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Theorem List for Intuitionistic Logic Explorer - 13201-13300   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremgcdi 13201 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
 
Theoremgcdmodi 13202 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
 
Theoremnumexp0 13203 Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0       (𝐴↑0) = 1
 
Theoremnumexp1 13204 Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0       (𝐴↑1) = 𝐴
 
Theoremnumexpp1 13205 Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0    &   𝑀 ∈ ℕ0    &   (𝑀 + 1) = 𝑁    &   ((𝐴𝑀) · 𝐴) = 𝐶       (𝐴𝑁) = 𝐶
 
Theoremnumexp2x 13206 Double an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0    &   𝑀 ∈ ℕ0    &   (2 · 𝑀) = 𝑁    &   (𝐴𝑀) = 𝐷    &   (𝐷 · 𝐷) = 𝐶       (𝐴𝑁) = 𝐶
 
Theoremdecsplit0b 13207 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)
 
Theoremdecsplit0 13208 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
 
Theoremdecsplit1 13209 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
 
Theoremdecsplit 13210 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
 
Theoremkaratsuba 13211 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 9843. (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
 
Theorem2exp4 13212 Two to the fourth power is 16. (Contributed by Mario Carneiro, 20-Apr-2015.)
(2↑4) = 16
 
Theorem2exp5 13213 Two to the fifth power is 32. (Contributed by AV, 16-Aug-2021.)
(2↑5) = 32
 
Theorem2exp6 13214 Two to the sixth power is 64. (Contributed by Mario Carneiro, 20-Apr-2015.) (Proof shortened by OpenAI, 25-Mar-2020.)
(2↑6) = 64
 
Theorem2exp7 13215 Two to the seventh power is 128. (Contributed by AV, 16-Aug-2021.)
(2↑7) = 128
 
Theorem2exp8 13216 Two to the eighth power is 256. (Contributed by Mario Carneiro, 20-Apr-2015.)
(2↑8) = 256
 
Theorem2exp11 13217 Two to the eleventh power is 2048. (Contributed by AV, 16-Aug-2021.)
(2↑11) = 2048
 
Theorem2exp16 13218 Two to the sixteenth power is 65536. (Contributed by Mario Carneiro, 20-Apr-2015.)
(2↑16) = 65536
 
Theorem3exp3 13219 Three to the third power is 27. (Contributed by Mario Carneiro, 20-Apr-2015.)
(3↑3) = 27
 
Theorem2expltfac 13220 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
 
Theoremballotfilemofi 13221* 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
 
Theoremballotfilem1 13222* 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)
 
Theoremballotfilemonn 13223* 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
 
Theoremballotfilemelo 13224* Elementhood in 𝑂. (Contributed by Thierry Arnoux, 17-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}       (𝐶𝑂 ↔ (𝐶 ⊆ (1...(𝑀 + 𝑁)) ∧ 𝐶 ∈ Fin ∧ (♯‘𝐶) = 𝑀))
 
Theoremballotfilemcdc 13225* 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}    &   (𝜑𝐶𝑂)    &   (𝜑𝐾 ∈ ℤ)       (𝜑DECID 𝐾𝐶)
 
Theoremballotfilemcinfi 13226* 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}    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℤ)       (𝜑 → ((1...𝐽) ∩ 𝐶) ∈ Fin)
 
Theoremballotfilemdifcfi 13227* 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}    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℤ)       (𝜑 → ((1...𝐽) ∖ 𝐶) ∈ Fin)
 
Theoremballotfilemcinfz 13228* 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}    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℤ)    &   (𝜑𝐾 ∈ ℤ)       (𝜑 → ((𝐽...𝐾) ∩ 𝐶) ∈ Fin)
 
Theoremballotfilemdifcfz 13229* 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}    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℤ)    &   (𝜑𝐾 ∈ ℤ)       (𝜑 → ((𝐽...𝐾) ∖ 𝐶) ∈ Fin)
 
Theoremballotfilem2 13230* 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‘{𝑐𝑂 ∣ ¬ 1 ∈ 𝑐}) = (𝑁 / (𝑀 + 𝑁))
 
