P. 64 of Theorem List - Intuitionistic Logic Explorer

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

Type	Label	Description
Statement
Theorem	tfr0dm 6301	Transfinite recursion is defined at the empty set. (Contributed by Jim Kingdon, 8-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    ⇒   ⊢ ((𝐺‘∅) ∈ 𝑉 → ∅ ∈ dom 𝐹)
Theorem	tfr0 6302	Transfinite recursion at the empty set. (Contributed by Jim Kingdon, 8-May-2020.)
⊢ 𝐹 = recs(𝐺)    ⇒   ⊢ ((𝐺‘∅) ∈ 𝑉 → (𝐹‘∅) = (𝐺‘∅))
Theorem	tfrlemisucfn 6303*	We can extend an acceptable function by one element to produce a function. Lemma for tfrlemi1 6311. (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 6304*	We can extend an acceptable function by one element to produce an acceptable function. Lemma for tfrlemi1 6311. (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 6305*	Each element of 𝐵 is an acceptable function. Lemma for tfrlemi1 6311. (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 6306*	The union of 𝐵 is defined on all ordinals. Lemma for tfrlemi1 6311. (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 6307*	The union of 𝐵 is a function defined on 𝑥. Lemma for tfrlemi1 6311. (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 6308*	The set 𝐵 exists. Lemma for tfrlemi1 6311. (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 6309*	The union of 𝐵 satisfies the recursion rule (lemma for tfrlemi1 6311). (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 6310*	Lemma for tfrlemi1 6311. (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 6311*	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 6312*	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 6313*	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 6314*	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 6315*	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 6344). 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 6316*	Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3ag 6288 but for tfr1on 6329 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 6317*	Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3 6290 but for tfr1on 6329 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 6318*	Lemma for tfr1on 6329. The union of functions acceptable for tfr1on 6329 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 6319*	We can extend an acceptable function by one element to produce a function. Lemma for tfr1on 6329. (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 6320*	Lemma for tfr1on 6329. 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 6321*	Lemma for tfr1on 6329. 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 6322*	Lemma for tfr1on 6329. 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 6323*	Lemma for tfr1on 6329. 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 6324*	Lemma for tfr1on 6329. 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 6325*	Lemma for tfr1on 6329. 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 6326*	Lemma for tfr1on 6329. (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 6327*	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 6328*	Lemma for tfr1on 6329. 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 6329*	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 6330*	Alternate proof of tfri1d 6314 in terms of tfr1on 6329. Although this does show that the tfr1on 6329 proof is general enough to also prove tfri1d 6314, the tfri1d 6314 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 6331*	Lemma for tfrcl 6343. The union of functions acceptable for tfrcl 6343 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 6332*	We can extend an acceptable function by one element to produce a function. Lemma for tfrcl 6343. (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 6333*	Lemma for tfrcl 6343. 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 6334*	Lemma for tfrcl 6343. 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 6335*	Lemma for tfrcl 6343. The union of 𝐵 is defined on all elements of 𝑋. (Contributed by Jim Kingdon, 25-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → 𝐷 ⊆ dom ∪ 𝐵)
Theorem	tfrcllembfn 6336*	Lemma for tfrcl 6343. The union of 𝐵 is a function defined on 𝑥. (Contributed by Jim Kingdon, 25-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∪ 𝐵:𝐷⟶𝑆)
