P. 84 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 8301-8400   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	undefnel 8301	The undefined value generated from a set is not a member of the set. (Contributed by NM, 15-Sep-2011.)
⊢ (𝑆 ∈ 𝑉 → (Undef‘𝑆) ∉ 𝑆)
Theorem	undefne0 8302	The undefined value generated from a set is not empty. (Contributed by NM, 3-Sep-2018.)
⊢ (𝑆 ∈ 𝑉 → (Undef‘𝑆) ≠ ∅)
2.4.17  Well-founded recursion
Syntax	cfrecs 8303	Declare the syntax for the well-founded recursion generator. See df-frecs 8304.
class frecs(𝑅, 𝐴, 𝐹)
Definition	df-frecs 8304*	This is the definition for the well-founded recursion generator. Similar to df-wrecs 8335 and df-recs 8409, 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 9675 has been introduced. (Contributed by Scott Fenton, 23-Dec-2021.)
⊢ frecs(𝑅, 𝐴, 𝐹) = ∪ {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐹(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}
Theorem	frecseq123 8305	Equality theorem for the well-founded recursion generator. (Contributed by Scott Fenton, 23-Dec-2021.)
⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵 ∧ 𝐹 = 𝐺) → frecs(𝑅, 𝐴, 𝐹) = frecs(𝑆, 𝐵, 𝐺))
Theorem	nffrecs 8306	Bound-variable hypothesis builder for the well-founded recursion generator. (Contributed by Scott Fenton, 23-Dec-2021.)
⊢ Ⅎ𝑥𝑅    &   ⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝐹    ⇒   ⊢ Ⅎ𝑥frecs(𝑅, 𝐴, 𝐹)
Theorem	csbfrecsg 8307	Move class substitution in and out of the well-founded recursive function generator. (Contributed by Scott Fenton, 18-Nov-2024.)
⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌frecs(𝑅, 𝐷, 𝐹) = frecs(⦋𝐴 / 𝑥⦌𝑅, ⦋𝐴 / 𝑥⦌𝐷, ⦋𝐴 / 𝑥⦌𝐹))
Theorem	fpr3g 8308*	Functions defined by well-founded recursion over a partial order are identical up to relation, domain, and characteristic function. This version of frr3g 9793 does not require infinity. (Contributed by Scott Fenton, 24-Aug-2022.)
⊢ (((𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴) ∧ (𝐹 Fn 𝐴 ∧ ∀𝑦 ∈ 𝐴 (𝐹‘𝑦) = (𝑦𝐻(𝐹 ↾ Pred(𝑅, 𝐴, 𝑦)))) ∧ (𝐺 Fn 𝐴 ∧ ∀𝑦 ∈ 𝐴 (𝐺‘𝑦) = (𝑦𝐻(𝐺 ↾ Pred(𝑅, 𝐴, 𝑦))))) → 𝐹 = 𝐺)
Theorem	frrlem1 8309*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    ⇒   ⊢ 𝐵 = {𝑔 ∣ ∃𝑧(𝑔 Fn 𝑧 ∧ (𝑧 ⊆ 𝐴 ∧ ∀𝑤 ∈ 𝑧 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑧) ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝑤𝐺(𝑔 ↾ Pred(𝑅, 𝐴, 𝑤))))}
Theorem	frrlem2 8310*	Lemma for well-founded recursion. An acceptable function is a function. (Contributed by Paul Chapman, 21-Apr-2012.)
