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Theorem List for Metamath Proof Explorer - 31701-31800   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremrrvf2 31701 A real-valued random variable is a function. (Contributed by Thierry Arnoux, 25-Jan-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))       (𝜑𝑋:dom 𝑋⟶ℝ)

Theoremrrvdmss 31702 The domain of a random variable. This is useful to shorten proofs. (Contributed by Thierry Arnoux, 25-Jan-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))       (𝜑 dom 𝑃 ⊆ dom 𝑋)

Theoremrrvfinvima 31703* For a real-value random variable 𝑋, any open interval in is the image of a measurable set. (Contributed by Thierry Arnoux, 25-Jan-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))       (𝜑 → ∀𝑦 ∈ 𝔅 (𝑋𝑦) ∈ dom 𝑃)

Theorem0rrv 31704* The constant function equal to zero is a random variable. (Contributed by Thierry Arnoux, 16-Jan-2017.) (Revised by Thierry Arnoux, 30-Jan-2017.)
(𝜑𝑃 ∈ Prob)       (𝜑 → (𝑥 dom 𝑃 ↦ 0) ∈ (rRndVar‘𝑃))

Theoremrrvadd 31705 The sum of two random variables is a random variable. (Contributed by Thierry Arnoux, 4-Jun-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝑌 ∈ (rRndVar‘𝑃))       (𝜑 → (𝑋f + 𝑌) ∈ (rRndVar‘𝑃))

Theoremrrvmulc 31706 A random variable multiplied by a constant is a random variable. (Contributed by Thierry Arnoux, 17-Jan-2017.) (Revised by Thierry Arnoux, 22-May-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐶 ∈ ℝ)       (𝜑 → (𝑋f/c · 𝐶) ∈ (rRndVar‘𝑃))

Theoremrrvsum 31707 An indexed sum of random variables is a random variable. (Contributed by Thierry Arnoux, 22-May-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋:ℕ⟶(rRndVar‘𝑃))    &   ((𝜑𝑁 ∈ ℕ) → 𝑆 = (seq1( ∘f + , 𝑋)‘𝑁))       ((𝜑𝑁 ∈ ℕ) → 𝑆 ∈ (rRndVar‘𝑃))

20.3.22.4  Preimage set mapping operator

Syntaxcorvc 31708 Extend class notation to include the preimage set mapping operator.
class RV/𝑐𝑅

Definitiondf-orvc 31709* Define the preimage set mapping operator. In probability theory, the notation 𝑃(𝑋 = 𝐴) denotes the probability that a random variable 𝑋 takes the value 𝐴. We introduce here an operator which enables to write this in Metamath as (𝑃‘(𝑋RV/𝑐 I 𝐴)), and keep a similar notation. Because with this notation (𝑋RV/𝑐 I 𝐴) is a set, we can also apply it to conditional probabilities, like in (𝑃‘(𝑋RV/𝑐 I 𝐴) ∣ (𝑌RV/𝑐 I 𝐵))).

The oRVC operator transforms a relation 𝑅 into an operation taking a random variable 𝑋 and a constant 𝐶, and returning the preimage through 𝑋 of the equivalence class of 𝐶.

The most commonly used relations are: - equality: {𝑋 = 𝐴} as (𝑋RV/𝑐 I 𝐴) cf. ideq 5718- elementhood: {𝑋𝐴} as (𝑋RV/𝑐 E 𝐴) cf. epel 5464- less-than: {𝑋𝐴} as (𝑋RV/𝑐𝐴)

Even though it is primarily designed to be used within probability theory and with random variables, this operator is defined on generic functions, and could be used in other fields, e.g., for continuous functions. (Contributed by Thierry Arnoux, 15-Jan-2017.)

RV/𝑐𝑅 = (𝑥 ∈ {𝑥 ∣ Fun 𝑥}, 𝑎 ∈ V ↦ (𝑥 “ {𝑦𝑦𝑅𝑎}))

Theoremorvcval 31710* Value of the preimage mapping operator applied on a given random variable and constant. (Contributed by Thierry Arnoux, 19-Jan-2017.)
(𝜑 → Fun 𝑋)    &   (𝜑𝑋𝑉)    &   (𝜑𝐴𝑊)       (𝜑 → (𝑋RV/𝑐𝑅𝐴) = (𝑋 “ {𝑦𝑦𝑅𝐴}))

Theoremorvcval2 31711* Another way to express the value of the preimage mapping operator. (Contributed by Thierry Arnoux, 19-Jan-2017.)
(𝜑 → Fun 𝑋)    &   (𝜑𝑋𝑉)    &   (𝜑𝐴𝑊)       (𝜑 → (𝑋RV/𝑐𝑅𝐴) = {𝑧 ∈ dom 𝑋 ∣ (𝑋𝑧)𝑅𝐴})

