P. 133 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 13201-13300   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	4sqexercise2 13201*	Exercise which may help in understanding the proof of 4sqlemsdc 13202. (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 13202*	Lemma for 4sq 13212. 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 13200 and 4sqexercise2 13201 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 13203*	Lemma for 4sq 13212. 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 13199 and fin0 7189. (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 13204*	Lemma for 4sq 13212. 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 13205*	Lemma for 4sq 13212. (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 13206*	Lemma for 4sq 13212. (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 13207*	Lemma for 4sq 13212. (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 13208*	Lemma for 4sq 13212. (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 13209*	Lemma for 4sq 13212. (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 13210*	Lemma for 4sq 13212. 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 13211*	Lemma for 4sq 13212. 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 13210. 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 13196 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 13212*	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 13213	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 13214	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 13215	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 13216	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 13217	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 13218	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 13219	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 13220	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	mod2xnegi 13221	Version of mod2xi 13219 where 𝐴↑𝐵 does not equal 𝐾 (mod 𝑁) but instead the negative of 𝐾 (mod 𝑁). (Contributed by Mario Carneiro, 21-Feb-2014.)
⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐾 ∈ ℕ    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ 𝐿 ∈ ℕ0    &   ⊢ ((𝐴↑𝐵) mod 𝑁) = (𝐿 mod 𝑁)    &   ⊢ (2 · 𝐵) = 𝐸    &   ⊢ (𝐿 + 𝐾) = 𝑁    &   ⊢ ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐾)    ⇒   ⊢ ((𝐴↑𝐸) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	modsubi 13222	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 13223	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 13224	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 13225	Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ (𝐴↑0) = 1
Theorem	numexp1 13226	Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ (𝐴↑1) = 𝐴
Theorem	numexpp1 13227	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 13228	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 13229	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 13230	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 13231	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 13232	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 13233	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 9851. (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 13234	Two to the fourth power is 16. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑4) = ;16
Theorem	2exp5 13235	Two to the fifth power is 32. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑5) = ;32
Theorem	2exp6 13236	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 13237	Two to the seventh power is 128. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑7) = ;;128
Theorem	2exp8 13238	Two to the eighth power is 256. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑8) = ;;256
Theorem	2exp11 13239	Two to the eleventh power is 2048. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑;11) = ;;;2048
Theorem	2exp16 13240	Two to the sixteenth power is 65536. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑;16) = ;;;;65536
Theorem	3exp3 13241	Three to the third power is 27. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (3↑3) = ;27
Theorem	2expltfac 13242	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  Specific prime numbers
Theorem	prmlem0 13243*	Lemma for prmlem1a 13244 and prmlem2 13257. (Contributed by Mario Carneiro, 18-Feb-2014.)
⊢ ((¬ 2 ∥ 𝑀 ∧ 𝑥 ∈ (ℤ≥‘𝑀)) → ((𝑥 ∈ (ℙ ∖ {2}) ∧ (𝑥↑2) ≤ 𝑁) → ¬ 𝑥 ∥ 𝑁))    &   ⊢ (𝐾 ∈ ℙ → ¬ 𝐾 ∥ 𝑁)    &   ⊢ (𝐾 + 2) = 𝑀    ⇒   ⊢ ((¬ 2 ∥ 𝐾 ∧ 𝑥 ∈ (ℤ≥‘𝐾)) → ((𝑥 ∈ (ℙ ∖ {2}) ∧ (𝑥↑2) ≤ 𝑁) → ¬ 𝑥 ∥ 𝑁))
Theorem	prmlem1a 13244*	Lemma for prmlem1 13245 and prmlem2 13257. (Contributed by Mario Carneiro, 18-Feb-2014.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 1 < 𝑁    &   ⊢ ¬ 2 ∥ 𝑁    &   ⊢ ¬ 3 ∥ 𝑁    &   ⊢ ((¬ 2 ∥ 5 ∧ 𝑥 ∈ (ℤ≥‘5)) → ((𝑥 ∈ (ℙ ∖ {2}) ∧ (𝑥↑2) ≤ 𝑁) → ¬ 𝑥 ∥ 𝑁))    ⇒   ⊢ 𝑁 ∈ ℙ
Theorem	prmlem1 13245	A quick proof skeleton to show that the numbers less than 25 are prime, by trial division. (Contributed by Mario Carneiro, 18-Feb-2014.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 1 < 𝑁    &   ⊢ ¬ 2 ∥ 𝑁    &   ⊢ ¬ 3 ∥ 𝑁    &   ⊢ 𝑁 < ;25    ⇒   ⊢ 𝑁 ∈ ℙ
Theorem	5prm 13246	5 is a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.)