Theoremballotfilemfval 13231* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℤ)       (𝜑 → ((𝐹𝐶)‘𝐽) = ((♯‘((1...𝐽) ∩ 𝐶)) − (♯‘((1...𝐽) ∖ 𝐶))))
 
Theoremballotfilemfelz 13232* (𝐹𝐶) has values in . (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℤ)       (𝜑 → ((𝐹𝐶)‘𝐽) ∈ ℤ)
 
Theoremballotfilemfp1 13233* If the 𝐽 th ballot is for A, (𝐹𝐶) goes up 1. If the 𝐽 th ballot is for B, (𝐹𝐶) 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℕ)       (𝜑 → ((¬ 𝐽𝐶 → ((𝐹𝐶)‘𝐽) = (((𝐹𝐶)‘(𝐽 − 1)) − 1)) ∧ (𝐽𝐶 → ((𝐹𝐶)‘𝐽) = (((𝐹𝐶)‘(𝐽 − 1)) + 1))))
 
Theoremballotfilemfc0 13234* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℕ)    &   (𝜑 → ∃𝑖 ∈ (1...𝐽)((𝐹𝐶)‘𝑖) ≤ 0)    &   (𝜑 → 0 < ((𝐹𝐶)‘𝐽))       (𝜑 → ∃𝑘 ∈ (1...𝐽)((𝐹𝐶)‘𝑘) = 0)
 
Theoremballotfilemfcc 13235* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℕ)    &   (𝜑 → ∃𝑖 ∈ (1...𝐽)0 ≤ ((𝐹𝐶)‘𝑖))    &   (𝜑 → ((𝐹𝐶)‘𝐽) < 0)       (𝜑 → ∃𝑘 ∈ (1...𝐽)((𝐹𝐶)‘𝑘) = 0)
 
Theoremballotfilemfmpn 13236* (𝐹𝐶) finishes counting at (𝑀𝑁). (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))       (𝐶𝑂 → ((𝐹𝐶)‘(𝑀 + 𝑁)) = (𝑀𝑁))
 
Theoremballotfilemfval0 13237* (𝐹𝐶) 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))       (𝐶𝑂 → ((𝐹𝐶)‘0) = 0)
 
Theoremballotfileme 13238* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}       (𝐶𝐸 ↔ (𝐶𝑂 ∧ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝐶)‘𝑖)))
 
Theoremballotfilemefi 13239* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}       𝐸 ∈ Fin
 
Theoremballotfilemafi 13240* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}       {𝑐 ∈ (𝑂𝐸) ∣ 1 ∈ 𝑐} ∈ Fin
 
Theoremballotfilembfi 13241* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}       {𝑐 ∈ (𝑂𝐸) ∣ ¬ 1 ∈ 𝑐} ∈ Fin
 
Theoremballotfilemodife 13242* Elements of (𝑂𝐸). (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}       (𝐶 ∈ (𝑂𝐸) ↔ (𝐶𝑂 ∧ ∃𝑖 ∈ (1...(𝑀 + 𝑁))((𝐹𝐶)‘𝑖) ≤ 0))
 
Theoremballotfilem4 13243* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}       (𝐶𝑂 → (¬ 1 ∈ 𝐶 → ¬ 𝐶𝐸))
 
Theoremballotfilem5 13244* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀       (𝐶 ∈ (𝑂𝐸) → ∃𝑘 ∈ (1...(𝑀 + 𝑁))((𝐹𝐶)‘𝑘) = 0)
 
Theoremballotfilemi 13245* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       (𝐶 ∈ (𝑂𝐸) → (𝐼𝐶) = inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝐶)‘𝑘) = 0}, ℝ, < ))
 
Theoremballotfilemiex 13246* Properties of (𝐼𝐶). (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       (𝐶 ∈ (𝑂𝐸) → ((𝐼𝐶) ∈ (1...(𝑀 + 𝑁)) ∧ ((𝐹𝐶)‘(𝐼𝐶)) = 0))
 
Theoremballotfilemi1 13247* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       ((𝐶 ∈ (𝑂𝐸) ∧ ¬ 1 ∈ 𝐶) → (𝐼𝐶) ≠ 1)
 
Theoremballotfilemii 13248* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       ((𝐶 ∈ (𝑂𝐸) ∧ 1 ∈ 𝐶) → (𝐼𝐶) ≠ 1)
 
