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Theorem List for Metamath Proof Explorer - 29701-29800   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremballotlemimin 29701* (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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)

Theoremballotlemic 29702* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)

Theoremballotlem1c 29703* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)

Theoremballotlemsval 29704* Value of 𝑆. (Contributed by Thierry Arnoux, 12-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)))

Theoremballotlemsv 29705* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))

Theoremballotlemsgt1 29706* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))

Theoremballotlemsdom 29707* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)))

Theoremballotlemsel1i 29708* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)))

Theoremballotlemsf1o 29709* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)))

Theoremballotlemsi 29710* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)

Theoremballotlemsima 29711* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)))

Theoremballotlemieq 29712* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))

Theoremballotlemrval 29713* Value of 𝑅. (Contributed by Thierry Arnoux, 14-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))

Theoremballotlemscr 29714* The image of (𝑅𝐶) by (𝑆𝐶). (Contributed by Thierry Arnoux, 21-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)

Theoremballotlemrv 29715* Value of 𝑅 evaluated at 𝐽. (Contributed by Thierry Arnoux, 17-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))

Theoremballotlemrv1 29716* Value of 𝑅 before the tie. (Contributed by Thierry Arnoux, 11-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))

Theoremballotlemrv2 29717* Value of 𝑅 after the tie. (Contributed by Thierry Arnoux, 11-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))

Theoremballotlemro 29718* Range of 𝑅 is included in 𝑂. (Contributed by Thierry Arnoux, 17-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)

Theoremballotlemgval 29719* Expand the value of . (Contributed by Thierry Arnoux, 21-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))       ((𝑈 ∈ Fin ∧ 𝑉 ∈ Fin) → (𝑈 𝑉) = ((#‘(𝑉𝑈)) − (#‘(𝑉𝑈))))

Theoremballotlemgun 29720* A property of the defined operator. (Contributed by Thierry Arnoux, 26-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))    &   (𝜑𝑈 ∈ Fin)    &   (𝜑𝑉 ∈ Fin)    &   (𝜑𝑊 ∈ Fin)    &   (𝜑 → (𝑉𝑊) = ∅)       (𝜑 → (𝑈 (𝑉𝑊)) = ((𝑈 𝑉) + (𝑈 𝑊)))

Theoremballotlemfg 29721* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (0...(𝑀 + 𝑁))) → ((𝐹𝐶)‘𝐽) = (𝐶 (1...𝐽)))

Theoremballotlemfrc 29722* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝐼𝐶))) → ((𝐹‘(𝑅𝐶))‘𝐽) = (𝐶 (((𝑆𝐶)‘𝐽)...(𝐼𝐶))))

Theoremballotlemfrci 29723* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))       (𝐶 ∈ (𝑂𝐸) → ((𝐹‘(𝑅𝐶))‘(𝐼𝐶)) = 0)

Theoremballotlemfrceq 29724* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝐼𝐶))) → ((𝐹𝐶)‘(((𝑆𝐶)‘𝐽) − 1)) = -((𝐹‘(𝑅𝐶))‘𝐽))

Theoremballotlemfrcn0 29725* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)

Theoremballotlemrc 29726* Range of 𝑅. (Contributed by Thierry Arnoux, 19-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))

Theoremballotlemirc 29727* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))

Theoremballotlemrinv0 29728* Lemma for ballotlemrinv 29729. (Contributed by Thierry Arnoux, 18-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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)))

Theoremballotlemrinv 29729* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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

Theoremballotlem1ri 29730* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))

Theoremballotlem7 29731* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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}

Theoremballotlem8 29732* 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)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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})

Theoremballotth 29733* Bertrand's ballot problem : the probability that A is ahead throughout the counting. This is Metamath 100 proof #30. (Contributed by Thierry Arnoux, 7-Dec-2016.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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))

TheoremballotlemiOLD 29734* Value of 𝐼 for a given counting 𝐶. (Contributed by Thierry Arnoux, 1-Dec-2016.) Obsolete version of ballotlemi 29696 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       (𝐶 ∈ (𝑂𝐸) → (𝐼𝐶) = sup({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝐶)‘𝑘) = 0}, ℝ, < ))

TheoremballotlemiexOLD 29735* Properties of (𝐼𝐶). (Contributed by Thierry Arnoux, 12-Dec-2016.) Obsolete version of ballotlemiex 29697 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       (𝐶 ∈ (𝑂𝐸) → ((𝐼𝐶) ∈ (1...(𝑀 + 𝑁)) ∧ ((𝐹𝐶)‘(𝐼𝐶)) = 0))

