P. 346 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 34501-34600   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	ballotlem4 34501*	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)) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    ⇒   ⊢ (𝐶 ∈ 𝑂 → (¬ 1 ∈ 𝐶 → ¬ 𝐶 ∈ 𝐸))
Theorem	ballotlem5 34502*	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)) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → ∃𝑘 ∈ (1...(𝑀 + 𝑁))((𝐹‘𝐶)‘𝑘) = 0)
Theorem	ballotlemi 34503*	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)) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → (𝐼‘𝐶) = inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝐶)‘𝑘) = 0}, ℝ, < ))
Theorem	ballotlemiex 34504*	Properties of (𝐼‘𝐶). (Contributed by Thierry Arnoux, 12-Dec-2016.) (Revised by AV, 6-Oct-2020.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → ((𝐼‘𝐶) ∈ (1...(𝑀 + 𝑁)) ∧ ((𝐹‘𝐶)‘(𝐼‘𝐶)) = 0))
Theorem	ballotlemi1 34505*	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)) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ ¬ 1 ∈ 𝐶) → (𝐼‘𝐶) ≠ 1)
Theorem	ballotlemii 34506*	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)) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 1 ∈ 𝐶) → (𝐼‘𝐶) ≠ 1)
Theorem	ballotlemsup 34507*	The set of zeroes of 𝐹 satisfies the conditions to have a supremum. (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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → ∃𝑧 ∈ ℝ (∀𝑤 ∈ {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝐶)‘𝑘) = 0} ¬ 𝑤 < 𝑧 ∧ ∀𝑤 ∈ ℝ (𝑧 < 𝑤 → ∃𝑦 ∈ {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝐶)‘𝑘) = 0}𝑦 < 𝑤)))
Theorem	ballotlemimin 34508*	(𝐼‘𝐶) 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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → ¬ ∃𝑘 ∈ (1...((𝐼‘𝐶) − 1))((𝐹‘𝐶)‘𝑘) = 0)
Theorem	ballotlemic 34509*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ ¬ 1 ∈ 𝐶) → (𝐼‘𝐶) ∈ 𝐶)
Theorem	ballotlem1c 34510*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 1 ∈ 𝐶) → ¬ (𝐼‘𝐶) ∈ 𝐶)
Theorem	ballotlemsval 34511*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → (𝑆‘𝐶) = (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝐶), (((𝐼‘𝐶) + 1) − 𝑖), 𝑖)))
Theorem	ballotlemsv 34512*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁))) → ((𝑆‘𝐶)‘𝐽) = if(𝐽 ≤ (𝐼‘𝐶), (((𝐼‘𝐶) + 1) − 𝐽), 𝐽))
Theorem	ballotlemsgt1 34513*	𝑆 maps values less than (𝐼‘𝐶) 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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁)) ∧ 𝐽 < (𝐼‘𝐶)) → 1 < ((𝑆‘𝐶)‘𝐽))
Theorem	ballotlemsdom 34514*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁))) → ((𝑆‘𝐶)‘𝐽) ∈ (1...(𝑀 + 𝑁)))
Theorem	ballotlemsel1i 34515*	The range (1...(𝐼‘𝐶)) is invariant under (𝑆‘𝐶). (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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (1...(𝐼‘𝐶))) → ((𝑆‘𝐶)‘𝐽) ∈ (1...(𝐼‘𝐶)))
Theorem	ballotlemsf1o 34516*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → ((𝑆‘𝐶):(1...(𝑀 + 𝑁))–1-1-onto→(1...(𝑀 + 𝑁)) ∧ ◡(𝑆‘𝐶) = (𝑆‘𝐶)))
