P. 65 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 6401-6500   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	tfr1onlembxssdm 6401*	Lemma for tfr1on 6408. 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 6402*	Lemma for tfr1on 6408. 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 6403*	Lemma for tfr1on 6408. 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 6404*	Lemma for tfr1on 6408. 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 6405*	Lemma for tfr1on 6408. (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 6406*	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 6407*	Lemma for tfr1on 6408. 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 6408*	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 6409*	Alternate proof of tfri1d 6393 in terms of tfr1on 6408. Although this does show that the tfr1on 6408 proof is general enough to also prove tfri1d 6393, the tfri1d 6393 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 6410*	Lemma for tfrcl 6422. The union of functions acceptable for tfrcl 6422 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 6411*	We can extend an acceptable function by one element to produce a function. Lemma for tfrcl 6422. (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 6412*	Lemma for tfrcl 6422. 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 6413*	Lemma for tfrcl 6422. 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 6414*	Lemma for tfrcl 6422. 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 6415*	Lemma for tfrcl 6422. 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 6416*	Lemma for tfrcl 6422. 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 6417*	Lemma for tfrcl 6422. 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 6418*	Lemma for tfrcl 6422. (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 6419*	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 6420*	Lemma for tfr1on 6408. 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 6421*	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 6422*	Closure for transfinite recursion. As with tfr1on 6408, 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 6423*	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 6424*	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 6423). 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 6425*	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 6423). 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 6426*	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 6427	Extend class notation with the recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class rec(𝐹, 𝐼)
Definition	df-irdg 6428*	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 6363 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 6449 and for suitable characteristic functions df-frec 6449 yields the same result as rec restricted to ω, as seen at frecrdg 6466. 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 6429	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 6430	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 6431	The recursive definition generator is a function. (Contributed by Mario Carneiro, 16-Nov-2014.)
⊢ Fun rec(𝐹, 𝐴)
Theorem	rdgtfr 6432*	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 6433*	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 6434*	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 6435	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 6436	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 6437	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 6444; in cases like df-oadd 6478 either presumably could work). (Contributed by Jim Kingdon, 13-Jul-2019.)
⊢ ((𝐹 Fn V ∧ 𝐴 ∈ 𝑉) → rec(𝐹, 𝐴) Fn On)
Theorem	rdgifnon2 6438*	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 6439*	Value of the recursive definition generator. Lemma for rdgival 6440 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 6440*	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 6441	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 6442*	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 6443. (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 6443*	Value of the recursive definition generator at a successor. This can be thought of as a generalization of oasuc 6522 and omsuc 6530. (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 6444*	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 6445	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 6446	The initial value of the recursive definition generator. (Contributed by NM, 25-Apr-1995.)
⊢ (𝐴 ∈ 𝐶 → (rec(𝐹, 𝐴)‘∅) = 𝐴)
Theorem	rdgexg 6447	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 6448	Extend class notation with the finite recursive definition generator, with characteristic function 𝐹 and initial value 𝐼.
class frec(𝐹, 𝐼)
Definition	df-frec 6449*	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 6363 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 6455 and frecsuc 6465. 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 4640. 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 6466, this definition and df-irdg 6428 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 6450	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 6451	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 6452	Finite recursion produces a set. (Contributed by Jim Kingdon, 20-Aug-2021.)
⊢ frec(𝐹, 𝐴) ∈ V
Theorem	frecfun 6453	Finite recursion produces a function. See also frecfnom 6459 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 6454	Bound-variable hypothesis builder for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)
⊢ Ⅎ𝑥𝐹    &   ⊢ Ⅎ𝑥𝐴    ⇒   ⊢ Ⅎ𝑥frec(𝐹, 𝐴)
Theorem	frec0g 6455	The initial value resulting from finite recursive definition generation. (Contributed by Jim Kingdon, 7-May-2020.)
⊢ (𝐴 ∈ 𝑉 → (frec(𝐹, 𝐴)‘∅) = 𝐴)
Theorem	frecabex 6456*	The class abstraction from df-frec 6449 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 6457*	The class abstraction from df-frec 6449 exists. Unlike frecabex 6456 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 6458*	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 6393 or tfri2d 6394, 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 6459*	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 6460*	Lemma for freccl 6461. 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 6461*	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 6462*	Lemma for frecfcl 6463. 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 6463*	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 6464*	Lemma for frecsuc 6465. 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 6465*	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 6466*	Transfinite recursion restricted to omega. Given a suitable characteristic function, df-frec 6449 produces the same results as df-irdg 6428 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 6467	Extend the definition of a class to include the ordinal number 1.
class 1o
Syntax	c2o 6468	Extend the definition of a class to include the ordinal number 2.
class 2o
Syntax	c3o 6469	Extend the definition of a class to include the ordinal number 3.
class 3o
Syntax	c4o 6470	Extend the definition of a class to include the ordinal number 4.
class 4o
Syntax	coa 6471	Extend the definition of a class to include the ordinal addition operation.
class +o
Syntax	comu 6472	Extend the definition of a class to include the ordinal multiplication operation.
class ·o
Syntax	coei 6473	Extend the definition of a class to include the ordinal exponentiation operation.
class ↑o
Definition	df-1o 6474	Define the ordinal number 1. (Contributed by NM, 29-Oct-1995.)