Theorem	tfrcllembex 6337*	Lemma for tfrcl 6343. The set 𝐵 exists. (Contributed by Jim Kingdon, 25-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → 𝐵 ∈ V)
Theorem	tfrcllemubacc 6338*	Lemma for tfrcl 6343. The union of 𝐵 satisfies the recursion rule. (Contributed by Jim Kingdon, 25-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∀𝑢 ∈ 𝐷 (∪ 𝐵‘𝑢) = (𝐺‘(∪ 𝐵 ↾ 𝑢)))
Theorem	tfrcllemex 6339*	Lemma for tfrcl 6343. (Contributed by Jim Kingdon, 26-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ 𝐵 = {ℎ ∣ ∃𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ 𝑔 ∈ 𝐴 ∧ ℎ = (𝑔 ∪ {⟨𝑧, (𝐺‘𝑔)⟩}))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝐷 ∈ 𝑋)    &   ⊢ (𝜑 → ∀𝑧 ∈ 𝐷 ∃𝑔(𝑔:𝑧⟶𝑆 ∧ ∀𝑤 ∈ 𝑧 (𝑔‘𝑤) = (𝐺‘(𝑔 ↾ 𝑤))))    ⇒   ⊢ (𝜑 → ∃𝑓(𝑓:𝐷⟶𝑆 ∧ ∀𝑢 ∈ 𝐷 (𝑓‘𝑢) = (𝐺‘(𝑓 ↾ 𝑢))))
Theorem	tfrcllemaccex 6340*	We can define an acceptable function on any element of 𝑋. As with many of the transfinite recursion theorems, we have hypotheses that state that 𝐹 is a function and that it is defined up to 𝑋. (Contributed by Jim Kingdon, 26-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    ⇒   ⊢ ((𝜑 ∧ 𝐶 ∈ 𝑋) → ∃𝑔(𝑔:𝐶⟶𝑆 ∧ ∀𝑢 ∈ 𝐶 (𝑔‘𝑢) = (𝐺‘(𝑔 ↾ 𝑢))))
Theorem	tfrcllemres 6341*	Lemma for tfr1on 6329. Recursion is defined on an ordinal if the characteristic function is defined up to a suitable point. (Contributed by Jim Kingdon, 18-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ 𝐴 = {𝑓 ∣ ∃𝑥 ∈ 𝑋 (𝑓:𝑥⟶𝑆 ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝐺‘(𝑓 ↾ 𝑦)))}    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑌 ∈ 𝑋)    ⇒   ⊢ (𝜑 → 𝑌 ⊆ dom 𝐹)
Theorem	tfrcldm 6342*	Recursion is defined on an ordinal if the characteristic function satisfies a closure hypothesis up to a suitable point. (Contributed by Jim Kingdon, 26-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑌 ∈ ∪ 𝑋)    ⇒   ⊢ (𝜑 → 𝑌 ∈ dom 𝐹)
Theorem	tfrcl 6343*	Closure for transfinite recursion. As with tfr1on 6329, the characteristic function must be defined up to a suitable point, not necessarily on all ordinals. (Contributed by Jim Kingdon, 25-Mar-2022.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → Fun 𝐺)    &   ⊢ (𝜑 → Ord 𝑋)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ 𝑋 ∧ 𝑓:𝑥⟶𝑆) → (𝐺‘𝑓) ∈ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ∪ 𝑋) → suc 𝑥 ∈ 𝑋)    &   ⊢ (𝜑 → 𝑌 ∈ ∪ 𝑋)    ⇒   ⊢ (𝜑 → (𝐹‘𝑌) ∈ 𝑆)
Theorem	tfri1 6344*	Principle of Transfinite Recursion, part 1 of 3. Theorem 7.41(1) of [TakeutiZaring] p. 47, with an additional condition. The condition is that 𝐺 is defined "everywhere", which is stated here as (𝐺‘𝑥) ∈ V. Alternately, ∀𝑥 ∈ On∀𝑓(𝑓 Fn 𝑥 → 𝑓 ∈ dom 𝐺) would suffice. Given a function 𝐺 satisfying that condition, we define a class 𝐴 of all "acceptable" functions. The final function we're interested in is the union 𝐹 = recs(𝐺) of them. 𝐹 is then said to be defined by transfinite recursion. The purpose of the 3 parts of this theorem is to demonstrate properties of 𝐹. In this first part we show that 𝐹 is a function whose domain is all ordinal numbers. (Contributed by Jim Kingdon, 4-May-2019.) (Revised by Mario Carneiro, 24-May-2019.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (Fun 𝐺 ∧ (𝐺‘𝑥) ∈ V)    ⇒   ⊢ 𝐹 Fn On
Theorem	tfri2 6345*	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 6344). Here we show that the function 𝐹 has the property that for any function 𝐺 satisfying that condition, the "next" value of 𝐹 is 𝐺 recursively applied to all "previous" values of 𝐹. (Contributed by Jim Kingdon, 4-May-2019.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (Fun 𝐺 ∧ (𝐺‘𝑥) ∈ V)    ⇒   ⊢ (𝐴 ∈ On → (𝐹‘𝐴) = (𝐺‘(𝐹 ↾ 𝐴)))
Theorem	tfri3 6346*	Principle of Transfinite Recursion, part 3 of 3. Theorem 7.41(3) of [TakeutiZaring] p. 47, with an additional condition on the recursion rule 𝐺 ( as described at tfri1 6344). Finally, we show that 𝐹 is unique. We do this by showing that any class 𝐵 with the same properties of 𝐹 that we showed in parts 1 and 2 is identical to 𝐹. (Contributed by Jim Kingdon, 4-May-2019.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (Fun 𝐺 ∧ (𝐺‘𝑥) ∈ V)    ⇒   ⊢ ((𝐵 Fn On ∧ ∀𝑥 ∈ On (𝐵‘𝑥) = (𝐺‘(𝐵 ↾ 𝑥))) → 𝐵 = 𝐹)
Theorem	tfrex 6347*	The transfinite recursion function is set-like if the input is. (Contributed by Mario Carneiro, 3-Jul-2019.)
