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Theorem List for Metamath Proof Explorer - 31901-32000   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremorvcgteel 31901 Preimage maps produced by the "greater than or equal to" relation are measurable sets. (Contributed by Thierry Arnoux, 5-Feb-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐴 ∈ ℝ)       (𝜑 → (𝑋RV/𝑐𝐴) ∈ dom 𝑃)

20.3.22.5  Distribution Functions

Theoremorvcelval 31902 Preimage maps produced by the membership relation. (Contributed by Thierry Arnoux, 6-Feb-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐴 ∈ 𝔅)       (𝜑 → (𝑋RV/𝑐 E 𝐴) = (𝑋𝐴))

Theoremorvcelel 31903 Preimage maps produced by the membership relation are measurable sets. (Contributed by Thierry Arnoux, 5-Feb-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐴 ∈ 𝔅)       (𝜑 → (𝑋RV/𝑐 E 𝐴) ∈ dom 𝑃)

Theoremdstrvval 31904* The value of the distribution of a random variable. (Contributed by Thierry Arnoux, 9-Feb-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐷 = (𝑎 ∈ 𝔅 ↦ (𝑃‘(𝑋RV/𝑐 E 𝑎))))    &   (𝜑𝐴 ∈ 𝔅)       (𝜑 → (𝐷𝐴) = (𝑃‘(𝑋𝐴)))

Theoremdstrvprob 31905* The distribution of a random variable is a probability law. (TODO: could be shortened using dstrvval 31904). (Contributed by Thierry Arnoux, 10-Feb-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐷 = (𝑎 ∈ 𝔅 ↦ (𝑃‘(𝑋RV/𝑐 E 𝑎))))       (𝜑𝐷 ∈ Prob)

20.3.22.6  Cumulative Distribution Functions

Theoremorvclteel 31906 Preimage maps produced by the "less than or equal to" relation are measurable sets. (Contributed by Thierry Arnoux, 4-Feb-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐴 ∈ ℝ)       (𝜑 → (𝑋RV/𝑐𝐴) ∈ dom 𝑃)

Theoremdstfrvel 31907 Elementhood of preimage maps produced by the "less than or equal to" relation. (Contributed by Thierry Arnoux, 13-Feb-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 dom 𝑃)    &   (𝜑 → (𝑋𝐵) ≤ 𝐴)       (𝜑𝐵 ∈ (𝑋RV/𝑐𝐴))

Theoremdstfrvunirn 31908* The limit of all preimage maps by the "less than or equal to" relation is the universe. (Contributed by Thierry Arnoux, 12-Feb-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))       (𝜑 ran (𝑛 ∈ ℕ ↦ (𝑋RV/𝑐𝑛)) = dom 𝑃)

Theoremorvclteinc 31909 Preimage maps produced by the "less than or equal to" relation are increasing. (Contributed by Thierry Arnoux, 11-Feb-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑𝐴𝐵)       (𝜑 → (𝑋RV/𝑐𝐴) ⊆ (𝑋RV/𝑐𝐵))

Theoremdstfrvinc 31910* A cumulative distribution function is nondecreasing. (Contributed by Thierry Arnoux, 11-Feb-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐹 = (𝑥 ∈ ℝ ↦ (𝑃‘(𝑋RV/𝑐𝑥))))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑𝐴𝐵)       (𝜑 → (𝐹𝐴) ≤ (𝐹𝐵))

Theoremdstfrvclim1 31911* The limit of the cumulative distribution function is one. (Contributed by Thierry Arnoux, 12-Feb-2017.) (Revised by Thierry Arnoux, 11-Jul-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐹 = (𝑥 ∈ ℝ ↦ (𝑃‘(𝑋RV/𝑐𝑥))))       (𝜑𝐹 ⇝ 1)

20.3.22.7  Probabilities - example

Theoremcoinfliplem 31912 Division in the extended real numbers can be used for the coin-flip example. (Contributed by Thierry Arnoux, 15-Jan-2017.)
𝐻 ∈ V    &   𝑇 ∈ V    &   𝐻𝑇    &   𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}       𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c /𝑒 2)

Theoremcoinflipprob 31913 The 𝑃 we defined for coin-flip is a probability law. (Contributed by Thierry Arnoux, 15-Jan-2017.)
𝐻 ∈ V    &   𝑇 ∈ V    &   𝐻𝑇    &   𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}       𝑃 ∈ Prob