⊢ 𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    ⇒   ⊢ (𝑔 ∈ 𝐵 → Fun 𝑔)
Theorem	frrlem3 8311*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    ⇒   ⊢ (𝑔 ∈ 𝐵 → dom 𝑔 ⊆ 𝐴)
Theorem	frrlem4 8312*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    ⇒   ⊢ ((𝑔 ∈ 𝐵 ∧ ℎ ∈ 𝐵) → ((𝑔 ↾ (dom 𝑔 ∩ dom ℎ)) Fn (dom 𝑔 ∩ dom ℎ) ∧ ∀𝑎 ∈ (dom 𝑔 ∩ dom ℎ)((𝑔 ↾ (dom 𝑔 ∩ dom ℎ))‘𝑎) = (𝑎𝐺((𝑔 ↾ (dom 𝑔 ∩ dom ℎ)) ↾ Pred(𝑅, (dom 𝑔 ∩ dom ℎ), 𝑎)))))
Theorem	frrlem5 8313*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    ⇒   ⊢ 𝐹 = ∪ 𝐵
Theorem	frrlem6 8314*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    ⇒   ⊢ Rel 𝐹
Theorem	frrlem7 8315*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    ⇒   ⊢ dom 𝐹 ⊆ 𝐴
Theorem	frrlem8 8316*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    ⇒   ⊢ (𝑧 ∈ dom 𝐹 → Pred(𝑅, 𝐴, 𝑧) ⊆ dom 𝐹)
Theorem	frrlem9 8317*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ⊢ ((𝜑 ∧ (𝑔 ∈ 𝐵 ∧ ℎ ∈ 𝐵)) → ((𝑥𝑔𝑢 ∧ 𝑥ℎ𝑣) → 𝑢 = 𝑣))    ⇒   ⊢ (𝜑 → Fun 𝐹)
Theorem	frrlem10 8318*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ⊢ ((𝜑 ∧ (𝑔 ∈ 𝐵 ∧ ℎ ∈ 𝐵)) → ((𝑥𝑔𝑢 ∧ 𝑥ℎ𝑣) → 𝑢 = 𝑣))    ⇒   ⊢ ((𝜑 ∧ 𝑦 ∈ dom 𝐹) → (𝐹‘𝑦) = (𝑦𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑦))))
Theorem	frrlem11 8319*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ⊢ ((𝜑 ∧ (𝑔 ∈ 𝐵 ∧ ℎ ∈ 𝐵)) → ((𝑥𝑔𝑢 ∧ 𝑥ℎ𝑣) → 𝑢 = 𝑣))    &   ⊢ 𝐶 = ((𝐹 ↾ 𝑆) ∪ {⟨𝑧, (𝑧𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})    ⇒   ⊢ ((𝜑 ∧ 𝑧 ∈ (𝐴 ∖ dom 𝐹)) → 𝐶 Fn ((𝑆 ∩ dom 𝐹) ∪ {𝑧}))
Theorem	frrlem12 8320*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ⊢ ((𝜑 ∧ (𝑔 ∈ 𝐵 ∧ ℎ ∈ 𝐵)) → ((𝑥𝑔𝑢 ∧ 𝑥ℎ𝑣) → 𝑢 = 𝑣))    &   ⊢ 𝐶 = ((𝐹 ↾ 𝑆) ∪ {⟨𝑧, (𝑧𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})    &   ⊢ (𝜑 → 𝑅 Fr 𝐴)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝐴) → Pred(𝑅, 𝐴, 𝑧) ⊆ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝐴) → ∀𝑤 ∈ 𝑆 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑆)    ⇒   ⊢ ((𝜑 ∧ 𝑧 ∈ (𝐴 ∖ dom 𝐹) ∧ 𝑤 ∈ ((𝑆 ∩ dom 𝐹) ∪ {𝑧})) → (𝐶‘𝑤) = (𝑤𝐺(𝐶 ↾ Pred(𝑅, 𝐴, 𝑤))))
Theorem	frrlem13 8321*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ⊢ ((𝜑 ∧ (𝑔 ∈ 𝐵 ∧ ℎ ∈ 𝐵)) → ((𝑥𝑔𝑢 ∧ 𝑥ℎ𝑣) → 𝑢 = 𝑣))    &   ⊢ 𝐶 = ((𝐹 ↾ 𝑆) ∪ {⟨𝑧, (𝑧𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})    &   ⊢ (𝜑 → 𝑅 Fr 𝐴)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝐴) → Pred(𝑅, 𝐴, 𝑧) ⊆ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝐴) → ∀𝑤 ∈ 𝑆 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝐴) → 𝑆 ∈ V)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝐴) → 𝑆 ⊆ 𝐴)    ⇒   ⊢ ((𝜑 ∧ (𝑧 ∈ (𝐴 ∖ dom 𝐹) ∧ Pred(𝑅, (𝐴 ∖ dom 𝐹), 𝑧) = ∅)) → 𝐶 ∈ 𝐵)
