P. 132 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 13101-13200   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	4sqlemafi 13101*	Lemma for 4sq 13116. 𝐴 is finite. (Contributed by Jim Kingdon, 24-May-2025.)
⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    ⇒   ⊢ (𝜑 → 𝐴 ∈ Fin)
Theorem	4sqlemffi 13102*	Lemma for 4sq 13116. ran 𝐹 is finite. (Contributed by Jim Kingdon, 24-May-2025.)
⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → ran 𝐹 ∈ Fin)
Theorem	4sqleminfi 13103*	Lemma for 4sq 13116. 𝐴 ∩ ran 𝐹 is finite. (Contributed by Jim Kingdon, 24-May-2025.)
⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → (𝐴 ∩ ran 𝐹) ∈ Fin)
Theorem	4sqexercise1 13104*	Exercise which may help in understanding the proof of 4sqlemsdc 13106. (Contributed by Jim Kingdon, 25-May-2025.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ 𝑛 = (𝑥↑2)}    ⇒   ⊢ (𝐴 ∈ ℕ0 → DECID 𝐴 ∈ 𝑆)
Theorem	4sqexercise2 13105*	Exercise which may help in understanding the proof of 4sqlemsdc 13106. (Contributed by Jim Kingdon, 30-May-2025.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑛 = ((𝑥↑2) + (𝑦↑2))}    ⇒   ⊢ (𝐴 ∈ ℕ0 → DECID 𝐴 ∈ 𝑆)
Theorem	4sqlemsdc 13106*	Lemma for 4sq 13116. The property of being the sum of four squares is decidable. The proof involves showing that (for a particular 𝐴) there are only a finite number of possible ways that it could be the sum of four squares, so checking each of those possibilities in turn decides whether the number is the sum of four squares. If this proof is hard to follow, especially because of its length, the simplified versions at 4sqexercise1 13104 and 4sqexercise2 13105 may help clarify, as they are using very much the same techniques on simplified versions of this lemma. (Contributed by Jim Kingdon, 25-May-2025.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ (𝐴 ∈ ℕ0 → DECID 𝐴 ∈ 𝑆)
Theorem	4sqlem11 13107*	Lemma for 4sq 13116. Use the pigeonhole principle to show that the sets {𝑚↑2 ∣ 𝑚 ∈ (0...𝑁)} and {-1 − 𝑛↑2 ∣ 𝑛 ∈ (0...𝑁)} have a common element, mod 𝑃. Note that although the conclusion is stated in terms of 𝐴 ∩ ran 𝐹 being nonempty, it is also inhabited by 4sqleminfi 13103 and fin0 7144. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → (𝐴 ∩ ran 𝐹) ≠ ∅)
Theorem	4sqlem12 13108*	Lemma for 4sq 13116. For any odd prime 𝑃, there is a 𝑘 < 𝑃 such that 𝑘𝑃 − 1 is a sum of two squares. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → ∃𝑘 ∈ (1...(𝑃 − 1))∃𝑢 ∈ ℤ[i] (((abs‘𝑢)↑2) + 1) = (𝑘 · 𝑃))
Theorem	4sqlem13m 13109*	Lemma for 4sq 13116. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    ⇒   ⊢ (𝜑 → (∃𝑗 𝑗 ∈ 𝑇 ∧ 𝑀 < 𝑃))
Theorem	4sqlem14 13110*	Lemma for 4sq 13116. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ (𝜑 → 𝑅 ∈ ℕ0)
Theorem	4sqlem15 13111*	Lemma for 4sq 13116. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ ((𝜑 ∧ 𝑅 = 𝑀) → ((((((𝑀↑2) / 2) / 2) − (𝐸↑2)) = 0 ∧ ((((𝑀↑2) / 2) / 2) − (𝐹↑2)) = 0) ∧ (((((𝑀↑2) / 2) / 2) − (𝐺↑2)) = 0 ∧ ((((𝑀↑2) / 2) / 2) − (𝐻↑2)) = 0)))
Theorem	4sqlem16 13112*	Lemma for 4sq 13116. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ (𝜑 → (𝑅 ≤ 𝑀 ∧ ((𝑅 = 0 ∨ 𝑅 = 𝑀) → (𝑀↑2) ∥ (𝑀 · 𝑃))))
Theorem	4sqlem17 13113*	Lemma for 4sq 13116. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ ¬ 𝜑
Theorem	4sqlem18 13114*	Lemma for 4sq 13116. Inductive step, odd prime case. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    ⇒   ⊢ (𝜑 → 𝑃 ∈ 𝑆)