Theoremballotlemi1OLD 29736* The first tie cannot be reached at the first pick. (Contributed by Thierry Arnoux, 12-Mar-2017.) Obsolete version of ballotlemi1 29698 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       ((𝐶 ∈ (𝑂𝐸) ∧ ¬ 1 ∈ 𝐶) → (𝐼𝐶) ≠ 1)

TheoremballotlemiiOLD 29737* The first tie cannot be reached at the first pick. (Contributed by Thierry Arnoux, 4-Apr-2017.) Obsolete version of ballotlemii 29699 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       ((𝐶 ∈ (𝑂𝐸) ∧ 1 ∈ 𝐶) → (𝐼𝐶) ≠ 1)

TheoremballotlemsupOLD 29738* The set of zeroes of 𝐹 satisfies the conditions to have a supremum. (Contributed by Thierry Arnoux, 1-Dec-2016.) Obsolete version of ballotlemsup 29700 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       (𝐶 ∈ (𝑂𝐸) → ∃𝑧 ∈ ℝ (∀𝑤 ∈ {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝐶)‘𝑘) = 0} ¬ 𝑧 < 𝑤 ∧ ∀𝑤 ∈ ℝ (𝑤 < 𝑧 → ∃𝑦 ∈ {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝐶)‘𝑘) = 0}𝑤 < 𝑦)))

TheoremballotlemiminOLD 29739* (𝐼𝐶) is the first tie. (Contributed by Thierry Arnoux, 1-Dec-2016.) Obsolete version of ballotlemimin 29701 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       (𝐶 ∈ (𝑂𝐸) → ¬ ∃𝑘 ∈ (1...((𝐼𝐶) − 1))((𝐹𝐶)‘𝑘) = 0)

TheoremballotlemicOLD 29740* If the first vote is for B, the vote on the first tie is for A. (Contributed by Thierry Arnoux, 1-Dec-2016.) Obsolete version of ballotlemic 29702 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       ((𝐶 ∈ (𝑂𝐸) ∧ ¬ 1 ∈ 𝐶) → (𝐼𝐶) ∈ 𝐶)

Theoremballotlem1cOLD 29741* If the first vote is for A, the vote on the first tie is for B. (Contributed by Thierry Arnoux, 4-Apr-2017.) Obsolete version of ballotlem1c 29703 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       ((𝐶 ∈ (𝑂𝐸) ∧ 1 ∈ 𝐶) → ¬ (𝐼𝐶) ∈ 𝐶)

TheoremballotlemsvalOLD 29742* Value of 𝑆. (Contributed by Thierry Arnoux, 12-Apr-2017.) Obsolete version of ballotlemsval 29704 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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)))

TheoremballotlemsvOLD 29743* Value of 𝑆 evaluated at 𝐽 for a given counting 𝐶. (Contributed by Thierry Arnoux, 12-Apr-2017.) Obsolete version of ballotlemsv 29705 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))

Theoremballotlemsgt1OLD 29744* 𝑆 maps values less than (𝐼𝐶) to values greater than 1. (Contributed by Thierry Arnoux, 28-Apr-2017.) Obsolete version of ballotlemsgt1 29706 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))

TheoremballotlemsdomOLD 29745* Domain of 𝑆 for a given counting 𝐶. (Contributed by Thierry Arnoux, 12-Apr-2017.) Obsolete version of ballotlemsdom 29707 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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)))

Theoremballotlemsel1iOLD 29746* The range (1...(𝐼𝐶)) is invariant under (𝑆𝐶). (Contributed by Thierry Arnoux, 28-Apr-2017.) Obsolete version of ballotlemsel1i 29708 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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)))

Theoremballotlemsf1oOLD 29747* The defined 𝑆 is a bijection, and an involution. (Contributed by Thierry Arnoux, 14-Apr-2017.) Obsolete version of ballotlemsf1o 29709 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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)))

TheoremballotlemsiOLD 29748* The image by 𝑆 of the first tie pick is the first pick. (Contributed by Thierry Arnoux, 14-Apr-2017.) Obsolete version of ballotlemsi 29710 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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)