Theoremelorvc 31712* Elementhood of a preimage. (Contributed by Thierry Arnoux, 21-Jan-2017.)
(𝜑 → Fun 𝑋)    &   (𝜑𝑋𝑉)    &   (𝜑𝐴𝑊)       ((𝜑𝑧 ∈ dom 𝑋) → (𝑧 ∈ (𝑋RV/𝑐𝑅𝐴) ↔ (𝑋𝑧)𝑅𝐴))

Theoremorvcval4 31713* The value of the preimage mapping operator can be restricted to preimages in the base set of the topology. Cf. orvcval 31710. (Contributed by Thierry Arnoux, 21-Jan-2017.)
(𝜑𝑆 ran sigAlgebra)    &   (𝜑𝐽 ∈ Top)    &   (𝜑𝑋 ∈ (𝑆MblFnM(sigaGen‘𝐽)))    &   (𝜑𝐴𝑉)       (𝜑 → (𝑋RV/𝑐𝑅𝐴) = (𝑋 “ {𝑦 𝐽𝑦𝑅𝐴}))

Theoremorvcoel 31714* If the relation produces open sets, preimage maps by a measurable function are measurable sets. (Contributed by Thierry Arnoux, 21-Jan-2017.)
(𝜑𝑆 ran sigAlgebra)    &   (𝜑𝐽 ∈ Top)    &   (𝜑𝑋 ∈ (𝑆MblFnM(sigaGen‘𝐽)))    &   (𝜑𝐴𝑉)    &   (𝜑 → {𝑦 𝐽𝑦𝑅𝐴} ∈ 𝐽)       (𝜑 → (𝑋RV/𝑐𝑅𝐴) ∈ 𝑆)

Theoremorvccel 31715* If the relation produces closed sets, preimage maps by a measurable function are measurable sets. (Contributed by Thierry Arnoux, 21-Jan-2017.)
(𝜑𝑆 ran sigAlgebra)    &   (𝜑𝐽 ∈ Top)    &   (𝜑𝑋 ∈ (𝑆MblFnM(sigaGen‘𝐽)))    &   (𝜑𝐴𝑉)    &   (𝜑 → {𝑦 𝐽𝑦𝑅𝐴} ∈ (Clsd‘𝐽))       (𝜑 → (𝑋RV/𝑐𝑅𝐴) ∈ 𝑆)

Theoremelorrvc 31716* Elementhood of a preimage for a real-valued random variable. (Contributed by Thierry Arnoux, 21-Jan-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐴𝑉)       ((𝜑𝑧 dom 𝑃) → (𝑧 ∈ (𝑋RV/𝑐𝑅𝐴) ↔ (𝑋𝑧)𝑅𝐴))

Theoremorrvcval4 31717* The value of the preimage mapping operator can be restricted to preimages of subsets of RR. (Contributed by Thierry Arnoux, 21-Jan-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐴𝑉)       (𝜑 → (𝑋RV/𝑐𝑅𝐴) = (𝑋 “ {𝑦 ∈ ℝ ∣ 𝑦𝑅𝐴}))

Theoremorrvcoel 31718* If the relation produces open sets, preimage maps of a random variable are measurable sets. (Contributed by Thierry Arnoux, 21-Jan-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐴𝑉)    &   (𝜑 → {𝑦 ∈ ℝ ∣ 𝑦𝑅𝐴} ∈ (topGen‘ran (,)))       (𝜑 → (𝑋RV/𝑐𝑅𝐴) ∈ dom 𝑃)

Theoremorrvccel 31719* If the relation produces closed sets, preimage maps are measurable sets. (Contributed by Thierry Arnoux, 21-Jan-2017.)
(𝜑𝑃 ∈ Prob)    &   (𝜑𝑋 ∈ (rRndVar‘𝑃))    &   (𝜑𝐴𝑉)    &   (𝜑 → {𝑦 ∈ ℝ ∣ 𝑦𝑅𝐴} ∈ (Clsd‘(topGen‘ran (,))))       (𝜑 → (𝑋RV/𝑐𝑅𝐴) ∈ dom 𝑃)