⊢ 𝐹 = recs(𝐺)    &   ⊢ (𝜑 → ∀𝑥(Fun 𝐺 ∧ (𝐺‘𝑥) ∈ V))    ⇒   ⊢ ((𝜑 ∧ 𝐴 ∈ 𝑉) → (𝐹‘𝐴) ∈ V)
2.6.21  Recursive definition generator
Syntax	crdg 6348	Extend class notation with the recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class rec(𝐹, 𝐼)
Definition	df-irdg 6349*	Define a recursive definition generator on On (the class of ordinal numbers) with characteristic function 𝐹 and initial value 𝐼. This rather amazing operation allows us to define, with compact direct definitions, functions that are usually defined in textbooks only with indirect self-referencing recursive definitions. A recursive definition requires advanced metalogic to justify - in particular, eliminating a recursive definition is very difficult and often not even shown in textbooks. On the other hand, the elimination of a direct definition is a matter of simple mechanical substitution. The price paid is the daunting complexity of our rec operation (especially when df-recs 6284 that it is built on is also eliminated). But once we get past this hurdle, definitions that would otherwise be recursive become relatively simple. In classical logic it would be easier to divide this definition into cases based on whether the domain of 𝑔 is zero, a successor, or a limit ordinal. Cases do not (in general) work that way in intuitionistic logic, so instead we choose a definition which takes the union of all the results of the characteristic function for ordinals in the domain of 𝑔. This means that this definition has the expected properties for increasing and continuous ordinal functions, which include ordinal addition and multiplication. For finite recursion we also define df-frec 6370 and for suitable characteristic functions df-frec 6370 yields the same result as rec restricted to ω, as seen at frecrdg 6387. Note: We introduce rec with the philosophical goal of being able to eliminate all definitions with direct mechanical substitution and to verify easily the soundness of definitions. Metamath itself has no built-in technical limitation that prevents multiple-part recursive definitions in the traditional textbook style. (Contributed by Jim Kingdon, 19-May-2019.)
⊢ rec(𝐹, 𝐼) = recs((𝑔 ∈ V ↦ (𝐼 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
Theorem	rdgeq1 6350	Equality theorem for the recursive definition generator. (Contributed by NM, 9-Apr-1995.) (Revised by Mario Carneiro, 9-May-2015.)
⊢ (𝐹 = 𝐺 → rec(𝐹, 𝐴) = rec(𝐺, 𝐴))
Theorem	rdgeq2 6351	Equality theorem for the recursive definition generator. (Contributed by NM, 9-Apr-1995.) (Revised by Mario Carneiro, 9-May-2015.)
⊢ (𝐴 = 𝐵 → rec(𝐹, 𝐴) = rec(𝐹, 𝐵))
Theorem	rdgfun 6352	The recursive definition generator is a function. (Contributed by Mario Carneiro, 16-Nov-2014.)
⊢ Fun rec(𝐹, 𝐴)
Theorem	rdgtfr 6353*	The recursion rule for the recursive definition generator is defined everywhere. (Contributed by Jim Kingdon, 14-May-2020.)
⊢ ((∀𝑧(𝐹‘𝑧) ∈ V ∧ 𝐴 ∈ 𝑉) → (Fun (𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) ∧ ((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))‘𝑓) ∈ V))
Theorem	rdgruledefgg 6354*	The recursion rule for the recursive definition generator is defined everywhere. (Contributed by Jim Kingdon, 4-Jul-2019.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉) → (Fun (𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) ∧ ((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))‘𝑓) ∈ V))
Theorem	rdgruledefg 6355*	The recursion rule for the recursive definition generator is defined everywhere. (Contributed by Jim Kingdon, 4-Jul-2019.)