Theorem	frrlem14 8322*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    &   ⊢ ((𝜑 ∧ (𝑔 ∈ 𝐵 ∧ ℎ ∈ 𝐵)) → ((𝑥𝑔𝑢 ∧ 𝑥ℎ𝑣) → 𝑢 = 𝑣))    &   ⊢ 𝐶 = ((𝐹 ↾ 𝑆) ∪ {⟨𝑧, (𝑧𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑧)))⟩})    &   ⊢ (𝜑 → 𝑅 Fr 𝐴)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝐴) → Pred(𝑅, 𝐴, 𝑧) ⊆ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝐴) → ∀𝑤 ∈ 𝑆 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝐴) → 𝑆 ∈ V)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝐴) → 𝑆 ⊆ 𝐴)    &   ⊢ ((𝜑 ∧ (𝐴 ∖ dom 𝐹) ≠ ∅) → ∃𝑧 ∈ (𝐴 ∖ dom 𝐹)Pred(𝑅, (𝐴 ∖ dom 𝐹), 𝑧) = ∅)    ⇒   ⊢ (𝜑 → dom 𝐹 = 𝐴)
Theorem	fprlem1 8323*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    ⇒   ⊢ (((𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴) ∧ (𝑔 ∈ 𝐵 ∧ ℎ ∈ 𝐵)) → ((𝑥𝑔𝑢 ∧ 𝑥ℎ𝑣) → 𝑢 = 𝑣))
Theorem	fprlem2 8324*	Lemma for well-founded recursion with a partial order. Establish a subset relation. (Contributed by Scott Fenton, 11-Sep-2023.)
⊢ (((𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴) ∧ 𝑧 ∈ 𝐴) → ∀𝑤 ∈ Pred (𝑅, 𝐴, 𝑧)Pred(𝑅, 𝐴, 𝑤) ⊆ Pred(𝑅, 𝐴, 𝑧))
Theorem	fpr2a 8325	Weak version of fpr2 8327 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 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴) ∧ 𝑋 ∈ dom 𝐹) → (𝐹‘𝑋) = (𝑋𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
Theorem	fpr1 8326	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 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴) → 𝐹 Fn 𝐴)
Theorem	fpr2 8327	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 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴) ∧ 𝑋 ∈ 𝐴) → (𝐹‘𝑋) = (𝑋𝐺(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
Theorem	fpr3 8328*	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 8326 and fpr2 8327 is identical to 𝐹. (Contributed by Scott Fenton, 11-Sep-2023.)
⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)    ⇒   ⊢ (((𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴) ∧ (𝐻 Fn 𝐴 ∧ ∀𝑧 ∈ 𝐴 (𝐻‘𝑧) = (𝑧𝐺(𝐻 ↾ Pred(𝑅, 𝐴, 𝑧))))) → 𝐹 = 𝐻)
Theorem	frrrel 8329	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
Theorem	frrdmss 8330	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 𝐹 ⊆ 𝐴
Theorem	frrdmcl 8331	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)
Theorem	fprfung 8332	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 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴) → Fun 𝐹)
Theorem	fprresex 8333	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 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴) ∧ 𝑋 ∈ dom 𝐹) → (𝐹 ↾ Pred(𝑅, 𝐴, 𝑋)) ∈ V)
2.4.18  Well-ordered recursion
Syntax	cwrecs 8334	Declare syntax for the well-ordered recursive function generator.
class wrecs(𝑅, 𝐴, 𝐹)
Definition	df-wrecs 8335	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 𝑥 ∈ 𝐴, (𝐺‘𝑥) = (𝐹‘(𝐺 ↾ Pred(𝑅, 𝐴, 𝑥))). See wfr1 8373, wfr2 8374, and wfr3 8375. (Contributed by Scott Fenton, 7-Jun-2018.) (Revised by BJ, 27-Oct-2024.)