Theorem	ballotlemsi 34517*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → ((𝑆‘𝐶)‘(𝐼‘𝐶)) = 1)
Theorem	ballotlemsima 34518*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (1...(𝐼‘𝐶))) → ((𝑆‘𝐶) “ (1...𝐽)) = (((𝑆‘𝐶)‘𝐽)...(𝐼‘𝐶)))
Theorem	ballotlemieq 34519*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐷 ∈ (𝑂 ∖ 𝐸) ∧ (𝐼‘𝐶) = (𝐼‘𝐷)) → (𝑆‘𝐶) = (𝑆‘𝐷))
Theorem	ballotlemrval 34520*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → (𝑅‘𝐶) = ((𝑆‘𝐶) “ 𝐶))
Theorem	ballotlemscr 34521*	The image of (𝑅‘𝐶) by (𝑆‘𝐶). (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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → ((𝑆‘𝐶) “ (𝑅‘𝐶)) = 𝐶)
Theorem	ballotlemrv 34522*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁))) → (𝐽 ∈ (𝑅‘𝐶) ↔ if(𝐽 ≤ (𝐼‘𝐶), (((𝐼‘𝐶) + 1) − 𝐽), 𝐽) ∈ 𝐶))
Theorem	ballotlemrv1 34523*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁)) ∧ 𝐽 ≤ (𝐼‘𝐶)) → (𝐽 ∈ (𝑅‘𝐶) ↔ (((𝐼‘𝐶) + 1) − 𝐽) ∈ 𝐶))
Theorem	ballotlemrv2 34524*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁)) ∧ (𝐼‘𝐶) < 𝐽) → (𝐽 ∈ (𝑅‘𝐶) ↔ 𝐽 ∈ 𝐶))
Theorem	ballotlemro 34525*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → (𝑅‘𝐶) ∈ 𝑂)
Theorem	ballotlemgval 34526*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    &   ⊢ ↑ = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((♯‘(𝑣 ∩ 𝑢)) − (♯‘(𝑣 ∖ 𝑢))))    ⇒   ⊢ ((𝑈 ∈ Fin ∧ 𝑉 ∈ Fin) → (𝑈 ↑ 𝑉) = ((♯‘(𝑉 ∩ 𝑈)) − (♯‘(𝑉 ∖ 𝑈))))
Theorem	ballotlemgun 34527*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    &   ⊢ ↑ = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((♯‘(𝑣 ∩ 𝑢)) − (♯‘(𝑣 ∖ 𝑢))))    &   ⊢ (𝜑 → 𝑈 ∈ Fin)    &   ⊢ (𝜑 → 𝑉 ∈ Fin)    &   ⊢ (𝜑 → 𝑊 ∈ Fin)    &   ⊢ (𝜑 → (𝑉 ∩ 𝑊) = ∅)    ⇒   ⊢ (𝜑 → (𝑈 ↑ (𝑉 ∪ 𝑊)) = ((𝑈 ↑ 𝑉) + (𝑈 ↑ 𝑊)))
Theorem	ballotlemfg 34528*	Express the value of (𝐹‘𝐶) 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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    &   ⊢ ↑ = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((♯‘(𝑣 ∩ 𝑢)) − (♯‘(𝑣 ∖ 𝑢))))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (0...(𝑀 + 𝑁))) → ((𝐹‘𝐶)‘𝐽) = (𝐶 ↑ (1...𝐽)))
Theorem	ballotlemfrc 34529*	Express the value of (𝐹‘(𝑅‘𝐶)) 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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    &   ⊢ ↑ = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((♯‘(𝑣 ∩ 𝑢)) − (♯‘(𝑣 ∖ 𝑢))))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (1...(𝐼‘𝐶))) → ((𝐹‘(𝑅‘𝐶))‘𝐽) = (𝐶 ↑ (((𝑆‘𝐶)‘𝐽)...(𝐼‘𝐶))))
Theorem	ballotlemfrci 34530*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    &   ⊢ ↑ = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((♯‘(𝑣 ∩ 𝑢)) − (♯‘(𝑣 ∖ 𝑢))))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → ((𝐹‘(𝑅‘𝐶))‘(𝐼‘𝐶)) = 0)
Theorem	ballotlemfrceq 34531*	Value of 𝐹 for a reverse counting (𝑅‘𝐶). (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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    &   ⊢ ↑ = (𝑢 ∈ Fin, 𝑣 ∈ Fin ↦ ((♯‘(𝑣 ∩ 𝑢)) − (♯‘(𝑣 ∖ 𝑢))))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (1...(𝐼‘𝐶))) → ((𝐹‘𝐶)‘(((𝑆‘𝐶)‘𝐽) − 1)) = -((𝐹‘(𝑅‘𝐶))‘𝐽))