Theorem	4sqlem19 13115*	Lemma for 4sq 13116. The proof is by strong induction - we show that if all the integers less than 𝑘 are in 𝑆, then 𝑘 is as well. In this part of the proof we do the induction argument and dispense with all the cases except the odd prime case, which is sent to 4sqlem18 13114. If 𝑘 is 0, 1, 2, we show 𝑘 ∈ 𝑆 directly; otherwise if 𝑘 is composite, 𝑘 is the product of two numbers less than it (and hence in 𝑆 by assumption), so by mul4sq 13100 𝑘 ∈ 𝑆. (Contributed by Mario Carneiro, 14-Jul-2014.) (Revised by Mario Carneiro, 20-Jun-2015.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ ℕ0 = 𝑆
Theorem	4sq 13116*	Lagrange's four-square theorem, or Bachet's conjecture: every nonnegative integer is expressible as a sum of four squares. This is Metamath 100 proof #19. (Contributed by Mario Carneiro, 16-Jul-2014.)
⊢ (𝐴 ∈ ℕ0 ↔ ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ ∃𝑐 ∈ ℤ ∃𝑑 ∈ ℤ 𝐴 = (((𝑎↑2) + (𝑏↑2)) + ((𝑐↑2) + (𝑑↑2))))
5.2.13  Decimal arithmetic (cont.)
Theorem	dec2dvds 13117	Divisibility by two is obvious in base 10. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ (𝐵 · 2) = 𝐶    &   ⊢ 𝐷 = (𝐶 + 1)    ⇒   ⊢ ¬ 2 ∥ ;𝐴𝐷
Theorem	dec5dvds 13118	Divisibility by five is obvious in base 10. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ    &   ⊢ 𝐵 < 5    ⇒   ⊢ ¬ 5 ∥ ;𝐴𝐵
Theorem	dec5dvds2 13119	Divisibility by five is obvious in base 10. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ    &   ⊢ 𝐵 < 5    &   ⊢ (5 + 𝐵) = 𝐶    ⇒   ⊢ ¬ 5 ∥ ;𝐴𝐶
Theorem	dec5nprm 13120	A decimal number greater than 10 and ending with five is not a prime number. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ    ⇒   ⊢ ¬ ;𝐴5 ∈ ℙ
Theorem	dec2nprm 13121	A decimal number greater than 10 and ending with an even digit is not a prime number. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ (𝐵 · 2) = 𝐶    ⇒   ⊢ ¬ ;𝐴𝐶 ∈ ℙ
Theorem	modxai 13122	Add exponents in a power mod calculation. (Contributed by Mario Carneiro, 21-Feb-2014.) (Revised by Mario Carneiro, 5-Feb-2015.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ 𝐶 ∈ ℕ0    &   ⊢ 𝐿 ∈ ℕ0    &   ⊢ ((𝐴↑𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ ((𝐴↑𝐶) mod 𝑁) = (𝐿 mod 𝑁)    &   ⊢ (𝐵 + 𝐶) = 𝐸    &   ⊢ ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐿)    ⇒   ⊢ ((𝐴↑𝐸) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	mod2xi 13123	Double exponents in a power mod calculation. (Contributed by Mario Carneiro, 21-Feb-2014.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ ((𝐴↑𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ (2 · 𝐵) = 𝐸    &   ⊢ ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐾)    ⇒   ⊢ ((𝐴↑𝐸) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	modxp1i 13124	Add one to an exponent in a power mod calculation. (Contributed by Mario Carneiro, 21-Feb-2014.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ ((𝐴↑𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ (𝐵 + 1) = 𝐸    &   ⊢ ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐴)    ⇒   ⊢ ((𝐴↑𝐸) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	modsubi 13125	Subtract from within a mod calculation. (Contributed by Mario Carneiro, 18-Feb-2014.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝐴 mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ (𝑀 + 𝐵) = 𝐾    ⇒   ⊢ ((𝐴 − 𝐵) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	gcdi 13126	Calculate a GCD via Euclid's algorithm. (Contributed by Mario Carneiro, 19-Feb-2014.)
⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑅 ∈ ℕ0    &   ⊢ 𝑁 ∈ ℕ0    &   ⊢ (𝑁 gcd 𝑅) = 𝐺    &   ⊢ ((𝐾 · 𝑁) + 𝑅) = 𝑀    ⇒   ⊢ (𝑀 gcd 𝑁) = 𝐺
Theorem	gcdmodi 13127	Calculate a GCD via Euclid's algorithm. Theorem 5.6 in [ApostolNT] p. 109. (Contributed by Mario Carneiro, 19-Feb-2014.)
⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑅 ∈ ℕ0    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ (𝐾 mod 𝑁) = (𝑅 mod 𝑁)    &   ⊢ (𝑁 gcd 𝑅) = 𝐺    ⇒   ⊢ (𝐾 gcd 𝑁) = 𝐺
Theorem	numexp0 13128	Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ (𝐴↑0) = 1
Theorem	numexp1 13129	Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ (𝐴↑1) = 𝐴
Theorem	numexpp1 13130	Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝑀 + 1) = 𝑁    &   ⊢ ((𝐴↑𝑀) · 𝐴) = 𝐶    ⇒   ⊢ (𝐴↑𝑁) = 𝐶
Theorem	numexp2x 13131	Double an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (2 · 𝑀) = 𝑁    &   ⊢ (𝐴↑𝑀) = 𝐷    &   ⊢ (𝐷 · 𝐷) = 𝐶    ⇒   ⊢ (𝐴↑𝑁) = 𝐶
Theorem	decsplit0b 13132	Split a decimal number into two parts. Base case: 𝑁 = 0. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ ((𝐴 · (;10↑0)) + 𝐵) = (𝐴 + 𝐵)
Theorem	decsplit0 13133	Split a decimal number into two parts. Base case: 𝑁 = 0. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ ((𝐴 · (;10↑0)) + 0) = 𝐴
Theorem	decsplit1 13134	Split a decimal number into two parts. Base case: 𝑁 = 1. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ ((𝐴 · (;10↑1)) + 𝐵) = ;𝐴𝐵
Theorem	decsplit 13135	Split a decimal number into two parts. Inductive step. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝑀 + 1) = 𝑁    &   ⊢ ((𝐴 · (;10↑𝑀)) + 𝐵) = 𝐶    ⇒   ⊢ ((𝐴 · (;10↑𝑁)) + ;𝐵𝐷) = ;𝐶𝐷
Theorem	karatsuba 13136	The Karatsuba multiplication algorithm. If 𝑋 and 𝑌 are decomposed into two groups of digits of length 𝑀 (only the lower group is known to be this size but the algorithm is most efficient when the partition is chosen near the middle of the digit string), then 𝑋𝑌 can be written in three groups of digits, where each group needs only one multiplication. Thus, we can halve both inputs with only three multiplications on the smaller operands, yielding an asymptotic improvement of n^(log2 3) instead of n^2 for the "naive" algorithm decmul1c 9779. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 9-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐶 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℕ0    &   ⊢ 𝑆 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝐴 · 𝐶) = 𝑅    &   ⊢ (𝐵 · 𝐷) = 𝑇    &   ⊢ ((𝐴 + 𝐵) · (𝐶 + 𝐷)) = ((𝑅 + 𝑆) + 𝑇)    &   ⊢ ((𝐴 · (;10↑𝑀)) + 𝐵) = 𝑋    &   ⊢ ((𝐶 · (;10↑𝑀)) + 𝐷) = 𝑌    &   ⊢ ((𝑅 · (;10↑𝑀)) + 𝑆) = 𝑊    &   ⊢ ((𝑊 · (;10↑𝑀)) + 𝑇) = 𝑍    ⇒   ⊢ (𝑋 · 𝑌) = 𝑍
Theorem	2exp4 13137	Two to the fourth power is 16. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑4) = ;16
Theorem	2exp5 13138	Two to the fifth power is 32. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑5) = ;32
Theorem	2exp6 13139	Two to the sixth power is 64. (Contributed by Mario Carneiro, 20-Apr-2015.) (Proof shortened by OpenAI, 25-Mar-2020.)