Theoremballotfilemscl 13249* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝐶)‘𝑘) = 0}       (𝐶 ∈ (𝑂𝐸) → inf(𝑆, ℝ, < ) ∈ 𝑆)
 
Theoremballotfilemsle 13250* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝐶)‘𝑘) = 0}       ((𝐶 ∈ (𝑂𝐸) ∧ 𝑋𝑆) → inf(𝑆, ℝ, < ) ≤ 𝑋)
 
Theoremballotfilemimin 13251* (𝐼𝐶) is the first tie. (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       (𝐶 ∈ (𝑂𝐸) → ¬ ∃𝑘 ∈ (1...((𝐼𝐶) − 1))((𝐹𝐶)‘𝑘) = 0)
 
Theoremballotfilemic 13252* If the first vote is for B, the vote on the first tie is for A. (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       ((𝐶 ∈ (𝑂𝐸) ∧ ¬ 1 ∈ 𝐶) → (𝐼𝐶) ∈ 𝐶)
 
Theoremballotfilem1c 13253* If the first vote is for A, the vote on the first tie is for B. (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       ((𝐶 ∈ (𝑂𝐸) ∧ 1 ∈ 𝐶) → ¬ (𝐼𝐶) ∈ 𝐶)
 
Theoremballotfilemsval 13254* Value of 𝑆. (Contributed by Thierry Arnoux, 12-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))       (𝐶 ∈ (𝑂𝐸) → (𝑆𝐶) = (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝐶), (((𝐼𝐶) + 1) − 𝑖), 𝑖)))
 
Theoremballotfilemsv 13255* Value of 𝑆 evaluated at 𝐽 for a given counting 𝐶. (Contributed by Thierry Arnoux, 12-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁))) → ((𝑆𝐶)‘𝐽) = if(𝐽 ≤ (𝐼𝐶), (((𝐼𝐶) + 1) − 𝐽), 𝐽))
 
Theoremballotfilemsgt1 13256* 𝑆 maps values less than (𝐼𝐶) to values greater than 1. (Contributed by Thierry Arnoux, 28-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁)) ∧ 𝐽 < (𝐼𝐶)) → 1 < ((𝑆𝐶)‘𝐽))
 
Theoremballotfilemsdom 13257* Domain of 𝑆 for a given counting 𝐶. (Contributed by Thierry Arnoux, 12-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁))) → ((𝑆𝐶)‘𝐽) ∈ (1...(𝑀 + 𝑁)))
 
Theoremballotfilemsel1i 13258* The range (1...(𝐼𝐶)) is invariant under (𝑆𝐶). (Contributed by Thierry Arnoux, 28-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝐼𝐶))) → ((𝑆𝐶)‘𝐽) ∈ (1...(𝐼𝐶)))
 
Theoremballotfilemsf1o 13259* The defined 𝑆 is a bijection, and an involution. (Contributed by Thierry Arnoux, 14-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))       (𝐶 ∈ (𝑂𝐸) → ((𝑆𝐶):(1...(𝑀 + 𝑁))–1-1-onto→(1...(𝑀 + 𝑁)) ∧ (𝑆𝐶) = (𝑆𝐶)))
 
Theoremballotfilemsi 13260* The image by 𝑆 of the first tie pick is the first pick. (Contributed by Thierry Arnoux, 14-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))       (𝐶 ∈ (𝑂𝐸) → ((𝑆𝐶)‘(𝐼𝐶)) = 1)
 
Theoremballotfilemsima 13261* The image by 𝑆 of an interval before the first pick. (Contributed by Thierry Arnoux, 5-May-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝐼𝐶))) → ((𝑆𝐶) “ (1...𝐽)) = (((𝑆𝐶)‘𝐽)...(𝐼𝐶)))
 
Theoremballotfilemieq 13262* If two countings share the same first tie, they also have the same swap function. (Contributed by Thierry Arnoux, 18-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐷 ∈ (𝑂𝐸) ∧ (𝐼𝐶) = (𝐼𝐷)) → (𝑆𝐶) = (𝑆𝐷))
 
Theoremballotfilemrval 13263* Value of 𝑅. (Contributed by Thierry Arnoux, 14-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       (𝐶 ∈ (𝑂𝐸) → (𝑅𝐶) = ((𝑆𝐶) “ 𝐶))
 