Theorem	ballotlemfrcn0 34532*	Value of 𝐹 for a reversed counting (𝑅‘𝐶), 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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐽 ∈ (1...(𝑀 + 𝑁)) ∧ 𝐽 < (𝐼‘𝐶)) → ((𝐹‘(𝑅‘𝐶))‘𝐽) ≠ 0)
Theorem	ballotlemrc 34533*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → (𝑅‘𝐶) ∈ (𝑂 ∖ 𝐸))
Theorem	ballotlemirc 34534*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → (𝐼‘(𝑅‘𝐶)) = (𝐼‘𝐶))
Theorem	ballotlemrinv0 34535*	Lemma for ballotlemrinv 34536. (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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝐷 = ((𝑆‘𝐶) “ 𝐶)) → (𝐷 ∈ (𝑂 ∖ 𝐸) ∧ 𝐶 = ((𝑆‘𝐷) “ 𝐷)))
Theorem	ballotlemrinv 34536*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ ◡𝑅 = 𝑅
Theorem	ballotlem1ri 34537*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → (1 ∈ (𝑅‘𝐶) ↔ (𝐼‘𝐶) ∈ 𝐶))
Theorem	ballotlem7 34538*	𝑅 is a bijection between two subsets of (𝑂 ∖ 𝐸): 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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ (𝑅 ↾ {𝑐 ∈ (𝑂 ∖ 𝐸) ∣ 1 ∈ 𝑐}):{𝑐 ∈ (𝑂 ∖ 𝐸) ∣ 1 ∈ 𝑐}–1-1-onto→{𝑐 ∈ (𝑂 ∖ 𝐸) ∣ ¬ 1 ∈ 𝑐}
Theorem	ballotlem8 34539*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ (♯‘{𝑐 ∈ (𝑂 ∖ 𝐸) ∣ 1 ∈ 𝑐}) = (♯‘{𝑐 ∈ (𝑂 ∖ 𝐸) ∣ ¬ 1 ∈ 𝑐})
Theorem	ballotth 34540*	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.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    &   ⊢ 𝑁 < 𝑀    &   ⊢ 𝐼 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝑐)‘𝑘) = 0}, ℝ, < ))    &   ⊢ 𝑆 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ (𝑖 ∈ (1...(𝑀 + 𝑁)) ↦ if(𝑖 ≤ (𝐼‘𝑐), (((𝐼‘𝑐) + 1) − 𝑖), 𝑖)))    &   ⊢ 𝑅 = (𝑐 ∈ (𝑂 ∖ 𝐸) ↦ ((𝑆‘𝑐) “ 𝑐))    ⇒   ⊢ (𝑃‘𝐸) = ((𝑀 − 𝑁) / (𝑀 + 𝑁))
21.3.24  Signum (sgn or sign) function - misc. additions
Theorem	sgncl 34541	Closure of the signum. (Contributed by Thierry Arnoux, 28-Sep-2018.)
⊢ (𝐴 ∈ ℝ* → (sgn‘𝐴) ∈ {-1, 0, 1})
Theorem	sgnclre 34542	Closure of the signum. (Contributed by Thierry Arnoux, 28-Sep-2018.)
⊢ (𝐴 ∈ ℝ → (sgn‘𝐴) ∈ ℝ)
Theorem	sgnneg 34543	Negation of the signum. (Contributed by Thierry Arnoux, 1-Oct-2018.)
⊢ (𝐴 ∈ ℝ → (sgn‘-𝐴) = -(sgn‘𝐴))
Theorem	sgn3da 34544	A conditional containing a signum is true if it is true in all three possible cases. (Contributed by Thierry Arnoux, 1-Oct-2018.)
⊢ (𝜑 → 𝐴 ∈ ℝ*)    &   ⊢ ((sgn‘𝐴) = 0 → (𝜓 ↔ 𝜒))    &   ⊢ ((sgn‘𝐴) = 1 → (𝜓 ↔ 𝜃))    &   ⊢ ((sgn‘𝐴) = -1 → (𝜓 ↔ 𝜏))    &   ⊢ ((𝜑 ∧ 𝐴 = 0) → 𝜒)    &   ⊢ ((𝜑 ∧ 0 < 𝐴) → 𝜃)    &   ⊢ ((𝜑 ∧ 𝐴 < 0) → 𝜏)    ⇒   ⊢ (𝜑 → 𝜓)
Theorem	sgnmul 34545	Signum of a product. (Contributed by Thierry Arnoux, 2-Oct-2018.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (sgn‘(𝐴 · 𝐵)) = ((sgn‘𝐴) · (sgn‘𝐵)))
Theorem	sgnmulrp2 34546	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))
Theorem	sgnsub 34547	Subtraction of a number of opposite sign. (Contributed by Thierry Arnoux, 2-Oct-2018.)
