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Theorem List for Intuitionistic Logic Explorer - 13201-13300   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theorem2expltfac 13201 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 13202* 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 13203* 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 13204* 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 13205* 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 13206* 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 13207* 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 13208* 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 13209* 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 13210* 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 13211* 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 13212* 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 13213* (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 13214* 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 13215* 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 13216* 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 13217* (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 13218* (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 13219* 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 13220* 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 13221* 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 13222* 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 13223* 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 13224* 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 13225* 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 13226* 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 13227* 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 13228* 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 13229* 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 13230* 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 13231* 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 13232* (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 13233* 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 13234* 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 13235* 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 13236* 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 13237* 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 13238* 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 13239* 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 13240* 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 13241* 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 13242* 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 13243* 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 13244* 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 13245* 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 13246* 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 13247* 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 13248* 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 13249* 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 13250* 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 13251* 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 13252* 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 13253* 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 13254* 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 13255* 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 13256* 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 13257* 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 13258* 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 13259* Lemma for ballotfilemrinv 13260. (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 13260* 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 13261* 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 13262* 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 13263* 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 13264* Lemma for ballotfi 13265. 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 13265* 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 13266 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 13267 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 13268 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 13269 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 13270 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 13271 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 13272 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 13273* Lemma for ennnfone 13299. A direct consequence of fidcenumlemrk 7265. (Contributed by Jim Kingdon, 15-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑𝑃 ∈ ω)       (𝜑DECID (𝐹𝑃) ∈ (𝐹𝑃))
 
Theoremennnfonelemk 13274* Lemma for ennnfone 13299. (Contributed by Jim Kingdon, 15-Jul-2023.)
(𝜑𝐹:ω–onto𝐴)    &   (𝜑𝐾 ∈ ω)    &   (𝜑𝑁 ∈ ω)    &   (𝜑 → ∀𝑗 ∈ suc 𝑁(𝐹𝐾) ≠ (𝐹𝑗))       (𝜑𝑁𝐾)
 
Theoremennnfonelemj0 13275* Lemma for ennnfone 13299. 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 13276* Lemma for ennnfone 13299. 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 13277* Lemma for ennnfone 13299. 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 13278* Lemma for ennnfone 13299. (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 13279* Lemma for ennnfone 13299. 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 13280* Lemma for ennnfone 13299. 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 13281* Lemma for ennnfone 13299. 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‘∅)⟩})
 
Theoremennnfonelemom 13282* Lemma for ennnfone 13299. 𝐻 yields finite sequences. (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)    &   (𝜑𝑃 ∈ ℕ0)       (𝜑 → dom (𝐻𝑃) ∈ ω)
 
Theoremennnfonelemhdmp1 13283* Lemma for ennnfone 13299. Domain at a successor where we need to add an element to the sequence. (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 → ¬ (𝐹‘(𝑁𝑃)) ∈ (𝐹 “ (𝑁𝑃)))       (𝜑 → dom (𝐻‘(𝑃 + 1)) = suc dom (𝐻𝑃))
 
Theoremennnfonelemss 13284* Lemma for ennnfone 13299. We only add elements to 𝐻 as the index increases. (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)    &   (𝜑𝑃 ∈ ℕ0)       (𝜑 → (𝐻𝑃) ⊆ (𝐻‘(𝑃 + 1)))
 
Theoremennnfoneleminc 13285* Lemma for ennnfone 13299. We only add elements to 𝐻 as the index increases. (Contributed by Jim Kingdon, 21-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)    &   (𝜑𝑄 ∈ ℕ0)    &   (𝜑𝑃𝑄)       (𝜑 → (𝐻𝑃) ⊆ (𝐻𝑄))
 
Theoremennnfonelemkh 13286* Lemma for ennnfone 13299. 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 → ∀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 → dom (𝐻𝑃) ⊆ (𝑁𝑃))
 
Theoremennnfonelemhf1o 13287* Lemma for ennnfone 13299. Each of the functions in 𝐻 is one to one and onto an image of 𝐹. (Contributed by Jim Kingdon, 17-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 → (𝐻𝑃):dom (𝐻𝑃)–1-1-onto→(𝐹 “ (𝑁𝑃)))
 
Theoremennnfonelemex 13288* Lemma for ennnfone 13299. Extending the sequence (𝐻𝑃) to include an additional element. (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)    &   (𝜑𝑃 ∈ ℕ0)       (𝜑 → ∃𝑖 ∈ ℕ0 dom (𝐻𝑃) ∈ dom (𝐻𝑖))
 