TheoremballotlemsimaOLD 29749* The image by 𝑆 of an interval before the first pick. (Contributed by Thierry Arnoux, 5-May-2017.) Obsolete version of ballotlemsima 29711 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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)))

TheoremballotlemieqOLD 29750* If two countings share the same first tie, they also have the same swap function. (Contributed by Thierry Arnoux, 18-Apr-2017.) Obsolete version of ballotlemieq 29712 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))

TheoremballotlemrvalOLD 29751* Value of 𝑅. (Contributed by Thierry Arnoux, 14-Apr-2017.) Obsolete version of ballotlemrval 29713 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))

TheoremballotlemscrOLD 29752* The image of (𝑅𝐶) by (𝑆𝐶). (Contributed by Thierry Arnoux, 21-Apr-2017.) Obsolete version of ballotlemscr 29714 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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)

TheoremballotlemrvOLD 29753* Value of 𝑅 evaluated at 𝐽. (Contributed by Thierry Arnoux, 17-Apr-2017.) Obsolete version of ballotlemrv 29715 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))

Theoremballotlemrv1OLD 29754* Value of 𝑅 before the tie. (Contributed by Thierry Arnoux, 11-Apr-2017.) Obsolete version of ballotlemrv1 29716 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))

Theoremballotlemrv2OLD 29755* Value of 𝑅 after the tie. (Contributed by Thierry Arnoux, 11-Apr-2017.) Obsolete version of ballotlemrv2 29717 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))

TheoremballotlemroOLD 29756* Range of 𝑅 is included in 𝑂. (Contributed by Thierry Arnoux, 17-Apr-2017.) Obsolete version of ballotlemro 29718 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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)

TheoremballotlemgvalOLD 29757* Expand the value of . (Contributed by Thierry Arnoux, 21-Apr-2017.) Obsolete version of ballotlemgval 29719 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))       ((𝑈 ∈ Fin ∧ 𝑉 ∈ Fin) → (𝑈 𝑉) = ((#‘(𝑉𝑈)) − (#‘(𝑉𝑈))))

TheoremballotlemgunOLD 29758* A property of the defined operator. (Contributed by Thierry Arnoux, 26-Apr-2017.) Obsolete version of ballotlemgun 29720 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))    &   (𝜑𝑈 ∈ Fin)    &   (𝜑𝑉 ∈ Fin)    &   (𝜑𝑊 ∈ Fin)    &   (𝜑 → (𝑉𝑊) = ∅)       (𝜑 → (𝑈 (𝑉𝑊)) = ((𝑈 𝑉) + (𝑈 𝑊)))

TheoremballotlemfgOLD 29759* Express the value of (𝐹𝐶) in terms of . (Contributed by Thierry Arnoux, 21-Apr-2017.) Obsolete version of ballotlemfg 29721 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (0...(𝑀 + 𝑁))) → ((𝐹𝐶)‘𝐽) = (𝐶 (1...𝐽)))

TheoremballotlemfrcOLD 29760* Express the value of (𝐹‘(𝑅𝐶)) in terms of the newly defined . (Contributed by Thierry Arnoux, 21-Apr-2017.) Obsolete version of ballotlemfrc 29722 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝐼𝐶))) → ((𝐹‘(𝑅𝐶))‘𝐽) = (𝐶 (((𝑆𝐶)‘𝐽)...(𝐼𝐶))))

TheoremballotlemfrciOLD 29761* Reverse counting preserves a tie at the first tie. (Contributed by Thierry Arnoux, 21-Apr-2017.) Obsolete version of ballotlemfrci 29723 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))       (𝐶 ∈ (𝑂𝐸) → ((𝐹‘(𝑅𝐶))‘(𝐼𝐶)) = 0)

TheoremballotlemfrceqOLD 29762* Value of 𝐹 for a reverse counting (𝑅𝐶). (Contributed by Thierry Arnoux, 27-Apr-2017.) Obsolete version of ballotlemfrceq 29724 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))    &    = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((#‘(𝑣𝑢)) − (#‘(𝑣𝑢))))       ((𝐶 ∈ (𝑂𝐸) ∧ 𝐽 ∈ (1...(𝐼𝐶))) → ((𝐹𝐶)‘(((𝑆𝐶)‘𝐽) − 1)) = -((𝐹‘(𝑅𝐶))‘𝐽))