⊢ (2↑6) = ;64
Theorem	2exp7 13140	Two to the seventh power is 128. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑7) = ;;128
Theorem	2exp8 13141	Two to the eighth power is 256. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑8) = ;;256
Theorem	2exp11 13142	Two to the eleventh power is 2048. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑;11) = ;;;2048
Theorem	2exp16 13143	Two to the sixteenth power is 65536. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑;16) = ;;;;65536
Theorem	3exp3 13144	Three to the third power is 27. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (3↑3) = ;27
Theorem	2expltfac 13145	The factorial grows faster than two to the power 𝑁. (Contributed by Mario Carneiro, 15-Sep-2016.)
⊢ (𝑁 ∈ (ℤ≥‘4) → (2↑𝑁) < (!‘𝑁))
5.2.14  Bertrand's Ballot Problem
Theorem	ballotfilemofi 13146*	𝑂 is finite. (Contributed by Jim Kingdon, 20-May-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    ⇒   ⊢ 𝑂 ∈ Fin
Theorem	ballotfilem1 13147*	The size of the universe is a binomial coefficient. (Contributed by Thierry Arnoux, 23-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    ⇒   ⊢ (♯‘𝑂) = ((𝑀 + 𝑁)C𝑀)
Theorem	ballotfilemonn 13148*	The size of the universe is at least one. (Contributed by Jim Kingdon, 4-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    ⇒   ⊢ (♯‘𝑂) ∈ ℕ
Theorem	ballotfilemelo 13149*	Elementhood in 𝑂. (Contributed by Thierry Arnoux, 17-Apr-2017.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    ⇒   ⊢ (𝐶 ∈ 𝑂 ↔ (𝐶 ⊆ (1...(𝑀 + 𝑁)) ∧ 𝐶 ∈ Fin ∧ (♯‘𝐶) = 𝑀))
Theorem	ballotfilemcdc 13150*	Lemma for ballotfi . It is decidable whether a given integer is an element of a particular element of 𝑂. (Contributed by Jim Kingdon, 7-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐾 ∈ ℤ)    ⇒   ⊢ (𝜑 → DECID 𝐾 ∈ 𝐶)
Theorem	ballotfilemcinfi 13151*	Lemma for ballotfi . The portion of a particular element of 𝑂 up to a specified integer is finite. (Contributed by Jim Kingdon, 8-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((1...𝐽) ∩ 𝐶) ∈ Fin)
Theorem	ballotfilemdifcfi 13152*	Lemma for ballotfi . The portion of an integer range which is not part of a particular element of 𝑂 is finite. (Contributed by Jim Kingdon, 8-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((1...𝐽) ∖ 𝐶) ∈ Fin)
Theorem	ballotfilem2 13153*	The probability that the first vote picked in a count is a B. (Contributed by Thierry Arnoux, 23-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    ⇒   ⊢ (𝑃‘{𝑐 ∈ 𝑂 ∣ ¬ 1 ∈ 𝑐}) = (𝑁 / (𝑀 + 𝑁))
Theorem	ballotfilemfval 13154*	The value of 𝐹. (Contributed by Thierry Arnoux, 23-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((𝐹‘𝐶)‘𝐽) = ((♯‘((1...𝐽) ∩ 𝐶)) − (♯‘((1...𝐽) ∖ 𝐶))))
Theorem	ballotfilemfelz 13155*	(𝐹‘𝐶) has values in ℤ. (Contributed by Thierry Arnoux, 23-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((𝐹‘𝐶)‘𝐽) ∈ ℤ)