⊢ (((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (𝐴 · 𝐵) < 0) → (sgn‘(𝐴 − 𝐵)) = (sgn‘𝐴))
Theorem	sgnnbi 34548	Negative signum. (Contributed by Thierry Arnoux, 2-Oct-2018.)
⊢ (𝐴 ∈ ℝ* → ((sgn‘𝐴) = -1 ↔ 𝐴 < 0))
Theorem	sgnpbi 34549	Positive signum. (Contributed by Thierry Arnoux, 2-Oct-2018.)
⊢ (𝐴 ∈ ℝ* → ((sgn‘𝐴) = 1 ↔ 0 < 𝐴))
Theorem	sgn0bi 34550	Zero signum. (Contributed by Thierry Arnoux, 10-Oct-2018.)
⊢ (𝐴 ∈ ℝ* → ((sgn‘𝐴) = 0 ↔ 𝐴 = 0))
Theorem	sgnsgn 34551	Signum is idempotent. (Contributed by Thierry Arnoux, 2-Oct-2018.)
⊢ (𝐴 ∈ ℝ* → (sgn‘(sgn‘𝐴)) = (sgn‘𝐴))
Theorem	sgnmulsgn 34552	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))
Theorem	sgnmulsgp 34553	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))))
Theorem	fzssfzo 34554	Condition for an integer interval to be a subset of a half-open integer interval. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ (𝐾 ∈ (𝑀..^𝑁) → (𝑀...𝐾) ⊆ (𝑀..^𝑁))
Theorem	gsumncl 34555*	Closure of a group sum in a non-commutative monoid. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ 𝐾 = (Base‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ Mnd)    &   ⊢ (𝜑 → 𝑃 ∈ (ℤ≥‘𝑁))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑁...𝑃)) → 𝐵 ∈ 𝐾)    ⇒   ⊢ (𝜑 → (𝑀 Σg (𝑘 ∈ (𝑁...𝑃) ↦ 𝐵)) ∈ 𝐾)
Theorem	gsumnunsn 34556*	Closure of a group sum in a non-commutative monoid. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ 𝐾 = (Base‘𝑀)    &   ⊢ (𝜑 → 𝑀 ∈ Mnd)    &   ⊢ (𝜑 → 𝑃 ∈ (ℤ≥‘𝑁))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (𝑁...𝑃)) → 𝐵 ∈ 𝐾)    &   ⊢ + = (+g‘𝑀)    &   ⊢ (𝜑 → 𝐶 ∈ 𝐾)    &   ⊢ ((𝜑 ∧ 𝑘 = (𝑃 + 1)) → 𝐵 = 𝐶)    ⇒   ⊢ (𝜑 → (𝑀 Σg (𝑘 ∈ (𝑁...(𝑃 + 1)) ↦ 𝐵)) = ((𝑀 Σg (𝑘 ∈ (𝑁...𝑃) ↦ 𝐵)) + 𝐶))
21.3.24.1  Operations on words
Theorem	ccatmulgnn0dir 34557	Concatenation of words follow the rule mulgnn0dir 19122 (although applying mulgnn0dir 19122 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})    &   ⊢ (𝜑 → 𝐾 ∈ 𝑆)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝐴 ++ 𝐵) = 𝐶)
Theorem	ofcccat 34558	Letterwise operations on word concatenations. (Contributed by Thierry Arnoux, 5-Oct-2018.)
⊢ (𝜑 → 𝐹 ∈ Word 𝑆)    &   ⊢ (𝜑 → 𝐺 ∈ Word 𝑆)    &   ⊢ (𝜑 → 𝐾 ∈ 𝑇)    ⇒   ⊢ (𝜑 → ((𝐹 ++ 𝐺) ∘f/c 𝑅𝐾) = ((𝐹 ∘f/c 𝑅𝐾) ++ (𝐺 ∘f/c 𝑅𝐾)))
Theorem	ofcs1 34559	Letterwise operations on a single letter word. (Contributed by Thierry Arnoux, 7-Oct-2018.)