Theoremcoinflipspace 31914 The space of our coin-flip probability. (Contributed by Thierry Arnoux, 15-Jan-2017.)
𝐻 ∈ V    &   𝑇 ∈ V    &   𝐻𝑇    &   𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}       dom 𝑃 = 𝒫 {𝐻, 𝑇}

Theoremcoinflipuniv 31915 The universe of our coin-flip probability is {𝐻, 𝑇}. (Contributed by Thierry Arnoux, 15-Jan-2017.)
𝐻 ∈ V    &   𝑇 ∈ V    &   𝐻𝑇    &   𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}        dom 𝑃 = {𝐻, 𝑇}

Theoremcoinfliprv 31916 The 𝑋 we defined for coin-flip is a random variable. (Contributed by Thierry Arnoux, 12-Jan-2017.)
𝐻 ∈ V    &   𝑇 ∈ V    &   𝐻𝑇    &   𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}       𝑋 ∈ (rRndVar‘𝑃)

Theoremcoinflippv 31917 The probability of heads is one-half. (Contributed by Thierry Arnoux, 15-Jan-2017.)
𝐻 ∈ V    &   𝑇 ∈ V    &   𝐻𝑇    &   𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}       (𝑃‘{𝐻}) = (1 / 2)

Theoremcoinflippvt 31918 The probability of tails is one-half. (Contributed by Thierry Arnoux, 5-Feb-2017.)
𝐻 ∈ V    &   𝑇 ∈ V    &   𝐻𝑇    &   𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}       (𝑃‘{𝑇}) = (1 / 2)

20.3.22.8  Bertrand's Ballot Problem

Theoremballotlemoex 31919* 𝑂 is a set. (Contributed by Thierry Arnoux, 7-Dec-2016.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}       𝑂 ∈ V

Theoremballotlem1 31920* The size of the universe is a binomial coefficient. (Contributed by Thierry Arnoux, 23-Nov-2016.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}       (♯‘𝑂) = ((𝑀 + 𝑁)C𝑀)

Theoremballotlemelo 31921* Elementhood in 𝑂. (Contributed by Thierry Arnoux, 17-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}       (𝐶𝑂 ↔ (𝐶 ⊆ (1...(𝑀 + 𝑁)) ∧ (♯‘𝐶) = 𝑀))

Theoremballotlem2 31922* The probability that the first vote picked in a count is a B. (Contributed by Thierry Arnoux, 23-Nov-2016.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((♯‘𝑥) / (♯‘𝑂)))       (𝑃‘{𝑐𝑂 ∣ ¬ 1 ∈ 𝑐}) = (𝑁 / (𝑀 + 𝑁))

Theoremballotlemfval 31923* The value of F. (Contributed by Thierry Arnoux, 23-Nov-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)))))    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℤ)       (𝜑 → ((𝐹𝐶)‘𝐽) = ((♯‘((1...𝐽) ∩ 𝐶)) − (♯‘((1...𝐽) ∖ 𝐶))))

Theoremballotlemfelz 31924* (𝐹𝐶) has values in . (Contributed by Thierry Arnoux, 23-Nov-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)))))    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℤ)       (𝜑 → ((𝐹𝐶)‘𝐽) ∈ ℤ)

Theoremballotlemfp1 31925* If the 𝐽 th ballot is for A, (𝐹𝐶) goes up 1. If the 𝐽 th ballot is for B, (𝐹𝐶) goes down 1. (Contributed by Thierry Arnoux, 24-Nov-2016.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℕ)       (𝜑 → ((¬ 𝐽𝐶 → ((𝐹𝐶)‘𝐽) = (((𝐹𝐶)‘(𝐽 − 1)) − 1)) ∧ (𝐽𝐶 → ((𝐹𝐶)‘𝐽) = (((𝐹𝐶)‘(𝐽 − 1)) + 1))))

Theoremballotlemfc0 31926* 𝐹 takes value 0 between negative and positive values. (Contributed by Thierry Arnoux, 24-Nov-2016.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℕ)    &   (𝜑 → ∃𝑖 ∈ (1...𝐽)((𝐹𝐶)‘𝑖) ≤ 0)    &   (𝜑 → 0 < ((𝐹𝐶)‘𝐽))       (𝜑 → ∃𝑘 ∈ (1...𝐽)((𝐹𝐶)‘𝑘) = 0)