Theorem	ballotfilemfp1 13156*	If the 𝐽 th ballot is for A, (𝐹‘𝐶) goes up 1. If the 𝐽 th ballot is for B, (𝐹‘𝐶) goes down 1. (Contributed by Thierry Arnoux, 24-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℕ)    ⇒   ⊢ (𝜑 → ((¬ 𝐽 ∈ 𝐶 → ((𝐹‘𝐶)‘𝐽) = (((𝐹‘𝐶)‘(𝐽 − 1)) − 1)) ∧ (𝐽 ∈ 𝐶 → ((𝐹‘𝐶)‘𝐽) = (((𝐹‘𝐶)‘(𝐽 − 1)) + 1))))
Theorem	ballotfilemfc0 13157*	𝐹 takes value 0 between negative and positive values. (Contributed by Thierry Arnoux, 24-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℕ)    &   ⊢ (𝜑 → ∃𝑖 ∈ (1...𝐽)((𝐹‘𝐶)‘𝑖) ≤ 0)    &   ⊢ (𝜑 → 0 < ((𝐹‘𝐶)‘𝐽))    ⇒   ⊢ (𝜑 → ∃𝑘 ∈ (1...𝐽)((𝐹‘𝐶)‘𝑘) = 0)
Theorem	ballotfilemfcc 13158*	𝐹 takes value 0 between positive and negative values. (Contributed by Thierry Arnoux, 2-Apr-2017.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℕ)    &   ⊢ (𝜑 → ∃𝑖 ∈ (1...𝐽)0 ≤ ((𝐹‘𝐶)‘𝑖))    &   ⊢ (𝜑 → ((𝐹‘𝐶)‘𝐽) < 0)    ⇒   ⊢ (𝜑 → ∃𝑘 ∈ (1...𝐽)((𝐹‘𝐶)‘𝑘) = 0)
Theorem	ballotfilemfmpn 13159*	(𝐹‘𝐶) finishes counting at (𝑀 − 𝑁). (Contributed by Thierry Arnoux, 25-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    ⇒   ⊢ (𝐶 ∈ 𝑂 → ((𝐹‘𝐶)‘(𝑀 + 𝑁)) = (𝑀 − 𝑁))
Theorem	ballotfilemfval0 13160*	(𝐹‘𝐶) always starts counting at 0 . (Contributed by Thierry Arnoux, 25-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    ⇒   ⊢ (𝐶 ∈ 𝑂 → ((𝐹‘𝐶)‘0) = 0)
Theorem	ballotfileme 13161*	Elements of 𝐸. (Contributed by Thierry Arnoux, 14-Dec-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    ⇒   ⊢ (𝐶 ∈ 𝐸 ↔ (𝐶 ∈ 𝑂 ∧ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝐶)‘𝑖)))
Theorem	ballotfilemodife 13162*	Elements of (𝑂 ∖ 𝐸). (Contributed by Thierry Arnoux, 7-Dec-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) ↔ (𝐶 ∈ 𝑂 ∧ ∃𝑖 ∈ (1...(𝑀 + 𝑁))((𝐹‘𝐶)‘𝑖) ≤ 0))
Theorem	ballotfilem4 13163*	If the first pick is a vote for B, A is not ahead throughout the count. (Contributed by Thierry Arnoux, 25-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    ⇒   ⊢ (𝐶 ∈ 𝑂 → (¬ 1 ∈ 𝐶 → ¬ 𝐶 ∈ 𝐸))
5.3  Cardinality of real and complex number subsets
5.3.1  Countability of integers and rationals
Theorem	oddennn 13164	There are as many odd positive integers as there are positive integers. (Contributed by Jim Kingdon, 11-May-2022.)
⊢ {𝑧 ∈ ℕ ∣ ¬ 2 ∥ 𝑧} ≈ ℕ
Theorem	evenennn 13165	There are as many even positive integers as there are positive integers. (Contributed by Jim Kingdon, 12-May-2022.)
⊢ {𝑧 ∈ ℕ ∣ 2 ∥ 𝑧} ≈ ℕ
Theorem	xpnnen 13166	The Cartesian product of the set of positive integers with itself is equinumerous to the set of positive integers. (Contributed by NM, 1-Aug-2004.)
⊢ (ℕ × ℕ) ≈ ℕ
Theorem	xpomen 13167	The Cartesian product of omega (the set of ordinal natural numbers) with itself is equinumerous to omega. Exercise 1 of [Enderton] p. 133. (Contributed by NM, 23-Jul-2004.)
⊢ (ω × ω) ≈ ω
Theorem	xpct 13168	The cartesian product of two sets dominated by ω is dominated by ω. (Contributed by Thierry Arnoux, 24-Sep-2017.)
⊢ ((𝐴 ≼ ω ∧ 𝐵 ≼ ω) → (𝐴 × 𝐵) ≼ ω)
Theorem	unennn 13169	The union of two disjoint countably infinite sets is countably infinite. (Contributed by Jim Kingdon, 13-May-2022.)