⊢ ((𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑇) → (⟨“𝐴”⟩ ∘f/c 𝑅𝐵) = ⟨“(𝐴𝑅𝐵)”⟩)
Theorem	ofcs2 34560	Letterwise operations on a double letter word. (Contributed by Thierry Arnoux, 9-Oct-2018.)
⊢ ((𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ∧ 𝐶 ∈ 𝑇) → (⟨“𝐴𝐵”⟩ ∘f/c 𝑅𝐶) = ⟨“(𝐴𝑅𝐶)(𝐵𝑅𝐶)”⟩)
21.3.25  Polynomials with real coefficients - misc additions
Theorem	plymul02 34561	Product of a polynomial with the zero polynomial. (Contributed by Thierry Arnoux, 26-Sep-2018.)
⊢ (𝐹 ∈ (Poly‘𝑆) → (0𝑝 ∘f · 𝐹) = 0𝑝)
Theorem	plymulx0 34562*	Coefficients of a polynomial multiplied by Xp. (Contributed by Thierry Arnoux, 25-Sep-2018.)
⊢ (𝐹 ∈ ((Poly‘ℝ) ∖ {0𝑝}) → (coeff‘(𝐹 ∘f · Xp)) = (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, 0, ((coeff‘𝐹)‘(𝑛 − 1)))))
Theorem	plymulx 34563*	Coefficients of a polynomial multiplied by Xp. (Contributed by Thierry Arnoux, 25-Sep-2018.)
⊢ (𝐹 ∈ (Poly‘ℝ) → (coeff‘(𝐹 ∘f · Xp)) = (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, 0, ((coeff‘𝐹)‘(𝑛 − 1)))))
Theorem	plyrecld 34564	Closure of a polynomial with real coefficients. (Contributed by Thierry Arnoux, 18-Sep-2018.)
⊢ (𝜑 → 𝐹 ∈ (Poly‘ℝ))    &   ⊢ (𝜑 → 𝑋 ∈ ℝ)    ⇒   ⊢ (𝜑 → (𝐹‘𝑋) ∈ ℝ)
Theorem	signsplypnf 34565*	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 = (𝐶‘𝐷)    &   ⊢ 𝐺 = (𝑥 ∈ ℝ+ ↦ (𝑥↑𝐷))    ⇒   ⊢ (𝐹 ∈ (Poly‘ℝ) → (𝐹 ∘f / 𝐺) ⇝𝑟 𝐵)
Theorem	signsply0 34566*	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 = (𝐶‘𝐷)    &   ⊢ 𝐴 = (𝐶‘0)    &   ⊢ (𝜑 → 𝐹 ∈ (Poly‘ℝ))    &   ⊢ (𝜑 → 𝐹 ≠ 0𝑝)    &   ⊢ (𝜑 → (𝐴 · 𝐵) < 0)    ⇒   ⊢ (𝜑 → ∃𝑧 ∈ ℝ+ (𝐹‘𝑧) = 0)
21.3.26  Descartes's rule of signs
21.3.26.1  Sign changes in a word over real numbers
Theorem	signspval 34567*	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))
Theorem	signsw0glem 34568*	Neutral element property of ⨣. (Contributed by Thierry Arnoux, 9-Sep-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    ⇒   ⊢ ∀𝑢 ∈ {-1, 0, 1} ((0 ⨣ 𝑢) = 𝑢 ∧ (𝑢 ⨣ 0) = 𝑢)
Theorem	signswbase 34569	The base of 𝑊 is the unordered triple reprensenting the possible signs. (Contributed by Thierry Arnoux, 9-Sep-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    ⇒   ⊢ {-1, 0, 1} = (Base‘𝑊)
Theorem	signswplusg 34570*	The operation of 𝑊. (Contributed by Thierry Arnoux, 9-Sep-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    ⇒   ⊢ ⨣ = (+g‘𝑊)
Theorem	signsw0g 34571*	The neutral element of 𝑊. (Contributed by Thierry Arnoux, 9-Sep-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    ⇒   ⊢ 0 = (0g‘𝑊)