Theoremballotlemfcc 31927* 𝐹 takes value 0 between positive and negative values. (Contributed by Thierry Arnoux, 2-Apr-2017.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}    &   𝑃 = (𝑥 ∈ 𝒫 𝑂 ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   𝐹 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   (𝜑𝐶𝑂)    &   (𝜑𝐽 ∈ ℕ)    &   (𝜑 → ∃𝑖 ∈ (1...𝐽)0 ≤ ((𝐹𝐶)‘𝑖))    &   (𝜑 → ((𝐹𝐶)‘𝐽) < 0)       (𝜑 → ∃𝑘 ∈ (1...𝐽)((𝐹𝐶)‘𝑘) = 0)

Theoremballotlemfmpn 31928* (𝐹𝐶) finishes counting at (𝑀𝑁). (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))       (𝐶𝑂 → ((𝐹𝐶)‘(𝑀 + 𝑁)) = (𝑀𝑁))

Theoremballotlemfval0 31929* (𝐹𝐶) always starts counting at 0 . (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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))       (𝐶𝑂 → ((𝐹𝐶)‘0) = 0)

Theoremballotleme 31930* Elements of 𝐸. (Contributed by Thierry Arnoux, 14-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)}       (𝐶𝐸 ↔ (𝐶𝑂 ∧ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝐶)‘𝑖)))

Theoremballotlemodife 31931* Elements of (𝑂𝐸). (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)}       (𝐶 ∈ (𝑂𝐸) ↔ (𝐶𝑂 ∧ ∃𝑖 ∈ (1...(𝑀 + 𝑁))((𝐹𝐶)‘𝑖) ≤ 0))

Theoremballotlem4 31932* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}       (𝐶𝑂 → (¬ 1 ∈ 𝐶 → ¬ 𝐶𝐸))

Theoremballotlem5 31933* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀       (𝐶 ∈ (𝑂𝐸) → ∃𝑘 ∈ (1...(𝑀 + 𝑁))((𝐹𝐶)‘𝑘) = 0)

Theoremballotlemi 31934* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       (𝐶 ∈ (𝑂𝐸) → (𝐼𝐶) = inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝐶)‘𝑘) = 0}, ℝ, < ))

Theoremballotlemiex 31935* Properties of (𝐼𝐶). (Contributed by Thierry Arnoux, 12-Dec-2016.) (Revised by AV, 6-Oct-2020.)
𝑀 ∈ ℕ    &   𝑁 ∈ ℕ    &   𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘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) → ((𝐼𝐶) ∈ (1...(𝑀 + 𝑁)) ∧ ((𝐹𝐶)‘(𝐼𝐶)) = 0))

Theoremballotlemi1 31936* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       ((𝐶 ∈ (𝑂𝐸) ∧ ¬ 1 ∈ 𝐶) → (𝐼𝐶) ≠ 1)

Theoremballotlemii 31937* 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 = (𝑐𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   𝐸 = {𝑐𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹𝑐)‘𝑖)}    &   𝑁 < 𝑀    &   𝐼 = (𝑐 ∈ (𝑂𝐸) ↦ inf({𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝑐)‘𝑘) = 0}, ℝ, < ))       ((𝐶 ∈ (𝑂𝐸) ∧ 1 ∈ 𝐶) → (𝐼𝐶) ≠ 1)

Theoremballotlemsup 31938* 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 = (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) → ∃𝑧 ∈ ℝ (∀𝑤 ∈ {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝐶)‘𝑘) = 0} ¬ 𝑤 < 𝑧 ∧ ∀𝑤 ∈ ℝ (𝑧 < 𝑤 → ∃𝑦 ∈ {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹𝐶)‘𝑘) = 0}𝑦 < 𝑤)))

Theoremballotlemimin 31939* (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 31940* 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 31941* 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 31942* 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 31943* 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 31944* 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 31945* 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 31946* 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 31947* 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 31948* 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 31949* 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 31950* 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 31951* 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 31952* 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 31953* 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 31954* 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 31955* 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 31956* 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 31957* 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 31958* 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 31959* 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 31960* 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 31961* 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 31962* 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 31963* 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 31964* 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 31965* 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 31966* Lemma for ballotlemrinv 31967. (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 31967* 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 31968* 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 31969* 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 31970* 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 31971* 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 = (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))