⊢ wrecs(𝑅, 𝐴, 𝐹) = frecs(𝑅, 𝐴, (𝐹 ∘ 2nd ))
Theorem	dfwrecsOLD 8336*	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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}
Theorem	wrecseq123 8337	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 = 𝑆 ∧ 𝐴 = 𝐵 ∧ 𝐹 = 𝐺) → wrecs(𝑅, 𝐴, 𝐹) = wrecs(𝑆, 𝐵, 𝐺))
Theorem	wrecseq123OLD 8338	Obsolete version of wrecseq123 8337 as of 17-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 7-Jun-2018.)
⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵 ∧ 𝐹 = 𝐺) → wrecs(𝑅, 𝐴, 𝐹) = wrecs(𝑆, 𝐵, 𝐺))
Theorem	nfwrecs 8339	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.)
⊢ Ⅎ𝑥𝑅    &   ⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝐹    ⇒   ⊢ Ⅎ𝑥wrecs(𝑅, 𝐴, 𝐹)
Theorem	nfwrecsOLD 8340	Obsolete version of nfwrecs 8339 as of 17-Nov-2024. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 9-Jun-2018.)
⊢ Ⅎ𝑥𝑅    &   ⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝐹    ⇒   ⊢ Ⅎ𝑥wrecs(𝑅, 𝐴, 𝐹)
Theorem	wrecseq1 8341	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))
Theorem	wrecseq2 8342	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))
Theorem	wrecseq3 8343	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))
Theorem	csbwrecsg 8344	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.)
⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌wrecs(𝑅, 𝐷, 𝐹) = wrecs(⦋𝐴 / 𝑥⦌𝑅, ⦋𝐴 / 𝑥⦌𝐷, ⦋𝐴 / 𝑥⦌𝐹))
Theorem	wfr3g 8345*	Functions defined by well-ordered recursion are identical up to relation, domain, and characteristic function. (Contributed by Scott Fenton, 11-Feb-2011.)
⊢ (((𝑅 We 𝐴 ∧ 𝑅 Se 𝐴) ∧ (𝐹 Fn 𝐴 ∧ ∀𝑦 ∈ 𝐴 (𝐹‘𝑦) = (𝐻‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑦)))) ∧ (𝐺 Fn 𝐴 ∧ ∀𝑦 ∈ 𝐴 (𝐺‘𝑦) = (𝐻‘(𝐺 ↾ Pred(𝑅, 𝐴, 𝑦))))) → 𝐹 = 𝐺)
Theorem	wfrlem1OLD 8346*	Obsolete version as of 18-Nov-2024. 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. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
⊢ 𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    ⇒   ⊢ 𝐵 = {𝑔 ∣ ∃𝑧(𝑔 Fn 𝑧 ∧ (𝑧 ⊆ 𝐴 ∧ ∀𝑤 ∈ 𝑧 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑧) ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐹‘(𝑔 ↾ Pred(𝑅, 𝐴, 𝑤))))}
Theorem	wfrlem2OLD 8347*	Obsolete version as of 18-Nov-2024. Lemma for well-ordered recursion. An acceptable function is a function. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
⊢ 𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    ⇒   ⊢ (𝑔 ∈ 𝐵 → Fun 𝑔)
Theorem	wfrlem3OLD 8348*	Obsolete version as of 18-Nov-2024. Lemma for well-ordered recursion. An acceptable function's domain is a subset of 𝐴. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 21-Apr-2011.)