⊢ 5 ∈ ℙ
Theorem	6nprm 13247	6 is not a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.)
⊢ ¬ 6 ∈ ℙ
Theorem	7prm 13248	7 is a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.)
⊢ 7 ∈ ℙ
Theorem	8nprm 13249	8 is not a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.)
⊢ ¬ 8 ∈ ℙ
Theorem	9nprm 13250	9 is not a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.)
⊢ ¬ 9 ∈ ℙ
Theorem	10nprm 13251	10 is not a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.) (Revised by AV, 6-Sep-2021.) (Proof shortened by Umit Teoman Dogan, 10-Jun-2026.)
⊢ ¬ ;10 ∈ ℙ
Theorem	11prm 13252	11 is a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.)
⊢ ;11 ∈ ℙ
Theorem	13prm 13253	13 is a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.)
⊢ ;13 ∈ ℙ
Theorem	17prm 13254	17 is a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.)
⊢ ;17 ∈ ℙ
Theorem	19prm 13255	19 is a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.)
⊢ ;19 ∈ ℙ
Theorem	23prm 13256	23 is a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.)
⊢ ;23 ∈ ℙ
Theorem	prmlem2 13257	Our last proving session got as far as 25 because we started with the two "bootstrap" primes 2 and 3, and the next prime is 5, so knowing that 2 and 3 are prime and 4 is not allows to cover the numbers less than 5↑2 = 25. Additionally, nonprimes are "easy", so we can extend this range of known prime/nonprimes all the way until 29, which is the first prime larger than 25. Thus, in this lemma we extend another blanket out to 29↑2 = 841, from which we can prove even more primes. If we wanted, we could keep doing this, but the goal is Bertrand's postulate, and for that we only need a few large primes - we don't need to find them all, as we have been doing thus far. So after this blanket runs out, we'll have to switch to another method (see 1259prm 13270). As a side note, you can see the pattern of the primes in the indentation pattern of this lemma! (Contributed by Mario Carneiro, 18-Feb-2014.) (Proof shortened by Mario Carneiro, 20-Apr-2015.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑁 < ;;841    &   ⊢ 1 < 𝑁    &   ⊢ ¬ 2 ∥ 𝑁    &   ⊢ ¬ 3 ∥ 𝑁    &   ⊢ ¬ 5 ∥ 𝑁    &   ⊢ ¬ 7 ∥ 𝑁    &   ⊢ ¬ ;11 ∥ 𝑁    &   ⊢ ¬ ;13 ∥ 𝑁    &   ⊢ ¬ ;17 ∥ 𝑁    &   ⊢ ¬ ;19 ∥ 𝑁    &   ⊢ ¬ ;23 ∥ 𝑁    ⇒   ⊢ 𝑁 ∈ ℙ
Theorem	37prm 13258	37 is a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.) (Proof shortened by Mario Carneiro, 20-Apr-2015.)
⊢ ;37 ∈ ℙ
Theorem	43prm 13259	43 is a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.) (Proof shortened by Mario Carneiro, 20-Apr-2015.)
⊢ ;43 ∈ ℙ
Theorem	83prm 13260	83 is a prime number. (Contributed by Mario Carneiro, 18-Feb-2014.) (Proof shortened by Mario Carneiro, 20-Apr-2015.)