Theoremballotfilemscr 13264* The image of (𝑅𝐶) by (𝑆𝐶). (Contributed by Thierry Arnoux, 21-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       (𝐶 ∈ (𝑂𝐸) → ((𝑆𝐶) “ (𝑅𝐶)) = 𝐶)
 
Theoremballotfilemrv 13265* Value of 𝑅 evaluated at 𝐽. (Contributed by Thierry Arnoux, 17-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁))) → (𝐽 ∈ (𝑅𝐶) ↔ if(𝐽 ≤ (𝐼𝐶), (((𝐼𝐶) + 1) − 𝐽), 𝐽) ∈ 𝐶))
 
Theoremballotfilemrv1 13266* Value of 𝑅 before the tie. (Contributed by Thierry Arnoux, 11-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁)) ∧ 𝐽 ≤ (𝐼𝐶)) → (𝐽 ∈ (𝑅𝐶) ↔ (((𝐼𝐶) + 1) − 𝐽) ∈ 𝐶))
 
Theoremballotfilemrv2 13267* Value of 𝑅 after the tie. (Contributed by Thierry Arnoux, 11-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁)) ∧ (𝐼𝐶) < 𝐽) → (𝐽 ∈ (𝑅𝐶) ↔ 𝐽𝐶))
 
Theoremballotfilemro 13268* Range of 𝑅 is included in 𝑂. (Contributed by Thierry Arnoux, 17-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       (𝐶 ∈ (𝑂𝐸) → (𝑅𝐶) ∈ 𝑂)
 
Theoremballotfilemgval 13269* Expand the value of . (Contributed by Thierry Arnoux, 21-Apr-2017.) (Revised by Jim Kingdon, 15-Jun-2026.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))    &    = (𝑢𝑂, 𝑣 ∈ Fin ↦ ((♯‘(𝑣𝑢)) − (♯‘(𝑣𝑢))))    &   (𝜑𝑈𝑂)    &   (𝜑𝐽 ∈ ℤ)    &   (𝜑𝐾 ∈ ℤ)    &   (𝜑𝑉 = (𝐽...𝐾))       (𝜑 → (𝑈 𝑉) = ((♯‘(𝑉𝑈)) − (♯‘(𝑉𝑈))))
 
Theoremballotfilemgun 13270* A property of the defined operator. (Contributed by Thierry Arnoux, 26-Apr-2017.) (Revised by Jim Kingdon, 15-Jun-2026.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))    &    = (𝑢𝑂, 𝑣 ∈ Fin ↦ ((♯‘(𝑣𝑢)) − (♯‘(𝑣𝑢))))    &   (𝜑𝑈𝑂)    &   (𝜑𝐿 ∈ (𝐽...𝐾))       (𝜑 → (𝑈 (𝐽...𝐾)) = ((𝑈 (𝐽...(𝐿 − 1))) + (𝑈 (𝐿...𝐾))))
 
Theoremballotfilemfg 13271* Express the value of (𝐹𝐶) in terms of . (Contributed by Thierry Arnoux, 21-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))    &    = (𝑢𝑂, 𝑣 ∈ Fin ↦ ((♯‘(𝑣𝑢)) − (♯‘(𝑣𝑢))))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (0...(𝑀 + 𝑁))) → ((𝐹𝐶)‘𝐽) = (𝐶 (1...𝐽)))
 
Theoremballotfilemfrc 13272* Express the value of (𝐹‘(𝑅𝐶)) in terms of the newly defined . (Contributed by Thierry Arnoux, 21-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))    &    = (𝑢𝑂, 𝑣 ∈ Fin ↦ ((♯‘(𝑣𝑢)) − (♯‘(𝑣𝑢))))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝐼𝐶))) → ((𝐹‘(𝑅𝐶))‘𝐽) = (𝐶 (((𝑆𝐶)‘𝐽)...(𝐼𝐶))))
 
Theoremballotfilemfrci 13273* Reverse counting preserves a tie at the first tie. (Contributed by Thierry Arnoux, 21-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))    &    = (𝑢𝑂, 𝑣 ∈ Fin ↦ ((♯‘(𝑣𝑢)) − (♯‘(𝑣𝑢))))       (𝐶 ∈ (𝑂𝐸) → ((𝐹‘(𝑅𝐶))‘(𝐼𝐶)) = 0)
 