⊢ 𝐹 Fn V    ⇒   ⊢ (𝐴 ∈ 𝑉 → (Fun (𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) ∧ ((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))‘𝑓) ∈ V))
Theorem	rdgexggg 6356	The recursive definition generator produces a set on a set input. (Contributed by Jim Kingdon, 4-Jul-2019.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (rec(𝐹, 𝐴)‘𝐵) ∈ V)
Theorem	rdgexgg 6357	The recursive definition generator produces a set on a set input. (Contributed by Jim Kingdon, 4-Jul-2019.)
⊢ 𝐹 Fn V    ⇒   ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (rec(𝐹, 𝐴)‘𝐵) ∈ V)
Theorem	rdgifnon 6358	The recursive definition generator is a function on ordinal numbers. The 𝐹 Fn V condition states that the characteristic function is defined for all sets (being defined for all ordinals might be enough if being used in a manner similar to rdgon 6365; in cases like df-oadd 6399 either presumably could work). (Contributed by Jim Kingdon, 13-Jul-2019.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉) → rec(𝐹, 𝐴) Fn On)
Theorem	rdgifnon2 6359*	The recursive definition generator is a function on ordinal numbers. (Contributed by Jim Kingdon, 14-May-2020.)
⊢ ((∀𝑧(𝐹‘𝑧) ∈ V ∧ 𝐴 ∈ 𝑉) → rec(𝐹, 𝐴) Fn On)
Theorem	rdgivallem 6360*	Value of the recursive definition generator. Lemma for rdgival 6361 which simplifies the value further. (Contributed by Jim Kingdon, 13-Jul-2019.) (New usage is discouraged.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ On) → (rec(𝐹, 𝐴)‘𝐵) = (𝐴 ∪ ∪ 𝑥 ∈ 𝐵 (𝐹‘((rec(𝐹, 𝐴) ↾ 𝐵)‘𝑥))))
Theorem	rdgival 6361*	Value of the recursive definition generator. (Contributed by Jim Kingdon, 26-Jul-2019.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ On) → (rec(𝐹, 𝐴)‘𝐵) = (𝐴 ∪ ∪ 𝑥 ∈ 𝐵 (𝐹‘(rec(𝐹, 𝐴)‘𝑥))))
Theorem	rdgss 6362	Subset and recursive definition generator. (Contributed by Jim Kingdon, 15-Jul-2019.)
⊢ (𝜑 → 𝐹 Fn V)    &   ⊢ (𝜑 → 𝐼 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐴 ∈ On)    &   ⊢ (𝜑 → 𝐵 ∈ On)    &   ⊢ (𝜑 → 𝐴 ⊆ 𝐵)    ⇒   ⊢ (𝜑 → (rec(𝐹, 𝐼)‘𝐴) ⊆ (rec(𝐹, 𝐼)‘𝐵))
Theorem	rdgisuc1 6363*	One way of describing the value of the recursive definition generator at a successor. There is no condition on the characteristic function 𝐹 other than 𝐹 Fn V. Given that, the resulting expression encompasses both the expected successor term (𝐹‘(rec(𝐹, 𝐴)‘𝐵)) but also terms that correspond to the initial value 𝐴 and to limit ordinals ∪ 𝑥 ∈ 𝐵(𝐹‘(rec(𝐹, 𝐴)‘𝑥)). If we add conditions on the characteristic function, we can show tighter results such as rdgisucinc 6364. (Contributed by Jim Kingdon, 9-Jun-2019.)
⊢ (𝜑 → 𝐹 Fn V)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐵 ∈ On)    ⇒   ⊢ (𝜑 → (rec(𝐹, 𝐴)‘suc 𝐵) = (𝐴 ∪ (∪ 𝑥 ∈ 𝐵 (𝐹‘(rec(𝐹, 𝐴)‘𝑥)) ∪ (𝐹‘(rec(𝐹, 𝐴)‘𝐵)))))
Theorem	rdgisucinc 6364*	Value of the recursive definition generator at a successor. This can be thought of as a generalization of oasuc 6443 and omsuc 6451. (Contributed by Jim Kingdon, 29-Aug-2019.)