⊢ ((𝐴 ≈ ℕ ∧ 𝐵 ≈ ℕ ∧ (𝐴 ∩ 𝐵) = ∅) → (𝐴 ∪ 𝐵) ≈ ℕ)
Theorem	znnen 13170	The set of integers and the set of positive integers are equinumerous. Corollary 8.1.23 of [AczelRathjen], p. 75. (Contributed by NM, 31-Jul-2004.)
⊢ ℤ ≈ ℕ
Theorem	ennnfonelemdc 13171*	Lemma for ennnfone 13197. A direct consequence of fidcenumlemrk 7226. (Contributed by Jim Kingdon, 15-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → 𝑃 ∈ ω)    ⇒   ⊢ (𝜑 → DECID (𝐹‘𝑃) ∈ (𝐹 “ 𝑃))
Theorem	ennnfonelemk 13172*	Lemma for ennnfone 13197. (Contributed by Jim Kingdon, 15-Jul-2023.)
⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → 𝐾 ∈ ω)    &   ⊢ (𝜑 → 𝑁 ∈ ω)    &   ⊢ (𝜑 → ∀𝑗 ∈ suc 𝑁(𝐹‘𝐾) ≠ (𝐹‘𝑗))    ⇒   ⊢ (𝜑 → 𝑁 ∈ 𝐾)
Theorem	ennnfonelemj0 13173*	Lemma for ennnfone 13197. Initial state for 𝐽. (Contributed by Jim Kingdon, 20-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ (𝜑 → (𝐽‘0) ∈ {𝑔 ∈ (𝐴 ↑pm ω) ∣ dom 𝑔 ∈ ω})
Theorem	ennnfonelemjn 13174*	Lemma for ennnfone 13197. Non-initial state for 𝐽. (Contributed by Jim Kingdon, 20-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ ((𝜑 ∧ 𝑓 ∈ (ℤ≥‘(0 + 1))) → (𝐽‘𝑓) ∈ ω)
Theorem	ennnfonelemg 13175*	Lemma for ennnfone 13197. Closure for 𝐺. (Contributed by Jim Kingdon, 20-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ ((𝜑 ∧ (𝑓 ∈ {𝑔 ∈ (𝐴 ↑pm ω) ∣ dom 𝑔 ∈ ω} ∧ 𝑗 ∈ ω)) → (𝑓𝐺𝑗) ∈ {𝑔 ∈ (𝐴 ↑pm ω) ∣ dom 𝑔 ∈ ω})
Theorem	ennnfonelemh 13176*	Lemma for ennnfone 13197. (Contributed by Jim Kingdon, 8-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ (𝜑 → 𝐻:ℕ0⟶(𝐴 ↑pm ω))
Theorem	ennnfonelem0 13177*	Lemma for ennnfone 13197. Initial value. (Contributed by Jim Kingdon, 15-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ (𝜑 → (𝐻‘0) = ∅)
Theorem	ennnfonelemp1 13178*	Lemma for ennnfone 13197. Value of 𝐻 at a successor. (Contributed by Jim Kingdon, 23-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝐻‘(𝑃 + 1)) = if((𝐹‘(◡𝑁‘𝑃)) ∈ (𝐹 “ (◡𝑁‘𝑃)), (𝐻‘𝑃), ((𝐻‘𝑃) ∪ {⟨dom (𝐻‘𝑃), (𝐹‘(◡𝑁‘𝑃))⟩})))
Theorem	ennnfonelem1 13179*	Lemma for ennnfone 13197. Second value. (Contributed by Jim Kingdon, 19-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    ⇒   ⊢ (𝜑 → (𝐻‘1) = {⟨∅, (𝐹‘∅)⟩})
Theorem	ennnfonelemom 13180*	Lemma for ennnfone 13197. 𝐻 yields finite sequences. (Contributed by Jim Kingdon, 19-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → dom (𝐻‘𝑃) ∈ ω)
Theorem	ennnfonelemhdmp1 13181*	Lemma for ennnfone 13197. Domain at a successor where we need to add an element to the sequence. (Contributed by Jim Kingdon, 23-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    &   ⊢ (𝜑 → ¬ (𝐹‘(◡𝑁‘𝑃)) ∈ (𝐹 “ (◡𝑁‘𝑃)))    ⇒   ⊢ (𝜑 → dom (𝐻‘(𝑃 + 1)) = suc dom (𝐻‘𝑃))
Theorem	ennnfonelemss 13182*	Lemma for ennnfone 13197. We only add elements to 𝐻 as the index increases. (Contributed by Jim Kingdon, 15-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝐻‘𝑃) ⊆ (𝐻‘(𝑃 + 1)))
Theorem	ennnfoneleminc 13183*	Lemma for ennnfone 13197. We only add elements to 𝐻 as the index increases. (Contributed by Jim Kingdon, 21-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑄 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑃 ≤ 𝑄)    ⇒   ⊢ (𝜑 → (𝐻‘𝑃) ⊆ (𝐻‘𝑄))
