑥𝑦) → (𝐴𝐹𝑥) ⊆ (𝐴𝐹𝑦))       ((𝐾𝑇 ∧ ∀𝑧𝐾 𝐿 ∈ On ∧ 𝐾 ≠ ∅) → (𝐴𝐹 𝑧𝐾 𝐿) = 𝑧𝐾 (𝐴𝐹𝐿))
 
Theoremonnseq 8290* There are no length ω decreasing sequences in the ordinals. See also noinfep 9596 for a stronger version assuming Regularity. (Contributed by Mario Carneiro, 19-May-2015.)
((𝐹‘∅) ∈ On → ∃𝑥 ∈ ω ¬ (𝐹‘suc 𝑥) ∈ (𝐹𝑥))
 
Syntaxwsmo 8291 Introduce the strictly monotone ordinal function. A strictly monotone function is one that is constantly increasing across the ordinals.
wff Smo 𝐴
 
Definitiondf-smo 8292* Definition of a strictly monotone ordinal function. Definition 7.46 in [TakeutiZaring] p. 50. (Contributed by Andrew Salmon, 15-Nov-2011.)
(Smo 𝐴 ↔ (𝐴:dom 𝐴⟶On ∧ Ord dom 𝐴 ∧ ∀𝑥 ∈ dom 𝐴𝑦 ∈ dom 𝐴(𝑥𝑦 → (𝐴𝑥) ∈ (𝐴𝑦))))
 
Theoremdfsmo2 8293* Alternate definition of a strictly monotone ordinal function. (Contributed by Mario Carneiro, 4-Mar-2013.)
(Smo 𝐹 ↔ (𝐹:dom 𝐹⟶On ∧ Ord dom 𝐹 ∧ ∀𝑥 ∈ dom 𝐹𝑦𝑥 (𝐹𝑦) ∈ (𝐹𝑥)))
 
Theoremissmo 8294* Conditions for which 𝐴 is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 15-Nov-2011.) Avoid ax-13 2370. (Revised by Gino Giotto, 19-May-2023.)
𝐴:𝐵⟶On    &   Ord 𝐵    &   ((𝑥𝐵𝑦𝐵) → (𝑥𝑦 → (𝐴𝑥) ∈ (𝐴𝑦)))    &   dom 𝐴 = 𝐵       Smo 𝐴
 
Theoremissmo2 8295* Alternate definition of a strictly monotone ordinal function. (Contributed by Mario Carneiro, 12-Mar-2013.)
(𝐹:𝐴𝐵 → ((𝐵 ⊆ On ∧ Ord 𝐴 ∧ ∀𝑥𝐴𝑦𝑥 (𝐹𝑦) ∈ (𝐹𝑥)) → Smo 𝐹))
 
Theoremsmoeq 8296 Equality theorem for strictly monotone functions. (Contributed by Andrew Salmon, 16-Nov-2011.)
(𝐴 = 𝐵 → (Smo 𝐴 ↔ Smo 𝐵))
 
Theoremsmodm 8297 The domain of a strictly monotone function is an ordinal. (Contributed by Andrew Salmon, 16-Nov-2011.)
(Smo 𝐴 → Ord dom 𝐴)
 
Theoremsmores 8298 A strictly monotone function restricted to an ordinal remains strictly monotone. (Contributed by Andrew Salmon, 16-Nov-2011.) (Proof shortened by Mario Carneiro, 5-Dec-2016.)
((Smo 𝐴𝐵 ∈ dom 𝐴) → Smo (𝐴𝐵))
 
Theoremsmores3 8299 A strictly monotone function restricted to an ordinal remains strictly monotone. (Contributed by Andrew Salmon, 19-Nov-2011.)
((Smo (𝐴𝐵) ∧ 𝐶 ∈ (dom 𝐴𝐵) ∧ Ord 𝐵) → Smo (𝐴𝐶))
 
Theoremsmores2 8300 A strictly monotone ordinal function restricted to an ordinal is still monotone. (Contributed by Mario Carneiro, 15-Mar-2013.)
((Smo 𝐹 ∧ Ord 𝐴) → Smo (𝐹𝐴))
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