⊢ 𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    ⇒   ⊢ (𝑔 ∈ 𝐵 → dom 𝑔 ⊆ 𝐴)
Theorem	wfrlem3OLDa 8349*	Obsolete version as of 18-Nov-2024. Lemma for well-ordered recursion. Show membership in the class of acceptable functions. (New usage is discouraged.) (Proof modification is discouraged.) (Contributed by Scott Fenton, 31-Jul-2020.)
⊢ 𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    &   ⊢ 𝐺 ∈ V    ⇒   ⊢ (𝐺 ∈ 𝐵 ↔ ∃𝑧(𝐺 Fn 𝑧 ∧ (𝑧 ⊆ 𝐴 ∧ ∀𝑤 ∈ 𝑧 Pred(𝑅, 𝐴, 𝑤) ⊆ 𝑧) ∧ ∀𝑤 ∈ 𝑧 (𝐺‘𝑤) = (𝐹‘(𝐺 ↾ Pred(𝑅, 𝐴, 𝑤)))))
Theorem	wfrlem4OLD 8350*	Obsolete version as of 18-Nov-2024. Lemma for well-ordered recursion. Properties of the restriction of an acceptable function to the domain of another one. (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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    ⇒   ⊢ ((𝑔 ∈ 𝐵 ∧ ℎ ∈ 𝐵) → ((𝑔 ↾ (dom 𝑔 ∩ dom ℎ)) Fn (dom 𝑔 ∩ dom ℎ) ∧ ∀𝑎 ∈ (dom 𝑔 ∩ dom ℎ)((𝑔 ↾ (dom 𝑔 ∩ dom ℎ))‘𝑎) = (𝐹‘((𝑔 ↾ (dom 𝑔 ∩ dom ℎ)) ↾ Pred(𝑅, (dom 𝑔 ∩ dom ℎ), 𝑎)))))
Theorem	wfrlem5OLD 8351*	Obsolete version as of 18-Nov-2024. Lemma for well-ordered recursion. The values of two acceptable functions agree within their domains. (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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐹‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}    ⇒   ⊢ ((𝑔 ∈ 𝐵 ∧ ℎ ∈ 𝐵) → ((𝑥𝑔𝑢 ∧ 𝑥ℎ𝑣) → 𝑢 = 𝑣))
Theorem	wfrrelOLD 8352	Obsolete version of wfrrel 8367 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
Theorem	wfrdmssOLD 8353	Obsolete version of wfrdmss 8368 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 𝐹 ⊆ 𝐴
Theorem	wfrlem8OLD 8354	Obsolete version as of 18-Nov-2024. Lemma for well-ordered recursion. Compute the predecessor class for an 𝑅 minimal element of (𝐴 ∖ dom 𝐹). (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))
Theorem	wfrdmclOLD 8355	Obsolete version of wfrdmcl 8369 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)
Theorem	wfrlem10OLD 8356*	Obsolete version as of 18-Nov-2024. 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. (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)
Theorem	wfrfunOLD 8357	Obsolete version of wfrfun 8370 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
Theorem	wfrlem12OLD 8358*	Obsolete version as of 18-Nov-2024. Lemma for well-ordered recursion. Here, we compute the value of the recursive definition generator. (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 → (𝐹‘𝑦) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑦))))
Theorem	wfrlem13OLD 8359*	Obsolete version as of 18-Nov-2024. Lemma for well-ordered recursion. From here through wfrlem16OLD 8362, 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. (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}))
Theorem	wfrlem14OLD 8360*	Obsolete version as of 18-Nov-2024. Lemma for well-ordered recursion. Compute the value of 𝐶. (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}) → (𝐶‘𝑦) = (𝐺‘(𝐶 ↾ Pred(𝑅, 𝐴, 𝑦)))))