Theoremennnfonelemhom 13289* Lemma for ennnfone 13299. The sequences in 𝐻 increase in length without bound if you go out far enough. (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)    &   (𝜑𝑀 ∈ ω)       (𝜑 → ∃𝑖 ∈ ℕ0 𝑀 ∈ dom (𝐻𝑖))
 
Theoremennnfonelemrnh 13290* Lemma for ennnfone 13299. A consequence of ennnfonelemss 13284. (Contributed by Jim Kingdon, 16-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)    &   (𝜑𝑋 ∈ ran 𝐻)    &   (𝜑𝑌 ∈ ran 𝐻)       (𝜑 → (𝑋𝑌𝑌𝑋))
 
Theoremennnfonelemfun 13291* Lemma for ennnfone 13299. 𝐿 is a function. (Contributed by Jim Kingdon, 16-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)    &   𝐿 = 𝑖 ∈ ℕ0 (𝐻𝑖)       (𝜑 → Fun 𝐿)
 
Theoremennnfonelemf1 13292* Lemma for ennnfone 13299. 𝐿 is one-to-one. (Contributed by Jim Kingdon, 16-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)    &   𝐿 = 𝑖 ∈ ℕ0 (𝐻𝑖)       (𝜑𝐿:dom 𝐿1-1𝐴)
 
Theoremennnfonelemrn 13293* Lemma for ennnfone 13299. 𝐿 is onto 𝐴. (Contributed by Jim Kingdon, 16-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)    &   𝐿 = 𝑖 ∈ ℕ0 (𝐻𝑖)       (𝜑 → ran 𝐿 = 𝐴)
 
Theoremennnfonelemdm 13294* Lemma for ennnfone 13299. The function 𝐿 is defined everywhere. (Contributed by Jim Kingdon, 16-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)    &   𝐿 = 𝑖 ∈ ℕ0 (𝐻𝑖)       (𝜑 → dom 𝐿 = ω)
 
Theoremennnfonelemen 13295* Lemma for ennnfone 13299. The result. (Contributed by Jim Kingdon, 16-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)    &   𝐿 = 𝑖 ∈ ℕ0 (𝐻𝑖)       (𝜑𝐴 ≈ ℕ)
 
Theoremennnfonelemnn0 13296* Lemma for ennnfone 13299. A version of ennnfonelemen 13295 expressed in terms of 0 instead of ω. (Contributed by Jim Kingdon, 27-Oct-2022.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ℕ0onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ℕ0𝑘 ∈ ℕ0𝑗 ∈ (0...𝑛)(𝐹𝑘) ≠ (𝐹𝑗))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)       (𝜑𝐴 ≈ ℕ)
 
Theoremennnfonelemr 13297* Lemma for ennnfone 13299. The interesting direction, expressed in deduction form. (Contributed by Jim Kingdon, 27-Oct-2022.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ℕ0onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ℕ0𝑘 ∈ ℕ0𝑗 ∈ (0...𝑛)(𝐹𝑘) ≠ (𝐹𝑗))       (𝜑𝐴 ≈ ℕ)
 
Theoremennnfonelemim 13298* Lemma for ennnfone 13299. The trivial direction. (Contributed by Jim Kingdon, 27-Oct-2022.)
(𝐴 ≈ ℕ → (∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓(𝑓:ℕ0onto𝐴 ∧ ∀𝑛 ∈ ℕ0𝑘 ∈ ℕ0𝑗 ∈ (0...𝑛)(𝑓𝑘) ≠ (𝑓𝑗))))
 
Theoremennnfone 13299* 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:ℕ0onto𝐴 part, as seen in theorems like ctm 7443), 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 ↔ (∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓(𝑓:ℕ0onto𝐴 ∧ ∀𝑛 ∈ ℕ0𝑘 ∈ ℕ0𝑗 ∈ (0...𝑛)(𝑓𝑘) ≠ (𝑓𝑗))))
 
Theoremexmidunben 13300* If any unbounded set of positive integers is equinumerous to , then the Limited Principle of Omniscience (LPO) implies excluded middle. (Contributed by Jim Kingdon, 29-Jul-2023.)
((∀𝑥((𝑥 ⊆ ℕ ∧ ∀𝑚 ∈ ℕ ∃𝑛𝑥 𝑚 < 𝑛) → 𝑥 ≈ ℕ) ∧ ω ∈ Omni) → EXMID)
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