⊢ (𝜑 → 𝐹 Fn V)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐵 ∈ On)    &   ⊢ (𝜑 → ∀𝑥 𝑥 ⊆ (𝐹‘𝑥))    ⇒   ⊢ (𝜑 → (rec(𝐹, 𝐴)‘suc 𝐵) = (𝐹‘(rec(𝐹, 𝐴)‘𝐵)))
Theorem	rdgon 6365*	Evaluating the recursive definition generator produces an ordinal. There is a hypothesis that the characteristic function produces ordinals on ordinal arguments. (Contributed by Jim Kingdon, 26-Jul-2019.) (Revised by Jim Kingdon, 13-Apr-2022.)
⊢ (𝜑 → 𝐴 ∈ On)    &   ⊢ (𝜑 → ∀𝑥 ∈ On (𝐹‘𝑥) ∈ On)    ⇒   ⊢ ((𝜑 ∧ 𝐵 ∈ On) → (rec(𝐹, 𝐴)‘𝐵) ∈ On)
Theorem	rdg0 6366	The initial value of the recursive definition generator. (Contributed by NM, 23-Apr-1995.) (Revised by Mario Carneiro, 14-Nov-2014.)
⊢ 𝐴 ∈ V    ⇒   ⊢ (rec(𝐹, 𝐴)‘∅) = 𝐴
Theorem	rdg0g 6367	The initial value of the recursive definition generator. (Contributed by NM, 25-Apr-1995.)
⊢ (𝐴 ∈ 𝐶 → (rec(𝐹, 𝐴)‘∅) = 𝐴)
Theorem	rdgexg 6368	The recursive definition generator produces a set on a set input. (Contributed by Mario Carneiro, 3-Jul-2019.)
⊢ 𝐴 ∈ V    &   ⊢ 𝐹 Fn V    ⇒   ⊢ (𝐵 ∈ 𝑉 → (rec(𝐹, 𝐴)‘𝐵) ∈ V)
2.6.22  Finite recursion
Syntax	cfrec 6369	Extend class notation with the finite recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class frec(𝐹, 𝐼)
Definition	df-frec 6370*	Define a recursive definition generator on ω (the class of finite ordinals) with characteristic function 𝐹 and initial value 𝐼. This rather amazing operation allows us to define, with compact direct definitions, functions that are usually defined in textbooks only with indirect self-referencing recursive definitions. A recursive definition requires advanced metalogic to justify - in particular, eliminating a recursive definition is very difficult and often not even shown in textbooks. On the other hand, the elimination of a direct definition is a matter of simple mechanical substitution. The price paid is the daunting complexity of our frec operation (especially when df-recs 6284 that it is built on is also eliminated). But once we get past this hurdle, definitions that would otherwise be recursive become relatively simple; see frec0g 6376 and frecsuc 6386. Unlike with transfinite recursion, finite recurson can readily divide definitions and proofs into zero and successor cases, because even without excluded middle we have theorems such as nn0suc 4588. The analogous situation with transfinite recursion - being able to say that an ordinal is zero, successor, or limit - is enabled by excluded middle and thus is not available to us. For the characteristic functions which satisfy the conditions given at frecrdg 6387, this definition and df-irdg 6349 restricted to ω produce the same result. Note: We introduce frec with the philosophical goal of being able to eliminate all definitions with direct mechanical substitution and to verify easily the soundness of definitions. Metamath itself has no built-in technical limitation that prevents multiple-part recursive definitions in the traditional textbook style. (Contributed by Mario Carneiro and Jim Kingdon, 10-Aug-2019.)
⊢ frec(𝐹, 𝐼) = (recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐼))})) ↾ ω)
Theorem	freceq1 6371	Equality theorem for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)
⊢ (𝐹 = 𝐺 → frec(𝐹, 𝐴) = frec(𝐺, 𝐴))
Theorem	freceq2 6372	Equality theorem for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)
⊢ (𝐴 = 𝐵 → frec(𝐹, 𝐴) = frec(𝐹, 𝐵))
Theorem	frecex 6373	Finite recursion produces a set. (Contributed by Jim Kingdon, 20-Aug-2021.)