⊢ ;83 ∈ ℙ
Theorem	139prm 13261	139 is a prime number. (Contributed by Mario Carneiro, 19-Feb-2014.) (Proof shortened by Mario Carneiro, 20-Apr-2015.)
⊢ ;;139 ∈ ℙ
Theorem	163prm 13262	163 is a prime number. (Contributed by Mario Carneiro, 19-Feb-2014.) (Proof shortened by Mario Carneiro, 20-Apr-2015.)
⊢ ;;163 ∈ ℙ
Theorem	317prm 13263	317 is a prime number. (Contributed by Mario Carneiro, 19-Feb-2014.) (Proof shortened by Mario Carneiro, 20-Apr-2015.)
⊢ ;;317 ∈ ℙ
Theorem	631prm 13264	631 is a prime number. (Contributed by Mario Carneiro, 1-Mar-2014.) (Proof shortened by Mario Carneiro, 20-Apr-2015.)
⊢ ;;631 ∈ ℙ
5.2.15  Very large primes
Theorem	1259lem1 13265	Lemma for 1259prm 13270. Calculate a power mod. In decimal, we calculate 2↑16 = 52𝑁 + 68≡68 and 2↑17≡68 · 2 = 136 in this lemma. (Contributed by Mario Carneiro, 22-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.) (Proof shortened by AV, 16-Sep-2021.)
⊢ 𝑁 = ;;;1259    ⇒   ⊢ ((2↑;17) mod 𝑁) = (;;136 mod 𝑁)
Theorem	1259lem2 13266	Lemma for 1259prm 13270. Calculate a power mod. In decimal, we calculate 2↑34 = (2↑17)↑2≡136↑2≡14𝑁 + 870. (Contributed by Mario Carneiro, 22-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.) (Proof shortened by AV, 15-Sep-2021.)
⊢ 𝑁 = ;;;1259    ⇒   ⊢ ((2↑;34) mod 𝑁) = (;;870 mod 𝑁)
Theorem	1259lem3 13267	Lemma for 1259prm 13270. Calculate a power mod. In decimal, we calculate 2↑38 = 2↑34 · 2↑4≡870 · 16 = 11𝑁 + 71 and 2↑76 = (2↑34)↑2≡71↑2 = 4𝑁 + 5≡5. (Contributed by Mario Carneiro, 22-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.) (Proof shortened by AV, 16-Sep-2021.)
⊢ 𝑁 = ;;;1259    ⇒   ⊢ ((2↑;76) mod 𝑁) = (5 mod 𝑁)
Theorem	1259lem4 13268	Lemma for 1259prm 13270. Calculate a power mod. In decimal, we calculate 2↑306 = (2↑76)↑4 · 4≡5↑4 · 4 = 2𝑁 − 18, 2↑612 = (2↑306)↑2≡18↑2 = 324, 2↑629 = 2↑612 · 2↑17≡324 · 136 = 35𝑁 − 1 and finally 2↑(𝑁 − 1) = (2↑629)↑2≡1↑2 = 1. (Contributed by Mario Carneiro, 22-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.) (Proof shortened by AV, 16-Sep-2021.)
⊢ 𝑁 = ;;;1259    ⇒   ⊢ ((2↑(𝑁 − 1)) mod 𝑁) = (1 mod 𝑁)
Theorem	1259lem5 13269	Lemma for 1259prm 13270. Calculate the GCD of 2↑34 − 1≡869 with 𝑁 = 1259. (Contributed by Mario Carneiro, 22-Feb-2014.) (Revised by Mario Carneiro, 20-Apr-2015.)
⊢ 𝑁 = ;;;1259    ⇒   ⊢ (((2↑;34) − 1) gcd 𝑁) = 1
Theorem	1259prm 13270	1259 is a prime number. (Contributed by Mario Carneiro, 22-Feb-2014.) (Proof shortened by Mario Carneiro, 20-Apr-2015.)