Theorem	ennnfonelemkh 13184*	Lemma for ennnfone 13197. Because we add zero or one entries for each new index, the length of each sequence is no greater than its index. (Contributed by Jim Kingdon, 19-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → dom (𝐻‘𝑃) ⊆ (◡𝑁‘𝑃))
Theorem	ennnfonelemhf1o 13185*	Lemma for ennnfone 13197. Each of the functions in 𝐻 is one to one and onto an image of 𝐹. (Contributed by Jim Kingdon, 17-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝐻‘𝑃):dom (𝐻‘𝑃)–1-1-onto→(𝐹 “ (◡𝑁‘𝑃)))
Theorem	ennnfonelemex 13186*	Lemma for ennnfone 13197. Extending the sequence (𝐻‘𝑃) to include an additional element. (Contributed by Jim Kingdon, 19-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ0)    ⇒   ⊢ (𝜑 → ∃𝑖 ∈ ℕ0 dom (𝐻‘𝑃) ∈ dom (𝐻‘𝑖))
Theorem	ennnfonelemhom 13187*	Lemma for ennnfone 13197. The sequences in 𝐻 increase in length without bound if you go out far enough. (Contributed by Jim Kingdon, 19-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑀 ∈ ω)    ⇒   ⊢ (𝜑 → ∃𝑖 ∈ ℕ0 𝑀 ∈ dom (𝐻‘𝑖))
Theorem	ennnfonelemrnh 13188*	Lemma for ennnfone 13197. A consequence of ennnfonelemss 13182. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ (𝜑 → 𝑋 ∈ ran 𝐻)    &   ⊢ (𝜑 → 𝑌 ∈ ran 𝐻)    ⇒   ⊢ (𝜑 → (𝑋 ⊆ 𝑌 ∨ 𝑌 ⊆ 𝑋))
Theorem	ennnfonelemfun 13189*	Lemma for ennnfone 13197. 𝐿 is a function. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ 𝐿 = ∪ 𝑖 ∈ ℕ0 (𝐻‘𝑖)    ⇒   ⊢ (𝜑 → Fun 𝐿)
Theorem	ennnfonelemf1 13190*	Lemma for ennnfone 13197. 𝐿 is one-to-one. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ 𝐿 = ∪ 𝑖 ∈ ℕ0 (𝐻‘𝑖)    ⇒   ⊢ (𝜑 → 𝐿:dom 𝐿–1-1→𝐴)
Theorem	ennnfonelemrn 13191*	Lemma for ennnfone 13197. 𝐿 is onto 𝐴. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ 𝐿 = ∪ 𝑖 ∈ ℕ0 (𝐻‘𝑖)    ⇒   ⊢ (𝜑 → ran 𝐿 = 𝐴)
Theorem	ennnfonelemdm 13192*	Lemma for ennnfone 13197. The function 𝐿 is defined everywhere. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ 𝐿 = ∪ 𝑖 ∈ ℕ0 (𝐻‘𝑖)    ⇒   ⊢ (𝜑 → dom 𝐿 = ω)
Theorem	ennnfonelemen 13193*	Lemma for ennnfone 13197. The result. (Contributed by Jim Kingdon, 16-Jul-2023.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝐺 = (𝑥 ∈ (𝐴 ↑pm ω), 𝑦 ∈ ω ↦ if((𝐹‘𝑦) ∈ (𝐹 “ 𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹‘𝑦)⟩})))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (◡𝑁‘(𝑥 − 1))))    &   ⊢ 𝐻 = seq0(𝐺, 𝐽)    &   ⊢ 𝐿 = ∪ 𝑖 ∈ ℕ0 (𝐻‘𝑖)    ⇒   ⊢ (𝜑 → 𝐴 ≈ ℕ)
Theorem	ennnfonelemnn0 13194*	Lemma for ennnfone 13197. A version of ennnfonelemen 13193 expressed in terms of ℕ0 instead of ω. (Contributed by Jim Kingdon, 27-Oct-2022.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ℕ0–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ0 ∃𝑘 ∈ ℕ0 ∀𝑗 ∈ (0...𝑛)(𝐹‘𝑘) ≠ (𝐹‘𝑗))    &   ⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    ⇒   ⊢ (𝜑 → 𝐴 ≈ ℕ)
Theorem	ennnfonelemr 13195*	Lemma for ennnfone 13197. The interesting direction, expressed in deduction form. (Contributed by Jim Kingdon, 27-Oct-2022.)