⊢ frec(𝐹, 𝐴) ∈ V
Theorem	frecfun 6374	Finite recursion produces a function. See also frecfnom 6380 which also states that the domain of that function is ω but which puts conditions on 𝐴 and 𝐹. (Contributed by Jim Kingdon, 13-Feb-2022.)
⊢ Fun frec(𝐹, 𝐴)
Theorem	nffrec 6375	Bound-variable hypothesis builder for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)
⊢ Ⅎ𝑥𝐹    &   ⊢ Ⅎ𝑥𝐴    ⇒   ⊢ Ⅎ𝑥frec(𝐹, 𝐴)
Theorem	frec0g 6376	The initial value resulting from finite recursive definition generation. (Contributed by Jim Kingdon, 7-May-2020.)
⊢ (𝐴 ∈ 𝑉 → (frec(𝐹, 𝐴)‘∅) = 𝐴)
Theorem	frecabex 6377*	The class abstraction from df-frec 6370 exists. This is a lemma for other finite recursion proofs. (Contributed by Jim Kingdon, 13-May-2020.)
⊢ (𝜑 → 𝑆 ∈ 𝑉)    &   ⊢ (𝜑 → ∀𝑦(𝐹‘𝑦) ∈ V)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑊)    ⇒   ⊢ (𝜑 → {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑆 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑆‘𝑚))) ∨ (dom 𝑆 = ∅ ∧ 𝑥 ∈ 𝐴))} ∈ V)
Theorem	frecabcl 6378*	The class abstraction from df-frec 6370 exists. Unlike frecabex 6377 the function 𝐹 only needs to be defined on 𝑆, not all sets. This is a lemma for other finite recursion proofs. (Contributed by Jim Kingdon, 21-Mar-2022.)
⊢ (𝜑 → 𝑁 ∈ ω)    &   ⊢ (𝜑 → 𝐺:𝑁⟶𝑆)    &   ⊢ (𝜑 → ∀𝑦 ∈ 𝑆 (𝐹‘𝑦) ∈ 𝑆)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑆)    ⇒   ⊢ (𝜑 → {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝐺 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝐺‘𝑚))) ∨ (dom 𝐺 = ∅ ∧ 𝑥 ∈ 𝐴))} ∈ 𝑆)
Theorem	frectfr 6379*	Lemma to connect transfinite recursion theorems with finite recursion. That is, given the conditions 𝐹 Fn V and 𝐴 ∈ 𝑉 on frec(𝐹, 𝐴), we want to be able to apply tfri1d 6314 or tfri2d 6315, and this lemma lets us satisfy hypotheses of those theorems. (Contributed by Jim Kingdon, 15-Aug-2019.)
⊢ 𝐺 = (𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴))})    ⇒   ⊢ ((∀𝑧(𝐹‘𝑧) ∈ V ∧ 𝐴 ∈ 𝑉) → ∀𝑦(Fun 𝐺 ∧ (𝐺‘𝑦) ∈ V))
Theorem	frecfnom 6380*	The function generated by finite recursive definition generation is a function on omega. (Contributed by Jim Kingdon, 13-May-2020.)
⊢ ((∀𝑧(𝐹‘𝑧) ∈ V ∧ 𝐴 ∈ 𝑉) → frec(𝐹, 𝐴) Fn ω)
Theorem	freccllem 6381*	Lemma for freccl 6382. Just giving a name to a common expression to simplify the proof. (Contributed by Jim Kingdon, 27-Mar-2022.)
⊢ (𝜑 → 𝐴 ∈ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝑆) → (𝐹‘𝑧) ∈ 𝑆)    &   ⊢ (𝜑 → 𝐵 ∈ ω)    &   ⊢ 𝐺 = recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴))}))    ⇒   ⊢ (𝜑 → (frec(𝐹, 𝐴)‘𝐵) ∈ 𝑆)
Theorem	freccl 6382*	Closure for finite recursion. (Contributed by Jim Kingdon, 27-Mar-2022.)
⊢ (𝜑 → 𝐴 ∈ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑧 ∈ 𝑆) → (𝐹‘𝑧) ∈ 𝑆)    &   ⊢ (𝜑 → 𝐵 ∈ ω)    ⇒   ⊢ (𝜑 → (frec(𝐹, 𝐴)‘𝐵) ∈ 𝑆)
Theorem	frecfcllem 6383*	Lemma for frecfcl 6384. Just giving a name to a common expression to simplify the proof. (Contributed by Jim Kingdon, 30-Mar-2022.)