⊢ 1o = suc ∅
Definition	df-2o 6475	Define the ordinal number 2. (Contributed by NM, 18-Feb-2004.)
⊢ 2o = suc 1o
Definition	df-3o 6476	Define the ordinal number 3. (Contributed by Mario Carneiro, 14-Jul-2013.)
⊢ 3o = suc 2o
Definition	df-4o 6477	Define the ordinal number 4. (Contributed by Mario Carneiro, 14-Jul-2013.)
⊢ 4o = suc 3o
Definition	df-oadd 6478*	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 6479*	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 6480*	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 6481	Ordinal 1 is an ordinal number. (Contributed by NM, 29-Oct-1995.)
⊢ 1o ∈ On
Theorem	1oex 6482	Ordinal 1 is a set. (Contributed by BJ, 4-Jul-2022.)
⊢ 1o ∈ V
Theorem	2on 6483	Ordinal 2 is an ordinal number. (Contributed by NM, 18-Feb-2004.) (Proof shortened by Andrew Salmon, 12-Aug-2011.)
⊢ 2o ∈ On
Theorem	2on0 6484	Ordinal two is not zero. (Contributed by Scott Fenton, 17-Jun-2011.)
⊢ 2o ≠ ∅
Theorem	3on 6485	Ordinal 3 is an ordinal number. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ 3o ∈ On
Theorem	4on 6486	Ordinal 3 is an ordinal number. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ 4o ∈ On
Theorem	df1o2 6487	Expanded value of the ordinal number 1. (Contributed by NM, 4-Nov-2002.)
⊢ 1o = {∅}
Theorem	df2o3 6488	Expanded value of the ordinal number 2. (Contributed by Mario Carneiro, 14-Aug-2015.)
⊢ 2o = {∅, 1o}
Theorem	df2o2 6489	Expanded value of the ordinal number 2. (Contributed by NM, 29-Jan-2004.)
⊢ 2o = {∅, {∅}}
Theorem	1n0 6490	Ordinal one is not equal to ordinal zero. (Contributed by NM, 26-Dec-2004.)
⊢ 1o ≠ ∅
Theorem	xp01disj 6491	Cartesian products with the singletons of ordinals 0 and 1 are disjoint. (Contributed by NM, 2-Jun-2007.)
⊢ ((𝐴 × {∅}) ∩ (𝐶 × {1o})) = ∅
Theorem	xp01disjl 6492	Cartesian products with the singletons of ordinals 0 and 1 are disjoint. (Contributed by Jim Kingdon, 11-Jul-2023.)
⊢ (({∅} × 𝐴) ∩ ({1o} × 𝐶)) = ∅
Theorem	ordgt0ge1 6493	Two ways to express that an ordinal class is positive. (Contributed by NM, 21-Dec-2004.)
⊢ (Ord 𝐴 → (∅ ∈ 𝐴 ↔ 1o ⊆ 𝐴))
Theorem	ordge1n0im 6494	An ordinal greater than or equal to 1 is nonzero. (Contributed by Jim Kingdon, 26-Jun-2019.)
⊢ (Ord 𝐴 → (1o ⊆ 𝐴 → 𝐴 ≠ ∅))
Theorem	el1o 6495	Membership in ordinal one. (Contributed by NM, 5-Jan-2005.)
⊢ (𝐴 ∈ 1o ↔ 𝐴 = ∅)
Theorem	dif1o 6496	Two ways to say that 𝐴 is a nonzero number of the set 𝐵. (Contributed by Mario Carneiro, 21-May-2015.)
⊢ (𝐴 ∈ (𝐵 ∖ 1o) ↔ (𝐴 ∈ 𝐵 ∧ 𝐴 ≠ ∅))
Theorem	2oconcl 6497	Closure of the pair swapping function on 2o. (Contributed by Mario Carneiro, 27-Sep-2015.)
⊢ (𝐴 ∈ 2o → (1o ∖ 𝐴) ∈ 2o)
Theorem	0lt1o 6498	Ordinal zero is less than ordinal one. (Contributed by NM, 5-Jan-2005.)
⊢ ∅ ∈ 1o
Theorem	0lt2o 6499	Ordinal zero is less than ordinal two. (Contributed by Jim Kingdon, 31-Jul-2022.)
⊢ ∅ ∈ 2o
Theorem	1lt2o 6500	Ordinal one is less than ordinal two. (Contributed by Jim Kingdon, 31-Jul-2022.)
⊢ 1o ∈ 2o