Theoremballotfilemfrceq 13274* Value of 𝐹 for a reverse counting (𝑅𝐶). (Contributed by Thierry Arnoux, 27-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))    &    = (𝑢𝑂, 𝑣 ∈ Fin ↦ ((♯‘(𝑣𝑢)) − (♯‘(𝑣𝑢))))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝐼𝐶))) → ((𝐹𝐶)‘(((𝑆𝐶)‘𝐽) − 1)) = -((𝐹‘(𝑅𝐶))‘𝐽))
 
Theoremballotfilemfrcn0 13275* Value of 𝐹 for a reversed counting (𝑅𝐶), before the first tie, cannot be zero. (Contributed by Thierry Arnoux, 25-Apr-2017.) (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁)) ∧ 𝐽 < (𝐼𝐶)) → ((𝐹‘(𝑅𝐶))‘𝐽) ≠ 0)
 
Theoremballotfilemrc 13276* Range of 𝑅. (Contributed by Thierry Arnoux, 19-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       (𝐶 ∈ (𝑂𝐸) → (𝑅𝐶) ∈ (𝑂𝐸))
 
Theoremballotfilemirc 13277* Applying 𝑅 does not change first ties. (Contributed by Thierry Arnoux, 19-Apr-2017.) (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       (𝐶 ∈ (𝑂𝐸) → (𝐼‘(𝑅𝐶)) = (𝐼𝐶))
 
Theoremballotfilemrinv0 13278* Lemma for ballotfilemrinv 13279. (Contributed by Thierry Arnoux, 18-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐷 = ((𝑆𝐶) “ 𝐶)) → (𝐷 ∈ (𝑂𝐸) ∧ 𝐶 = ((𝑆𝐷) “ 𝐷)))
 
Theoremballotfilemrinv 13279* 𝑅 is its own inverse : it is an involution. (Contributed by Thierry Arnoux, 10-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       𝑅 = 𝑅
 
Theoremballotfilem1ri 13280* When the vote on the first tie is for A, the first vote is also for A on the reverse counting. (Contributed by Thierry Arnoux, 18-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       (𝐶 ∈ (𝑂𝐸) → (1 ∈ (𝑅𝐶) ↔ (𝐼𝐶) ∈ 𝐶))
 
Theoremballotfilem7 13281* 𝑅 is a bijection between two subsets of (𝑂𝐸): one where a vote for A is picked first, and one where a vote for B is picked first. (Contributed by Thierry Arnoux, 12-Dec-2016.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       (𝑅 ↾ {𝑐 ∈ (𝑂𝐸) ∣ 1 ∈ 𝑐}):{𝑐 ∈ (𝑂𝐸) ∣ 1 ∈ 𝑐}–1-1-onto→{𝑐 ∈ (𝑂𝐸) ∣ ¬ 1 ∈ 𝑐}
 
Theoremballotfilem8 13282* There are as many countings with ties starting with a ballot for 𝐴 as there are starting with a ballot for 𝐵. (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       (♯‘{𝑐 ∈ (𝑂𝐸) ∣ 1 ∈ 𝑐}) = (♯‘{𝑐 ∈ (𝑂𝐸) ∣ ¬ 1 ∈ 𝑐})
 
Theoremballotfilemth 13283* Lemma for ballotfi 13284. The result, with several additional hypotheses which are for use during the proof. (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))    &   𝑆 = (𝑐 ∈ (𝑂𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼𝑐), (((𝐼𝑐) + 1) − 𝑖), 𝑖)))    &   𝑅 = (𝑐 ∈ (𝑂𝐸) ↦ ((𝑆𝑐) “ 𝑐))       (𝑃𝐸) = ((𝑀𝑁) / (𝑀 + 𝑁))
 
Theoremballotfi 13284* Bertrand's ballot problem : the probability that A is ahead throughout the counting. The proof formalized here is a proof "by reflection", as opposed to other known proofs "by induction" or "by permutation". This is Metamath 100 proof #30. (Contributed by Thierry Arnoux, 7-Dec-2016.) (Revised 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀       (𝑃𝐸) = ((𝑀𝑁) / (𝑀 + 𝑁))
 