⊢ (𝜑 → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦)    &   ⊢ (𝜑 → 𝐹:ℕ0–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ0 ∃𝑘 ∈ ℕ0 ∀𝑗 ∈ (0...𝑛)(𝐹‘𝑘) ≠ (𝐹‘𝑗))    ⇒   ⊢ (𝜑 → 𝐴 ≈ ℕ)
Theorem	ennnfonelemim 13196*	Lemma for ennnfone 13197. The trivial direction. (Contributed by Jim Kingdon, 27-Oct-2022.)
⊢ (𝐴 ≈ ℕ → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓(𝑓:ℕ0–onto→𝐴 ∧ ∀𝑛 ∈ ℕ0 ∃𝑘 ∈ ℕ0 ∀𝑗 ∈ (0...𝑛)(𝑓‘𝑘) ≠ (𝑓‘𝑗))))
Theorem	ennnfone 13197*	A condition for a set being countably infinite. Corollary 8.1.13 of [AczelRathjen], p. 73. Roughly speaking, the condition says that 𝐴 is countable (that's the 𝑓:ℕ0–onto→𝐴 part, as seen in theorems like ctm 7402), infinite (that's the part about being able to find an element of 𝐴 distinct from any mapping of a natural number via 𝑓), and has decidable equality. (Contributed by Jim Kingdon, 27-Oct-2022.)
⊢ (𝐴 ≈ ℕ ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓(𝑓:ℕ0–onto→𝐴 ∧ ∀𝑛 ∈ ℕ0 ∃𝑘 ∈ ℕ0 ∀𝑗 ∈ (0...𝑛)(𝑓‘𝑘) ≠ (𝑓‘𝑗))))
Theorem	exmidunben 13198*	If any unbounded set of positive integers is equinumerous to ℕ, then the Limited Principle of Omniscience (LPO) implies excluded middle. (Contributed by Jim Kingdon, 29-Jul-2023.)
⊢ ((∀𝑥((𝑥 ⊆ ℕ ∧ ∀𝑚 ∈ ℕ ∃𝑛 ∈ 𝑥 𝑚 < 𝑛) → 𝑥 ≈ ℕ) ∧ ω ∈ Omni) → EXMID)
Theorem	ctinfomlemom 13199*	Lemma for ctinfom 13200. Converting between ω and ℕ0. (Contributed by Jim Kingdon, 10-Aug-2023.)
⊢ 𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   ⊢ 𝐺 = (𝐹 ∘ ◡𝑁)    &   ⊢ (𝜑 → 𝐹:ω–onto→𝐴)    &   ⊢ (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ¬ (𝐹‘𝑘) ∈ (𝐹 “ 𝑛))    ⇒   ⊢ (𝜑 → (𝐺:ℕ0–onto→𝐴 ∧ ∀𝑚 ∈ ℕ0 ∃𝑗 ∈ ℕ0 ∀𝑖 ∈ (0...𝑚)(𝐺‘𝑗) ≠ (𝐺‘𝑖)))
Theorem	ctinfom 13200*	A condition for a set being countably infinite. Restates ennnfone 13197 in terms of ω and function image. Like ennnfone 13197 the condition can be summarized as 𝐴 being countable, infinite, and having decidable equality. (Contributed by Jim Kingdon, 7-Aug-2023.)
⊢ (𝐴 ≈ ℕ ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓(𝑓:ω–onto→𝐴 ∧ ∀𝑛 ∈ ω ∃𝑘 ∈ ω ¬ (𝑓‘𝑘) ∈ (𝑓 “ 𝑛))))