Theorem	wfrlem15OLD 8361*	Obsolete version as of 18-Nov-2024. 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. (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 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))})
Theorem	wfrlem16OLD 8362*	Obsolete version as of 18-Nov-2024. 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 8353), dom 𝐹 = 𝐴. (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
Theorem	wfrlem17OLD 8363	Obsolete version as of 18-Nov-2024. Without using ax-rep 5284, show that all restrictions of wrecs are sets. (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)
Theorem	wfr2aOLD 8364	Obsolete version of wfr2a 8372 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 → (𝐹‘𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
Theorem	wfr1OLD 8365	Obsolete version of wfr1 8373 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
Theorem	wfr2OLD 8366	Obsolete version of wfr2 8374 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)    ⇒   ⊢ (𝑋 ∈ 𝐴 → (𝐹‘𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
Theorem	wfrrel 8367	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
Theorem	wfrdmss 8368	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 𝐹 ⊆ 𝐴
Theorem	wfrdmcl 8369	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)
Theorem	wfrfun 8370	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 𝐴 ∧ 𝑅 Se 𝐴) → Fun 𝐹)
Theorem	wfrresex 8371	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 𝐴 ∧ 𝑅 Se 𝐴) ∧ 𝑋 ∈ dom 𝐹) → (𝐹 ↾ Pred(𝑅, 𝐴, 𝑋)) ∈ V)
Theorem	wfr2a 8372	A weak version of wfr2 8374 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 𝐴 ∧ 𝑅 Se 𝐴) ∧ 𝑋 ∈ dom 𝐹) → (𝐹‘𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
Theorem	wfr1 8373	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 𝐴 ∧ 𝑅 Se 𝐴) → 𝐹 Fn 𝐴)
Theorem	wfr2 8374	The Principle of Well-Ordered Recursion, part 2 of 3. Next, we show that the value of 𝐹 at any 𝑋 ∈ 𝐴 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 𝐴 ∧ 𝑅 Se 𝐴) ∧ 𝑋 ∈ 𝐴) → (𝐹‘𝑋) = (𝐺‘(𝐹 ↾ Pred(𝑅, 𝐴, 𝑋))))
Theorem	wfr3 8375*	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 8373 and wfr2 8374 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 𝐴 ∧ 𝑅 Se 𝐴) ∧ (𝐻 Fn 𝐴 ∧ ∀𝑧 ∈ 𝐴 (𝐻‘𝑧) = (𝐺‘(𝐻 ↾ Pred(𝑅, 𝐴, 𝑧))))) → 𝐹 = 𝐻)
Theorem	wfr3OLD 8376*	Obsolete form of wfr3 8375 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 ∧ ∀𝑧 ∈ 𝐴 (𝐻‘𝑧) = (𝐺‘(𝐻 ↾ Pred(𝑅, 𝐴, 𝑧)))) → 𝐹 = 𝐻)
2.4.19  Functions on ordinals; strictly monotone ordinal functions
Theorem	iunon 8377*	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)
Theorem	iinon 8378*	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.)
⊢ ((∀𝑥 ∈ 𝐴 𝐵 ∈ On ∧ 𝐴 ≠ ∅) → ∩ 𝑥 ∈ 𝐴 𝐵 ∈ On)
Theorem	onfununi 8379*	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 → (𝐹‘𝑦) = ∪ 𝑥 ∈ 𝑦 (𝐹‘𝑥))    &   ⊢ ((𝑥 ∈ On ∧ 𝑦 ∈ On ∧ 𝑥 ⊆ 𝑦) → (𝐹‘𝑥) ⊆ (𝐹‘𝑦))    ⇒   ⊢ ((𝑆 ∈ 𝑇 ∧ 𝑆 ⊆ On ∧ 𝑆 ≠ ∅) → (𝐹‘∪ 𝑆) = ∪ 𝑥 ∈ 𝑆 (𝐹‘𝑥))
Theorem	onovuni 8380*	A variant of onfununi 8379 for operations. (Contributed by Eric Schmidt, 26-May-2009.) (Revised by Mario Carneiro, 11-Sep-2015.)