Theoremballotlemfrcn0OLD 29763* Value of 𝐹 for a reversed counting (𝑅𝐶), before the first tie, cannot be zero . (Contributed by Thierry Arnoux, 25-Apr-2017.) Obsolete version of ballotlemfrcn0 29725 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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)

TheoremballotlemrcOLD 29764* Range of 𝑅. (Contributed by Thierry Arnoux, 19-Apr-2017.) Obsolete version of ballotlemrc 29726 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))

TheoremballotlemircOLD 29765* Applying 𝑅 does not change first ties. (Contributed by Thierry Arnoux, 19-Apr-2017.) Obsolete version of ballotlemirc 29727 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))

Theoremballotlemrinv0OLD 29766* Lemma for ballotlemrinv 29729. (Contributed by Thierry Arnoux, 18-Apr-2017.) Obsolete version of ballotlemrinv0 29728 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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)))

TheoremballotlemrinvOLD 29767* 𝑅 is its own inverse : it is an involution. (Contributed by Thierry Arnoux, 10-Apr-2017.) Obsolete version of ballotlemrinv 29729 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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

Theoremballotlem1riOLD 29768* 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.) Obsolete version of ballotlem1ri 29730 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))

Theoremballotlem7OLD 29769* 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.) Obsolete version of ballotlem7 29731 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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}

Theoremballotlem8OLD 29770* 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.) Obsolete version of ballotlem8 29732 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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})

TheoremballotthOLD 29771* Bertrand's ballot problem : the probability that A is ahead throughout the counting. This is Metamath 100 proof #30. (Contributed by Thierry Arnoux, 7-Dec-2016.) Obsolete version of ballotth 29733 as of 6-Oct-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (#‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((#‘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) ↦ sup({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))

20.3.21  Signum (sgn or sign) function - misc. additions

Theoremsgncl 29772 Closure of the signum. (Contributed by Thierry Arnoux, 28-Sep-2018.)
(𝐴 ∈ ℝ* → (sgn‘𝐴) ∈ {-1, 0, 1})

Theoremsgnclre 29773 Closure of the signum. (Contributed by Thierry Arnoux, 28-Sep-2018.)
(𝐴 ∈ ℝ → (sgn‘𝐴) ∈ ℝ)

Theoremsgnneg 29774 Negation of the signum. (Contributed by Thierry Arnoux, 1-Oct-2018.)
(𝐴 ∈ ℝ → (sgn‘-𝐴) = -(sgn‘𝐴))

Theoremsgn3da 29775 A conditional containing a signum is true if it is true in all three possible cases. (Contributed by Thierry Arnoux, 1-Oct-2018.)
(𝜑𝐴 ∈ ℝ*)    &   ((sgn‘𝐴) = 0 → (𝜓𝜒))    &   ((sgn‘𝐴) = 1 → (𝜓𝜃))    &   ((sgn‘𝐴) = -1 → (𝜓𝜏))    &   ((𝜑𝐴 = 0) → 𝜒)    &   ((𝜑 ∧ 0 < 𝐴) → 𝜃)    &   ((𝜑𝐴 < 0) → 𝜏)       (𝜑𝜓)

Theoremsgnmul 29776 Signum of a product. (Contributed by Thierry Arnoux, 2-Oct-2018.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (sgn‘(𝐴 · 𝐵)) = ((sgn‘𝐴) · (sgn‘𝐵)))

Theoremsgnmulrp2 29777 Multiplication by a positive number does not affect signum. (Contributed by Thierry Arnoux, 2-Oct-2018.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+) → (sgn‘(𝐴 · 𝐵)) = (sgn‘𝐴))

Theoremsgnsub 29778 Subtraction of a number of opposite sign. (Contributed by Thierry Arnoux, 2-Oct-2018.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (𝐴 · 𝐵) < 0) → (sgn‘(𝐴𝐵)) = (sgn‘𝐴))

Theoremsgnnbi 29779 Negative signum. (Contributed by Thierry Arnoux, 2-Oct-2018.)
(𝐴 ∈ ℝ* → ((sgn‘𝐴) = -1 ↔ 𝐴 < 0))

Theoremsgnpbi 29780 Positive signum. (Contributed by Thierry Arnoux, 2-Oct-2018.)
(𝐴 ∈ ℝ* → ((sgn‘𝐴) = 1 ↔ 0 < 𝐴))