Theorem	signswmnd 34572*	𝑊 is a monoid structure on {-1, 0, 1} which operation retains the right side, but skips zeroes. This will be used for skipping zeroes when counting sign changes. (Contributed by Thierry Arnoux, 9-Sep-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    ⇒   ⊢ 𝑊 ∈ Mnd
Theorem	signswrid 34573*	The zero-skipping operation propagages nonzeros. (Contributed by Thierry Arnoux, 11-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    ⇒   ⊢ (𝑋 ∈ {-1, 0, 1} → (𝑋 ⨣ 0) = 𝑋)
Theorem	signswlid 34574*	The zero-skipping operation keeps nonzeros. (Contributed by Thierry Arnoux, 12-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    ⇒   ⊢ (((𝑋 ∈ {-1, 0, 1} ∧ 𝑌 ∈ {-1, 0, 1}) ∧ 𝑌 ≠ 0) → (𝑋 ⨣ 𝑌) = 𝑌)
Theorem	signswn0 34575*	The zero-skipping operation propagages nonzeros. (Contributed by Thierry Arnoux, 11-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    ⇒   ⊢ (((𝑋 ∈ {-1, 0, 1} ∧ 𝑌 ∈ {-1, 0, 1}) ∧ 𝑋 ≠ 0) → (𝑋 ⨣ 𝑌) ≠ 0)
Theorem	signswch 34576*	The zero-skipping operation changes value when the operands change signs. (Contributed by Thierry Arnoux, 9-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    ⇒   ⊢ ((𝑋 ∈ {-1, 1} ∧ 𝑌 ∈ {-1, 0, 1}) → ((𝑋 ⨣ 𝑌) ≠ 𝑋 ↔ (𝑋 · 𝑌) < 0))
21.3.26.2  Counting sign changes in a word over real numbers
Theorem	signslema 34577	Computational part of ~? signwlemn . (Contributed by Thierry Arnoux, 29-Sep-2018.)
⊢ (𝜑 → 𝐸 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐹 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐺 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐻 ∈ ℕ0)    &   ⊢ (𝜑 → (𝐸 < 𝐺 ∧ ¬ 2 ∥ (𝐺 − 𝐸)))    &   ⊢ (𝜑 → ((𝐻 − 𝐺) − (𝐹 − 𝐸)) ∈ {0, 2})    ⇒   ⊢ (𝜑 → (𝐹 < 𝐻 ∧ ¬ 2 ∥ (𝐻 − 𝐹)))
Theorem	signstfv 34578*	Value of the zero-skipping sign word. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ (𝐹 ∈ Word ℝ → (𝑇‘𝐹) = (𝑛 ∈ (0..^(♯‘𝐹)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝐹‘𝑖))))))
Theorem	signstfval 34579*	Value of the zero-skipping sign word. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ ((𝐹 ∈ Word ℝ ∧ 𝑁 ∈ (0..^(♯‘𝐹))) → ((𝑇‘𝐹)‘𝑁) = (𝑊 Σg (𝑖 ∈ (0...𝑁) ↦ (sgn‘(𝐹‘𝑖)))))
Theorem	signstcl 34580*	Closure of the zero skipping sign word. (Contributed by Thierry Arnoux, 9-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ ((𝐹 ∈ Word ℝ ∧ 𝑁 ∈ (0..^(♯‘𝐹))) → ((𝑇‘𝐹)‘𝑁) ∈ {-1, 0, 1})
Theorem	signstf 34581*	The zero skipping sign word is a word. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ (𝐹 ∈ Word ℝ → (𝑇‘𝐹) ∈ Word ℝ)
Theorem	signstlen 34582*	Length of the zero skipping sign word. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ (𝐹 ∈ Word ℝ → (♯‘(𝑇‘𝐹)) = (♯‘𝐹))
