P. 67 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 6601-6700   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	tfrlemiubacc 6601*	The union of 𝐵 satisfies the recursion rule (lemma for tfrlemi1 6603). (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 6602*	Lemma for tfrlemi1 6603. (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 6603*	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 6604*	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 6605*	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 6606*	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 6607*	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 6636). 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 6608*	Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3ag 6580 but for tfr1on 6621 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 6609*	Lemma for transfinite recursion. This lemma changes some bound variables in 𝐴 (version of tfrlem3 6582 but for tfr1on 6621 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 6610*	Lemma for tfr1on 6621. The union of functions acceptable for tfr1on 6621 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 6611*	We can extend an acceptable function by one element to produce a function. Lemma for tfr1on 6621. (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 6612*	Lemma for tfr1on 6621. 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 6613*	Lemma for tfr1on 6621. 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 6614*	Lemma for tfr1on 6621. 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 6615*	Lemma for tfr1on 6621. 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 6616*	Lemma for tfr1on 6621. 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 6617*	Lemma for tfr1on 6621. 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 6618*	Lemma for tfr1on 6621. (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 6619*	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 6620*	Lemma for tfr1on 6621. 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 6621*	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 6622*	Alternate proof of tfri1d 6606 in terms of tfr1on 6621. Although this does show that the tfr1on 6621 proof is general enough to also prove tfri1d 6606, the tfri1d 6606 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 6623*	Lemma for tfrcl 6635. The union of functions acceptable for tfrcl 6635 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 6624*	We can extend an acceptable function by one element to produce a function. Lemma for tfrcl 6635. (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 6625*	Lemma for tfrcl 6635. 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 6626*	Lemma for tfrcl 6635. 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 6627*	Lemma for tfrcl 6635. 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 6628*	Lemma for tfrcl 6635. 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 6629*	Lemma for tfrcl 6635. 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 6630*	Lemma for tfrcl 6635. 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 6631*	Lemma for tfrcl 6635. (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 6632*	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 6633*	Lemma for tfr1on 6621. 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 6634*	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 6635*	Closure for transfinite recursion. As with tfr1on 6621, 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 6636*	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 6637*	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 6636). 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 6638*	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 6636). 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 6639*	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)
2.6.22  Recursive definition generator
Syntax	crdg 6640	Extend class notation with the recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class rec(𝐹, 𝐼)
Definition	df-irdg 6641*	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 6576 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 6662 and for suitable characteristic functions df-frec 6662 yields the same result as rec restricted to ω, as seen at frecrdg 6679. 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 6642	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 6643	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 6644	The recursive definition generator is a function. (Contributed by Mario Carneiro, 16-Nov-2014.)
⊢ Fun rec(𝐹, 𝐴)
Theorem	rdgtfr 6645*	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 6646*	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 6647*	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 6648	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 6649	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 6650	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 6657; in cases like df-oadd 6691 either presumably could work). (Contributed by Jim Kingdon, 13-Jul-2019.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉) → rec(𝐹, 𝐴) Fn On)
Theorem	rdgifnon2 6651*	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 6652*	Value of the recursive definition generator. Lemma for rdgival 6653 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 6653*	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 6654	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 6655*	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 6656. (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 6656*	Value of the recursive definition generator at a successor. This can be thought of as a generalization of oasuc 6737 and omsuc 6745. (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 6657*	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 6658	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 6659	The initial value of the recursive definition generator. (Contributed by NM, 25-Apr-1995.)
⊢ (𝐴 ∈ 𝐶 → (rec(𝐹, 𝐴)‘∅) = 𝐴)
Theorem	rdgexg 6660	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)
2.6.23  Finite recursion
Syntax	cfrec 6661	Extend class notation with the finite recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class frec(𝐹, 𝐼)
Definition	df-frec 6662*	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 6576 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 6668 and frecsuc 6678. 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 4751. 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 6679, this definition and df-irdg 6641 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 6663	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 6664	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 6665	Finite recursion produces a set. (Contributed by Jim Kingdon, 20-Aug-2021.)
⊢ frec(𝐹, 𝐴) ∈ V
Theorem	frecfun 6666	Finite recursion produces a function. See also frecfnom 6672 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 6667	Bound-variable hypothesis builder for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)
⊢ Ⅎ𝑥𝐹    &   ⊢ Ⅎ𝑥𝐴    ⇒   ⊢ Ⅎ𝑥frec(𝐹, 𝐴)
Theorem	frec0g 6668	The initial value resulting from finite recursive definition generation. (Contributed by Jim Kingdon, 7-May-2020.)