⊢ 𝑁 = ;;;1259    ⇒   ⊢ 𝑁 ∈ ℙ
5.2.16  Bertrand's Ballot Problem
Theorem	ballotfilemofi 13271*	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 13272*	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 13273*	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 13274*	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 13275*	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 13276*	Lemma for ballotfi . The portion of a counting representing votes for A 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 13277*	Lemma for ballotfi . The portion of a counting representing votes for B 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	ballotfilemcinfz 13278*	Lemma for ballotfi . The portion of a counting representing votes for A within a specified integer range is finite. (Contributed by Jim Kingdon, 15-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((𝐽...𝐾) ∩ 𝐶) ∈ Fin)
Theorem	ballotfilemdifcfz 13279*	Lemma for ballotfi . The portion of a counting representing votes for B within a specified integer range is finite. (Contributed by Jim Kingdon, 15-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((𝐽...𝐾) ∖ 𝐶) ∈ Fin)
Theorem	ballotfilem2 13280*	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 13281*	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 13282*	(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 13283*	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 13284*	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 13285*	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 13286*	(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 13287*	(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 13288*	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	ballotfilemefi 13289*	𝐸 is finite. (Contributed by Jim Kingdon, 17-Jun-2026.)
⊢ 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)}    ⇒   ⊢ 𝐸 ∈ Fin
Theorem	ballotfilemafi 13290*	The set of countings where A got the first vote, but does not stay strictly ahead throughout, is finite. (Contributed by Jim Kingdon, 17-Jun-2026.)
⊢ 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) ∣ 1 ∈ 𝑐} ∈ Fin
Theorem	ballotfilembfi 13291*	The set of countings where B got the first vote is finite. (Contributed by Jim Kingdon, 17-Jun-2026.)
⊢ 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) ∣ ¬ 1 ∈ 𝑐} ∈ Fin
Theorem	ballotfilemodife 13292*	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 13293*	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))
Theorem	ballotfilem5 13294*	If A is not ahead throughout, there is a 𝑘 where votes are tied. (Contributed by Thierry Arnoux, 1-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)}    &   ⊢ 𝑁 < 𝑀    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → ∃𝑘 ∈ (1...(𝑀 + 𝑁))((𝐹‘𝐶)‘𝑘) = 0)
Theorem	ballotfilemi 13295*	Value of 𝐼 for a given counting 𝐶. (Contributed by Thierry Arnoux, 1-Dec-2016.) (Revised by AV, 6-Oct-2020.)
⊢ 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)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → (𝐼‘𝐶) = inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝐶)‘𝑘) = 0}, ℝ, < ))
Theorem	ballotfilemiex 13296*	Properties of (𝐼‘𝐶). (Contributed by Thierry Arnoux, 12-Dec-2016.) (Revised by AV, 6-Oct-2020.)
⊢ 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)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → ((𝐼‘𝐶) ∈ (1...(𝑀 + 𝑁)) ∧ ((𝐹‘𝐶)‘(𝐼‘𝐶)) = 0))
Theorem	ballotfilemi1 13297*	The first tie cannot be reached at the first pick. (Contributed by Thierry Arnoux, 12-Mar-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)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ ¬ 1 ∈ 𝐶) → (𝐼‘𝐶) ≠ 1)
Theorem	ballotfilemii 13298*	The first tie cannot be reached at the first pick. (Contributed by Thierry Arnoux, 4-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)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 1 ∈ 𝐶) → (𝐼‘𝐶) ≠ 1)
Theorem	ballotfilemscl 13299*	The set of zeroes of 𝐹 has an infimum. (Contributed by Jim Kingdon, 12-Jun-2026.)
⊢ 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)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝐶)‘𝑘) = 0}    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → inf(𝑆, ℝ, < ) ∈ 𝑆)
Theorem	ballotfilemsle 13300*	The infimum of the set of zeroes of 𝐹 is a lower bound. (Contributed by Jim Kingdon, 12-Jun-2026.)
⊢ 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)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝐶)‘𝑘) = 0}    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝑋 ∈ 𝑆) → inf(𝑆, ℝ, < ) ≤ 𝑋)