Theorem	signstf0 34583*	Sign of a single letter word. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ (𝐾 ∈ ℝ → (𝑇‘⟨“𝐾”⟩) = ⟨“(sgn‘𝐾)”⟩)
Theorem	signstfvn 34584*	Zero-skipping sign in a word compared to a shorter word. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ ((𝐹 ∈ (Word ℝ ∖ {∅}) ∧ 𝐾 ∈ ℝ) → ((𝑇‘(𝐹 ++ ⟨“𝐾”⟩))‘(♯‘𝐹)) = (((𝑇‘𝐹)‘((♯‘𝐹) − 1)) ⨣ (sgn‘𝐾)))
Theorem	signsvtn0 34585*	If the last letter is nonzero, then this is the zero-skipping sign. (Contributed by Thierry Arnoux, 8-Oct-2018.) (Proof shortened by AV, 3-Nov-2022.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    &   ⊢ 𝑁 = (♯‘𝐹)    ⇒   ⊢ ((𝐹 ∈ (Word ℝ ∖ {∅}) ∧ (𝐹‘(𝑁 − 1)) ≠ 0) → ((𝑇‘𝐹)‘(𝑁 − 1)) = (sgn‘(𝐹‘(𝑁 − 1))))
Theorem	signstfvp 34586*	Zero-skipping sign in a word compared to a shorter word. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ ((𝐹 ∈ Word ℝ ∧ 𝐾 ∈ ℝ ∧ 𝑁 ∈ (0..^(♯‘𝐹))) → ((𝑇‘(𝐹 ++ ⟨“𝐾”⟩))‘𝑁) = ((𝑇‘𝐹)‘𝑁))
Theorem	signstfvneq0 34587*	In case the first letter is not zero, the zero skipping sign is never zero. (Contributed by Thierry Arnoux, 10-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ (((𝐹 ∈ (Word ℝ ∖ {∅}) ∧ (𝐹‘0) ≠ 0) ∧ 𝑁 ∈ (0..^(♯‘𝐹))) → ((𝑇‘𝐹)‘𝑁) ≠ 0)
Theorem	signstfvcl 34588*	Closure of the zero skipping sign in case the first letter is not zero. (Contributed by Thierry Arnoux, 10-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ (((𝐹 ∈ (Word ℝ ∖ {∅}) ∧ (𝐹‘0) ≠ 0) ∧ 𝑁 ∈ (0..^(♯‘𝐹))) → ((𝑇‘𝐹)‘𝑁) ∈ {-1, 1})
Theorem	signstfvc 34589*	Zero-skipping sign in a word compared to a shorter word. (Contributed by Thierry Arnoux, 11-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ ((𝐹 ∈ Word ℝ ∧ 𝐺 ∈ Word ℝ ∧ 𝑁 ∈ (0..^(♯‘𝐹))) → ((𝑇‘(𝐹 ++ 𝐺))‘𝑁) = ((𝑇‘𝐹)‘𝑁))
Theorem	signstres 34590*	Restriction of a zero skipping sign to a subword. (Contributed by Thierry Arnoux, 11-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ ((𝐹 ∈ Word ℝ ∧ 𝑁 ∈ (0...(♯‘𝐹))) → ((𝑇‘𝐹) ↾ (0..^𝑁)) = (𝑇‘(𝐹 ↾ (0..^𝑁))))
Theorem	signstfveq0a 34591*	Lemma for signstfveq0 34592. (Contributed by Thierry Arnoux, 11-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    &   ⊢ 𝑁 = (♯‘𝐹)    ⇒   ⊢ (((𝐹 ∈ (Word ℝ ∖ {∅}) ∧ (𝐹‘0) ≠ 0) ∧ (𝐹‘(𝑁 − 1)) = 0) → 𝑁 ∈ (ℤ≥‘2))
Theorem	signstfveq0 34592*	In case the last letter is zero, the zero skipping sign is the same as the previous letter. (Contributed by Thierry Arnoux, 11-Oct-2018.) (Proof shortened by AV, 4-Nov-2022.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    &   ⊢ 𝑁 = (♯‘𝐹)    ⇒   ⊢ (((𝐹 ∈ (Word ℝ ∖ {∅}) ∧ (𝐹‘0) ≠ 0) ∧ (𝐹‘(𝑁 − 1)) = 0) → ((𝑇‘𝐹)‘(𝑁 − 1)) = ((𝑇‘𝐹)‘(𝑁 − 2)))