Theoremsgn0bi 29781 Zero signum. (Contributed by Thierry Arnoux, 10-Oct-2018.)
(𝐴 ∈ ℝ* → ((sgn‘𝐴) = 0 ↔ 𝐴 = 0))

Theoremsgnsgn 29782 Signum is idempotent. (Contributed by Thierry Arnoux, 2-Oct-2018.)
(𝐴 ∈ ℝ* → (sgn‘(sgn‘𝐴)) = (sgn‘𝐴))

Theoremsgnmulsgn 29783 If two real numbers are of different signs, so are their signs. (Contributed by Thierry Arnoux, 12-Oct-2018.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((𝐴 · 𝐵) < 0 ↔ ((sgn‘𝐴) · (sgn‘𝐵)) < 0))

Theoremsgnmulsgp 29784 If two real numbers are of different signs, so are their signs. (Contributed by Thierry Arnoux, 12-Oct-2018.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (0 < (𝐴 · 𝐵) ↔ 0 < ((sgn‘𝐴) · (sgn‘𝐵))))

Theoremfzssfzo 29785 Condition for an integer interval to be a subset of an half-open integer interval. (Contributed by Thierry Arnoux, 8-Oct-2018.)
(𝐾 ∈ (𝑀..^𝑁) → (𝑀...𝐾) ⊆ (𝑀..^𝑁))

Theoremgsumncl 29786* Closure of a group sum in a non-commutative monoid. (Contributed by Thierry Arnoux, 8-Oct-2018.)
𝐾 = (Base‘𝑀)    &   (𝜑𝑀 ∈ Mnd)    &   (𝜑𝑃 ∈ (ℤ𝑁))    &   ((𝜑𝑘 ∈ (𝑁...𝑃)) → 𝐵𝐾)       (𝜑 → (𝑀 Σg (𝑘 ∈ (𝑁...𝑃) ↦ 𝐵)) ∈ 𝐾)

Theoremgsumnunsn 29787* Closure of a group sum in a non-commutative monoid. (Contributed by Thierry Arnoux, 8-Oct-2018.)
𝐾 = (Base‘𝑀)    &   (𝜑𝑀 ∈ Mnd)    &   (𝜑𝑃 ∈ (ℤ𝑁))    &   ((𝜑𝑘 ∈ (𝑁...𝑃)) → 𝐵𝐾)    &    + = (+g𝑀)    &   (𝜑𝐶𝐾)    &   ((𝜑𝑘 = (𝑃 + 1)) → 𝐵 = 𝐶)       (𝜑 → (𝑀 Σg (𝑘 ∈ (𝑁...(𝑃 + 1)) ↦ 𝐵)) = ((𝑀 Σg (𝑘 ∈ (𝑁...𝑃) ↦ 𝐵)) + 𝐶))

20.3.22  Words over a set - misc additions

Theoremwrdres 29788 Condition for the restriction of a word to be a word itself. (Contributed by Thierry Arnoux, 5-Oct-2018.)
((𝑊 ∈ Word 𝑆𝑁 ∈ (0...(#‘𝑊))) → (𝑊 ↾ (0..^𝑁)) ∈ Word 𝑆)

Theoremwrdsplex 29789* Existence of a split of a word at a given index. (Contributed by Thierry Arnoux, 11-Oct-2018.)
((𝑊 ∈ Word 𝑆𝑁 ∈ (0...(#‘𝑊))) → ∃𝑣 ∈ Word 𝑆𝑊 = ((𝑊 ↾ (0..^𝑁)) ++ 𝑣))

20.3.22.1  Operations on words

Theoremccatmulgnn0dir 29790 Concatenation of words follow the rule mulgnn0dir 17286 (although applying mulgnn0dir 17286 would require 𝑆 to be a set). In this case 𝐴 is ⟨“𝐾”⟩ to the power 𝑀 in the free monoid. (Contributed by Thierry Arnoux, 5-Oct-2018.)
𝐴 = ((0..^𝑀) × {𝐾})    &   𝐵 = ((0..^𝑁) × {𝐾})    &   𝐶 = ((0..^(𝑀 + 𝑁)) × {𝐾})    &   (𝜑𝐾𝑆)    &   (𝜑𝑀 ∈ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝐴 ++ 𝐵) = 𝐶)