Theoremorvcgteel 31720 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 31721 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 31722 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 31723* 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 31724* The distribution of a random variable is a probability law. (TODO: could be shortened using dstrvval 31723) (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 31725 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 31726 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 31727* 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 31728 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 31729* 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 31730* 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 31731 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 31732 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 31733 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 31734 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 31735 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 31736 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 31737 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 31738* 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 31739* 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 31740* 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 31741* 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 31742* 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 31743* (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 31744* 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 31745* 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 31746* 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 31747* (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 31748* (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 31749* 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 31750* 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 31751* 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 31752* 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 31753* 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 31754* 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 31755* 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 31756* 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 31757* 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 31758* (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 31759* 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 31760* 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 31761* 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 31762* 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 31763* 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 31764* 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 31765* 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 31766* 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 31767* 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 31768* 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 31769* 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 31770* 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 31771* 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 31772* 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 31773* 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 31774* 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 31775* 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 31776* 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 31777* 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 31778* 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 31779* 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 31780* 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 31781* 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 31782* 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 31783* 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 31784* 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 31785* Lemma for ballotlemrinv 31786. (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 31786* 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 31787* 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 31788* 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 31789* 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 31790* 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 31791 Closure of the signum. (Contributed by Thierry Arnoux, 28-Sep-2018.)
(𝐴 ∈ ℝ* → (sgn‘𝐴) ∈ {-1, 0, 1})

Theoremsgnclre 31792 Closure of the signum. (Contributed by Thierry Arnoux, 28-Sep-2018.)
(𝐴 ∈ ℝ → (sgn‘𝐴) ∈ ℝ)

Theoremsgnneg 31793 Negation of the signum. (Contributed by Thierry Arnoux, 1-Oct-2018.)
(𝐴 ∈ ℝ → (sgn‘-𝐴) = -(sgn‘𝐴))

Theoremsgn3da 31794 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 31795 Signum of a product. (Contributed by Thierry Arnoux, 2-Oct-2018.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (sgn‘(𝐴 · 𝐵)) = ((sgn‘𝐴) · (sgn‘𝐵)))

Theoremsgnmulrp2 31796 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 31797 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 31798 Negative signum. (Contributed by Thierry Arnoux, 2-Oct-2018.)
(𝐴 ∈ ℝ* → ((sgn‘𝐴) = -1 ↔ 𝐴 < 0))

Theoremsgnpbi 31799 Positive signum. (Contributed by Thierry Arnoux, 2-Oct-2018.)
(𝐴 ∈ ℝ* → ((sgn‘𝐴) = 1 ↔ 0 < 𝐴))

Theoremsgn0bi 31800 Zero signum. (Contributed by Thierry Arnoux, 10-Oct-2018.)
(𝐴 ∈ ℝ* → ((sgn‘𝐴) = 0 ↔ 𝐴 = 0))

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268 26701-26800 269 26801-26900 270 26901-27000 271 27001-27100 272 27101-27200 273 27201-27300 274 27301-27400 275 27401-27500 276 27501-27600 277 27601-27700 278 27701-27800 279 27801-27900 280 27901-28000 281 28001-28100 282 28101-28200 283 28201-28300 284 28301-28400 285 28401-28500 286 28501-28600 287 28601-28700 288 28701-28800 289 28801-28900 290 28901-29000 291 29001-29100 292 29101-29200 293 29201-29300 294 29301-29400 295 29401-29500 296 29501-29600 297 29601-29700 298 29701-29800 299 29801-29900 300 29901-30000 301 30001-30100 302 30101-30200 303 30201-30300 304 30301-30400 305 30401-30500 306 30501-30600 307 30601-30700 308 30701-30800 309 30801-30900 310 30901-31000 311 31001-31100 312 31101-31200 313 31201-31300 314 31301-31400 315 31401-31500 316 31501-31600 317 31601-31700 318 31701-31800 319 31801-31900 320 31901-32000 321 32001-32100 322 32101-32200 323 32201-32300 324 32301-32400 325 32401-32500 326 32501-32600 327 32601-32700 328 32701-32800 329 32801-32900 330 32901-33000 331 33001-33100 332 33101-33200 333 33201-33300 334 33301-33400 335 33401-33500 336 33501-33600 337 33601-33700 338 33701-33800 339 33801-33900 340 33901-34000 341 34001-34100 342 34101-34200 343 34201-34300 344 34301-34400 345 34401-34500 346 34501-34600 347 34601-34700 348 34701-34800 349 34801-34900 350 34901-35000 351 35001-35100 352 35101-35200 353 35201-35300 354 35301-35400 355 35401-35500 356 35501-35600 357 35601-35700 358 35701-35800 359 35801-35900 360 35901-36000 361 36001-36100 362 36101-36200 363 36201-36300 364 36301-36400 365 36401-36500 366 36501-36600 367 36601-36700 368 36701-36800 369 36801-36900 370 36901-37000 371 37001-37100 372 37101-37200 373 37201-37300 374 37301-37400 375 37401-37500 376 37501-37600 377 37601-37700 378 37701-37800 379 37801-37900 380 37901-38000 381 38001-38100 382 38101-38200 383 38201-38300 384 38301-38400 385 38401-38500 386 38501-38600 387 38601-38700 388 38701-38800 389 38801-38900 390 38901-39000 391 39001-39100 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