5.3  Cardinality of real and complex number subsets
 
5.3.1  Countability of integers and rationals
 
Theoremoddennn 13285 There are as many odd positive integers as there are positive integers. (Contributed by Jim Kingdon, 11-May-2022.)
{𝑧 ∈ ℕ ∣ ¬ 2 ∥ 𝑧} ≈ ℕ
 
Theoremevenennn 13286 There are as many even positive integers as there are positive integers. (Contributed by Jim Kingdon, 12-May-2022.)
{𝑧 ∈ ℕ ∣ 2 ∥ 𝑧} ≈ ℕ
 
Theoremxpnnen 13287 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
 
Theoremxpomen 13288 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
 
Theoremxpct 13289 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)
 
Theoremunennn 13290 The union of two disjoint countably infinite sets is countably infinite. (Contributed by Jim Kingdon, 13-May-2022.)
((𝐴 ≈ ℕ ∧ 𝐵 ≈ ℕ ∧ (𝐴𝐵) = ∅) → (𝐴𝐵) ≈ ℕ)
 
Theoremznnen 13291 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
 
Theoremennnfonelemdc 13292* Lemma for ennnfone 13318. A direct consequence of fidcenumlemrk 7271. (Contributed by Jim Kingdon, 15-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑𝑃 ∈ ω)       (𝜑DECID (𝐹𝑃) ∈ (𝐹𝑃))
 
Theoremennnfonelemk 13293* Lemma for ennnfone 13318. (Contributed by Jim Kingdon, 15-Jul-2023.)
(𝜑𝐹:ω–onto𝐴)    &   (𝜑𝐾 ∈ ω)    &   (𝜑𝑁 ∈ ω)    &   (𝜑 → ∀𝑗 ∈ suc 𝑁(𝐹𝐾) ≠ (𝐹𝑗))       (𝜑𝑁𝐾)
 
Theoremennnfonelemj0 13294* Lemma for ennnfone 13318. Initial state for 𝐽. (Contributed by Jim Kingdon, 20-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       (𝜑 → (𝐽‘0) ∈ {𝑔 ∈ (𝐴pm ω) ∣ dom 𝑔 ∈ ω})
 
Theoremennnfonelemjn 13295* Lemma for ennnfone 13318. Non-initial state for 𝐽. (Contributed by Jim Kingdon, 20-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       ((𝜑𝑓 ∈ (ℤ‘(0 + 1))) → (𝐽𝑓) ∈ ω)
 
Theoremennnfonelemg 13296* Lemma for ennnfone 13318. Closure for 𝐺. (Contributed by Jim Kingdon, 20-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       ((𝜑 ∧ (𝑓 ∈ {𝑔 ∈ (𝐴pm ω) ∣ dom 𝑔 ∈ ω} ∧ 𝑗 ∈ ω)) → (𝑓𝐺𝑗) ∈ {𝑔 ∈ (𝐴pm ω) ∣ dom 𝑔 ∈ ω})
 
Theoremennnfonelemh 13297* Lemma for ennnfone 13318. (Contributed by Jim Kingdon, 8-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       (𝜑𝐻:ℕ0⟶(𝐴pm ω))
 
Theoremennnfonelem0 13298* Lemma for ennnfone 13318. Initial value. (Contributed by Jim Kingdon, 15-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       (𝜑 → (𝐻‘0) = ∅)
 
Theoremennnfonelemp1 13299* Lemma for ennnfone 13318. Value of 𝐻 at a successor. (Contributed by Jim Kingdon, 23-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   (𝜑𝑃 ∈ ℕ0)       (𝜑 → (𝐻‘(𝑃 + 1)) = if((𝐹‘(𝑁𝑃)) ∈ (𝐹 “ (𝑁𝑃)), (𝐻𝑃), ((𝐻𝑃) ∪ {⟨dom (𝐻𝑃), (𝐹‘(𝑁𝑃))⟩})))
 
Theoremennnfonelem1 13300* Lemma for ennnfone 13318. Second value. (Contributed by Jim Kingdon, 19-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       (𝜑 → (𝐻‘1) = {⟨∅, (𝐹‘∅)⟩})
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