Theoremofcccat 29791 Letterwise operations on word concatenations. (Contributed by Thierry Arnoux, 5-Oct-2018.)
(𝜑𝐹 ∈ Word 𝑆)    &   (𝜑𝐺 ∈ Word 𝑆)    &   (𝜑𝐾𝑇)       (𝜑 → ((𝐹 ++ 𝐺)∘𝑓/𝑐𝑅𝐾) = ((𝐹𝑓/𝑐𝑅𝐾) ++ (𝐺𝑓/𝑐𝑅𝐾)))

Theoremofcs1 29792 Letterwise operations on a single letter word. (Contributed by Thierry Arnoux, 7-Oct-2018.)
((𝐴𝑆𝐵𝑇) → (⟨“𝐴”⟩∘𝑓/𝑐𝑅𝐵) = ⟨“(𝐴𝑅𝐵)”⟩)

Theoremofcs2 29793 Letterwise operations on a double letter word. (Contributed by Thierry Arnoux, 9-Oct-2018.)
((𝐴𝑆𝐵𝑆𝐶𝑇) → (⟨“𝐴𝐵”⟩∘𝑓/𝑐𝑅𝐶) = ⟨“(𝐴𝑅𝐶)(𝐵𝑅𝐶)”⟩)

20.3.23  Polynomials with real coefficients - misc additions

Theoremplymul02 29794 Product of a polynomial with the zero polynomial. (Contributed by Thierry Arnoux, 26-Sep-2018.)
(𝐹 ∈ (Poly‘𝑆) → (0𝑝𝑓 · 𝐹) = 0𝑝)

Theoremplymulx0 29795* Coefficients of a polynomial multiplyed by Xp. (Contributed by Thierry Arnoux, 25-Sep-2018.)
(𝐹 ∈ ((Poly‘ℝ) ∖ {0𝑝}) → (coeff‘(𝐹𝑓 · Xp)) = (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, 0, ((coeff‘𝐹)‘(𝑛 − 1)))))

Theoremplymulx 29796* Coefficients of a polynomial multiplyed by Xp. (Contributed by Thierry Arnoux, 25-Sep-2018.)
(𝐹 ∈ (Poly‘ℝ) → (coeff‘(𝐹𝑓 · Xp)) = (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, 0, ((coeff‘𝐹)‘(𝑛 − 1)))))

Theoremplyrecld 29797 Closure of a polynomial with real coefficients. (Contributed by Thierry Arnoux, 18-Sep-2018.)
(𝜑𝐹 ∈ (Poly‘ℝ))    &   (𝜑𝑋 ∈ ℝ)       (𝜑 → (𝐹𝑋) ∈ ℝ)

Theoremsignsplypnf 29798* The quotient of a polynomial 𝐹 by a monic monomial of same degree 𝐺 converges to the highest coefficient of 𝐹. (Contributed by Thierry Arnoux, 18-Sep-2018.)
𝐷 = (deg‘𝐹)    &   𝐶 = (coeff‘𝐹)    &   𝐵 = (𝐶𝐷)    &   𝐺 = (𝑥 ∈ ℝ+ ↦ (𝑥𝐷))       (𝐹 ∈ (Poly‘ℝ) → (𝐹𝑓 / 𝐺) ⇝𝑟 𝐵)

Theoremsignsply0 29799* Lemma for the rule of signs, based on Bolzano's intermediate value theorem for polynomials : If the lowest and highest coefficient 𝐴 and 𝐵 are of opposite signs, the polynomial admits a positive root. (Contributed by Thierry Arnoux, 19-Sep-2018.)
𝐷 = (deg‘𝐹)    &   𝐶 = (coeff‘𝐹)    &   𝐵 = (𝐶𝐷)    &   𝐴 = (𝐶‘0)    &   (𝜑𝐹 ∈ (Poly‘ℝ))    &   (𝜑𝐹 ≠ 0𝑝)    &   (𝜑 → (𝐴 · 𝐵) < 0)       (𝜑 → ∃𝑧 ∈ ℝ+ (𝐹𝑧) = 0)

20.3.24  Descartes's rule of signs

20.3.24.1  Sign changes in a word over real numbers

Theoremsignspval 29800* The value of the skipping 0 sign operation. (Contributed by Thierry Arnoux, 9-Sep-2018.)
= (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))       ((𝑋 ∈ {-1, 0, 1} ∧ 𝑌 ∈ {-1, 0, 1}) → (𝑋 𝑌) = if(𝑌 = 0, 𝑋, 𝑌))

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