⊢ (Lim 𝑦 → (𝐴𝐹𝑦) = ∪ 𝑥 ∈ 𝑦 (𝐴𝐹𝑥))    &   ⊢ ((𝑥 ∈ On ∧ 𝑦 ∈ On ∧ 𝑥 ⊆ 𝑦) → (𝐴𝐹𝑥) ⊆ (𝐴𝐹𝑦))    ⇒   ⊢ ((𝑆 ∈ 𝑇 ∧ 𝑆 ⊆ On ∧ 𝑆 ≠ ∅) → (𝐴𝐹∪ 𝑆) = ∪ 𝑥 ∈ 𝑆 (𝐴𝐹𝑥))
Theorem	onoviun 8381*	A variant of onovuni 8380 with indexed unions. (Contributed by Eric Schmidt, 26-May-2009.) (Proof shortened by Mario Carneiro, 5-Dec-2016.)
⊢ (Lim 𝑦 → (𝐴𝐹𝑦) = ∪ 𝑥 ∈ 𝑦 (𝐴𝐹𝑥))    &   ⊢ ((𝑥 ∈ On ∧ 𝑦 ∈ On ∧ 𝑥 ⊆ 𝑦) → (𝐴𝐹𝑥) ⊆ (𝐴𝐹𝑦))    ⇒   ⊢ ((𝐾 ∈ 𝑇 ∧ ∀𝑧 ∈ 𝐾 𝐿 ∈ On ∧ 𝐾 ≠ ∅) → (𝐴𝐹∪ 𝑧 ∈ 𝐾 𝐿) = ∪ 𝑧 ∈ 𝐾 (𝐴𝐹𝐿))
Theorem	onnseq 8382*	There are no length ω decreasing sequences in the ordinals. See also noinfep 9697 for a stronger version assuming Regularity. (Contributed by Mario Carneiro, 19-May-2015.)
⊢ ((𝐹‘∅) ∈ On → ∃𝑥 ∈ ω ¬ (𝐹‘suc 𝑥) ∈ (𝐹‘𝑥))
Syntax	wsmo 8383	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 8384*	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 8385*	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 8386*	Conditions for which 𝐴 is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 15-Nov-2011.) Avoid ax-13 2374. (Revised by GG, 19-May-2023.)
⊢ 𝐴:𝐵⟶On    &   ⊢ Ord 𝐵    &   ⊢ ((𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵) → (𝑥 ∈ 𝑦 → (𝐴‘𝑥) ∈ (𝐴‘𝑦)))    &   ⊢ dom 𝐴 = 𝐵    ⇒   ⊢ Smo 𝐴
Theorem	issmo2 8387*	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 8388	Equality theorem for strictly monotone functions. (Contributed by Andrew Salmon, 16-Nov-2011.)
⊢ (𝐴 = 𝐵 → (Smo 𝐴 ↔ Smo 𝐵))
Theorem	smodm 8389	The domain of a strictly monotone function is an ordinal. (Contributed by Andrew Salmon, 16-Nov-2011.)
⊢ (Smo 𝐴 → Ord dom 𝐴)
Theorem	smores 8390	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 8391	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 8392	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 8393	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	smofvon2 8394	The function values of a strictly monotone ordinal function are ordinals. (Contributed by Mario Carneiro, 12-Mar-2013.)
⊢ (Smo 𝐹 → (𝐹‘𝐵) ∈ On)
Theorem	iordsmo 8395	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 8396	The null set is a strictly monotone ordinal function. (Contributed by Andrew Salmon, 20-Nov-2011.)
⊢ Smo ∅
Theorem	smofvon 8397	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 8398	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 8399*	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 8400	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)