⊢ (𝐴 ∈ 𝑉 → (frec(𝐹, 𝐴)‘∅) = 𝐴)
Theorem	frecabex 6669*	The class abstraction from df-frec 6662 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 6670*	The class abstraction from df-frec 6662 exists. Unlike frecabex 6669 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 6671*	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 6606 or tfri2d 6607, 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 6672*	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 6673*	Lemma for freccl 6674. 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 6674*	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 6675*	Lemma for frecfcl 6676. 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 6676*	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 6677*	Lemma for frecsuc 6678. 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 6678*	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 6679*	Transfinite recursion restricted to omega. Given a suitable characteristic function, df-frec 6662 produces the same results as df-irdg 6641 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))
2.6.24  Ordinal arithmetic
Syntax	c1o 6680	Extend the definition of a class to include the ordinal number 1.
class 1o
Syntax	c2o 6681	Extend the definition of a class to include the ordinal number 2.
class 2o
Syntax	c3o 6682	Extend the definition of a class to include the ordinal number 3.
class 3o
Syntax	c4o 6683	Extend the definition of a class to include the ordinal number 4.
class 4o
Syntax	coa 6684	Extend the definition of a class to include the ordinal addition operation.
class +o
Syntax	comu 6685	Extend the definition of a class to include the ordinal multiplication operation.
class ·o
Syntax	coei 6686	Extend the definition of a class to include the ordinal exponentiation operation.
class ↑o
Definition	df-1o 6687	Define the ordinal number 1. (Contributed by NM, 29-Oct-1995.)
⊢ 1o = suc ∅
Definition	df-2o 6688	Define the ordinal number 2. (Contributed by NM, 18-Feb-2004.)
⊢ 2o = suc 1o
Definition	df-3o 6689	Define the ordinal number 3. (Contributed by Mario Carneiro, 14-Jul-2013.)
⊢ 3o = suc 2o
Definition	df-4o 6690	Define the ordinal number 4. (Contributed by Mario Carneiro, 14-Jul-2013.)
⊢ 4o = suc 3o
Definition	df-oadd 6691*	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 6692*	Define the ordinal multiplication operation. (Contributed by NM, 26-Aug-1995.)
⊢ ·o = (𝑥 ∈ On, 𝑦 ∈ On ↦ (rec((𝑧 ∈ V ↦ (𝑧 +o 𝑥)), ∅)‘𝑦))
Definition	df-oexpi 6693*	Define the ordinal exponentiation operation. This definition is similar to a conventional definition of exponentiation except that it defines ∅ ↑o 𝐴 to be 1o for all 𝐴 ∈ On, in order to avoid having different cases for whether the base is ∅ or not. We do not yet have an extensive development of ordinal exponentiation. For background on ordinal exponentiation without excluded middle, see Tom de Jong, Nicolai Kraus, Fredrik Nordvall Forsberg, and Chuangjie Xu (2025), "Ordinal Exponentiation in Homotopy Type Theory", arXiv:2501.14542 , https://arxiv.org/abs/2501.14542 which is formalized in the TypeTopology proof library at https://ordinal-exponentiation-hott.github.io/. (Contributed by Mario Carneiro, 4-Jul-2019.)
⊢ ↑o = (𝑥 ∈ On, 𝑦 ∈ On ↦ (rec((𝑧 ∈ V ↦ (𝑧 ·o 𝑥)), 1o)‘𝑦))
Theorem	1on 6694	Ordinal 1 is an ordinal number. (Contributed by NM, 29-Oct-1995.)
⊢ 1o ∈ On
Theorem	1oex 6695	Ordinal 1 is a set. (Contributed by BJ, 4-Jul-2022.)
⊢ 1o ∈ V
Theorem	2on 6696	Ordinal 2 is an ordinal number. (Contributed by NM, 18-Feb-2004.) (Proof shortened by Andrew Salmon, 12-Aug-2011.)
⊢ 2o ∈ On
Theorem	2on0 6697	Ordinal two is not zero. (Contributed by Scott Fenton, 17-Jun-2011.)
⊢ 2o ≠ ∅
Theorem	3on 6698	Ordinal 3 is an ordinal number. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ 3o ∈ On
Theorem	ord3 6699	Ordinal 3 is an ordinal class. (Contributed by BTernaryTau, 6-Jan-2025.)
⊢ Ord 3o
Theorem	4on 6700	Ordinal 4 is an ordinal number. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ 4o ∈ On