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

Theoremsgncl 31972 Closure of the signum. (Contributed by Thierry Arnoux, 28-Sep-2018.)
(𝐴 ∈ ℝ* → (sgn‘𝐴) ∈ {-1, 0, 1})

Theoremsgnclre 31973 Closure of the signum. (Contributed by Thierry Arnoux, 28-Sep-2018.)
(𝐴 ∈ ℝ → (sgn‘𝐴) ∈ ℝ)

Theoremsgnneg 31974 Negation of the signum. (Contributed by Thierry Arnoux, 1-Oct-2018.)
(𝐴 ∈ ℝ → (sgn‘-𝐴) = -(sgn‘𝐴))

Theoremsgn3da 31975 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 31976 Signum of a product. (Contributed by Thierry Arnoux, 2-Oct-2018.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (sgn‘(𝐴 · 𝐵)) = ((sgn‘𝐴) · (sgn‘𝐵)))

Theoremsgnmulrp2 31977 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 31978 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 31979 Negative signum. (Contributed by Thierry Arnoux, 2-Oct-2018.)
(𝐴 ∈ ℝ* → ((sgn‘𝐴) = -1 ↔ 𝐴 < 0))

Theoremsgnpbi 31980 Positive signum. (Contributed by Thierry Arnoux, 2-Oct-2018.)
(𝐴 ∈ ℝ* → ((sgn‘𝐴) = 1 ↔ 0 < 𝐴))

Theoremsgn0bi 31981 Zero signum. (Contributed by Thierry Arnoux, 10-Oct-2018.)
(𝐴 ∈ ℝ* → ((sgn‘𝐴) = 0 ↔ 𝐴 = 0))

Theoremsgnsgn 31982 Signum is idempotent. (Contributed by Thierry Arnoux, 2-Oct-2018.)
(𝐴 ∈ ℝ* → (sgn‘(sgn‘𝐴)) = (sgn‘𝐴))

Theoremsgnmulsgn 31983 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 31984 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 31985 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))

Theoremgsumncl 31986* 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 31987* 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))

20.3.23.1  Operations on words

Theoremccatmulgnn0dir 31988 Concatenation of words follow the rule mulgnn0dir 18270 (although applying mulgnn0dir 18270 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 31989 Letterwise operations on word concatenations. (Contributed by Thierry Arnoux, 5-Oct-2018.)
(𝜑𝐹 ∈ Word 𝑆)    &   (𝜑𝐺 ∈ Word 𝑆)    &   (𝜑𝐾𝑇)       (𝜑 → ((𝐹 ++ 𝐺) ∘f/c 𝑅𝐾) = ((𝐹f/c 𝑅𝐾) ++ (𝐺f/c 𝑅𝐾)))

Theoremofcs1 31990 Letterwise operations on a single letter word. (Contributed by Thierry Arnoux, 7-Oct-2018.)
((𝐴𝑆𝐵𝑇) → (⟨“𝐴”⟩ ∘f/c 𝑅𝐵) = ⟨“(𝐴𝑅𝐵)”⟩)

Theoremofcs2 31991 Letterwise operations on a double letter word. (Contributed by Thierry Arnoux, 9-Oct-2018.)
((𝐴𝑆𝐵𝑆𝐶𝑇) → (⟨“𝐴𝐵”⟩ ∘f/c 𝑅𝐶) = ⟨“(𝐴𝑅𝐶)(𝐵𝑅𝐶)”⟩)

20.3.24  Polynomials with real coefficients - misc additions

Theoremplymul02 31992 Product of a polynomial with the zero polynomial. (Contributed by Thierry Arnoux, 26-Sep-2018.)
(𝐹 ∈ (Poly‘𝑆) → (0𝑝f · 𝐹) = 0𝑝)

Theoremplymulx0 31993* Coefficients of a polynomial multiplied 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 31994* Coefficients of a polynomial multiplied 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 31995 Closure of a polynomial with real coefficients. (Contributed by Thierry Arnoux, 18-Sep-2018.)
(𝜑𝐹 ∈ (Poly‘ℝ))    &   (𝜑𝑋 ∈ ℝ)       (𝜑 → (𝐹𝑋) ∈ ℝ)

Theoremsignsplypnf 31996* 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 31997* 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)

20.3.25  Descartes's rule of signs

20.3.25.1  Sign changes in a word over real numbers

Theoremsignspval 31998* 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))

Theoremsignsw0glem 31999* 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)

Theoremsignswbase 32000 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)

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