⊢ 𝐺 = recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴))}))    ⇒   ⊢ ((∀𝑧 ∈ 𝑆 (𝐹‘𝑧) ∈ 𝑆 ∧ 𝐴 ∈ 𝑆) → frec(𝐹, 𝐴):ω⟶𝑆)
Theorem	frecfcl 6384*	Finite recursion yields a function on the natural numbers. (Contributed by Jim Kingdon, 30-Mar-2022.)
⊢ ((∀𝑧 ∈ 𝑆 (𝐹‘𝑧) ∈ 𝑆 ∧ 𝐴 ∈ 𝑆) → frec(𝐹, 𝐴):ω⟶𝑆)
Theorem	frecsuclem 6385*	Lemma for frecsuc 6386. Just giving a name to a common expression to simplify the proof. (Contributed by Jim Kingdon, 29-Mar-2022.)
⊢ 𝐺 = (𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴))})    ⇒   ⊢ ((∀𝑧 ∈ 𝑆 (𝐹‘𝑧) ∈ 𝑆 ∧ 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ ω) → (frec(𝐹, 𝐴)‘suc 𝐵) = (𝐹‘(frec(𝐹, 𝐴)‘𝐵)))
Theorem	frecsuc 6386*	The successor value resulting from finite recursive definition generation. (Contributed by Jim Kingdon, 31-Mar-2022.)
⊢ ((∀𝑧 ∈ 𝑆 (𝐹‘𝑧) ∈ 𝑆 ∧ 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ ω) → (frec(𝐹, 𝐴)‘suc 𝐵) = (𝐹‘(frec(𝐹, 𝐴)‘𝐵)))
Theorem	frecrdg 6387*	Transfinite recursion restricted to omega. Given a suitable characteristic function, df-frec 6370 produces the same results as df-irdg 6349 restricted to ω. Presumably the theorem would also hold if 𝐹 Fn V were changed to ∀𝑧(𝐹‘𝑧) ∈ V. (Contributed by Jim Kingdon, 29-Aug-2019.)
⊢ (𝜑 → 𝐹 Fn V)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → ∀𝑥 𝑥 ⊆ (𝐹‘𝑥))    ⇒   ⊢ (𝜑 → frec(𝐹, 𝐴) = (rec(𝐹, 𝐴) ↾ ω))
2.6.23  Ordinal arithmetic
Syntax	c1o 6388	Extend the definition of a class to include the ordinal number 1.
class 1o
Syntax	c2o 6389	Extend the definition of a class to include the ordinal number 2.
class 2o
Syntax	c3o 6390	Extend the definition of a class to include the ordinal number 3.
class 3o
Syntax	c4o 6391	Extend the definition of a class to include the ordinal number 4.
class 4o
Syntax	coa 6392	Extend the definition of a class to include the ordinal addition operation.
class +o
Syntax	comu 6393	Extend the definition of a class to include the ordinal multiplication operation.
class ·o
Syntax	coei 6394	Extend the definition of a class to include the ordinal exponentiation operation.
class ↑o
Definition	df-1o 6395	Define the ordinal number 1. (Contributed by NM, 29-Oct-1995.)
⊢ 1o = suc ∅
Definition	df-2o 6396	Define the ordinal number 2. (Contributed by NM, 18-Feb-2004.)
⊢ 2o = suc 1o
Definition	df-3o 6397	Define the ordinal number 3. (Contributed by Mario Carneiro, 14-Jul-2013.)
⊢ 3o = suc 2o
Definition	df-4o 6398	Define the ordinal number 4. (Contributed by Mario Carneiro, 14-Jul-2013.)
⊢ 4o = suc 3o
Definition	df-oadd 6399*	Define the ordinal addition operation. (Contributed by NM, 3-May-1995.)
⊢ +o = (𝑥 ∈ On, 𝑦 ∈ On ↦ (rec((𝑧 ∈ V ↦ suc 𝑧), 𝑥)‘𝑦))
Definition	df-omul 6400*	Define the ordinal multiplication operation. (Contributed by NM, 26-Aug-1995.)
⊢ ·o = (𝑥 ∈ On, 𝑦 ∈ On ↦ (rec((𝑧 ∈ V ↦ (𝑧 +o 𝑥)), ∅)‘𝑦))