Theorem	signsvvfval 34593*	The value of 𝑉, which represents the number of times the sign changes in a word. (Contributed by Thierry Arnoux, 7-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ (𝐹 ∈ Word ℝ → (𝑉‘𝐹) = Σ𝑗 ∈ (1..^(♯‘𝐹))if(((𝑇‘𝐹)‘𝑗) ≠ ((𝑇‘𝐹)‘(𝑗 − 1)), 1, 0))
Theorem	signsvvf 34594*	𝑉 is a function. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ 𝑉:Word ℝ⟶ℕ0
Theorem	signsvf0 34595*	There is no change of sign in the empty word. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ (𝑉‘∅) = 0
Theorem	signsvf1 34596*	In a single-letter word, which represents a constant polynomial, there is no change of sign. (Contributed by Thierry Arnoux, 8-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ (𝐾 ∈ ℝ → (𝑉‘⟨“𝐾”⟩) = 0)
Theorem	signsvfn 34597*	Number of changes in a word compared to a shorter word. (Contributed by Thierry Arnoux, 12-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    ⇒   ⊢ (((𝐹 ∈ (Word ℝ ∖ {∅}) ∧ (𝐹‘0) ≠ 0) ∧ 𝐾 ∈ ℝ) → (𝑉‘(𝐹 ++ ⟨“𝐾”⟩)) = ((𝑉‘𝐹) + if((((𝑇‘𝐹)‘((♯‘𝐹) − 1)) · 𝐾) < 0, 1, 0)))
Theorem	signsvtp 34598*	Adding a letter of the same sign as the highest coefficient does not change the sign. (Contributed by Thierry Arnoux, 12-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    &   ⊢ (𝜑 → 𝐸 ∈ (Word ℝ ∖ {∅}))    &   ⊢ (𝜑 → (𝐸‘0) ≠ 0)    &   ⊢ (𝜑 → 𝐹 = (𝐸 ++ ⟨“𝐴”⟩))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ 𝑁 = (♯‘𝐸)    &   ⊢ 𝐵 = ((𝑇‘𝐸)‘(𝑁 − 1))    ⇒   ⊢ ((𝜑 ∧ 0 < (𝐴 · 𝐵)) → (𝑉‘𝐹) = (𝑉‘𝐸))
Theorem	signsvtn 34599*	Adding a letter of a different sign as the highest coefficient changes the sign. (Contributed by Thierry Arnoux, 12-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    &   ⊢ (𝜑 → 𝐸 ∈ (Word ℝ ∖ {∅}))    &   ⊢ (𝜑 → (𝐸‘0) ≠ 0)    &   ⊢ (𝜑 → 𝐹 = (𝐸 ++ ⟨“𝐴”⟩))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ 𝑁 = (♯‘𝐸)    &   ⊢ 𝐵 = ((𝑇‘𝐸)‘(𝑁 − 1))    ⇒   ⊢ ((𝜑 ∧ (𝐴 · 𝐵) < 0) → ((𝑉‘𝐹) − (𝑉‘𝐸)) = 1)
Theorem	signsvfpn 34600*	Adding a letter of the same sign as the highest coefficient does not change the sign. (Contributed by Thierry Arnoux, 12-Oct-2018.)
⊢ ⨣ = (𝑎 ∈ {-1, 0, 1}, 𝑏 ∈ {-1, 0, 1} ↦ if(𝑏 = 0, 𝑎, 𝑏))    &   ⊢ 𝑊 = {⟨(Base‘ndx), {-1, 0, 1}⟩, ⟨(+g‘ndx), ⨣ ⟩}    &   ⊢ 𝑇 = (𝑓 ∈ Word ℝ ↦ (𝑛 ∈ (0..^(♯‘𝑓)) ↦ (𝑊 Σg (𝑖 ∈ (0...𝑛) ↦ (sgn‘(𝑓‘𝑖))))))    &   ⊢ 𝑉 = (𝑓 ∈ Word ℝ ↦ Σ𝑗 ∈ (1..^(♯‘𝑓))if(((𝑇‘𝑓)‘𝑗) ≠ ((𝑇‘𝑓)‘(𝑗 − 1)), 1, 0))    &   ⊢ (𝜑 → 𝐸 ∈ (Word ℝ ∖ {∅}))    &   ⊢ (𝜑 → (𝐸‘0) ≠ 0)    &   ⊢ (𝜑 → 𝐹 = (𝐸 ++ ⟨“𝐴”⟩))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ 𝑁 = (♯‘𝐸)    &   ⊢ 𝐵 = (𝐸‘(𝑁 − 1))    ⇒   ⊢ ((𝜑 ∧ 0 < (𝐵 · 𝐴)) → (𝑉‘𝐹) = (𝑉‘𝐸))