P. 347 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 34601-34700   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	rrvvf 34601	A real-valued random variable is a function. (Contributed by Thierry Arnoux, 25-Jan-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    ⇒   ⊢ (𝜑 → 𝑋:∪ dom 𝑃⟶ℝ)
Theorem	rrvfn 34602	A real-valued random variable is a function over the universe. (Contributed by Thierry Arnoux, 25-Jan-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    ⇒   ⊢ (𝜑 → 𝑋 Fn ∪ dom 𝑃)
Theorem	rrvdm 34603	The domain of a random variable is the universe. (Contributed by Thierry Arnoux, 25-Jan-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    ⇒   ⊢ (𝜑 → dom 𝑋 = ∪ dom 𝑃)
Theorem	rrvrnss 34604	The range of a random variable as a subset of ℝ. (Contributed by Thierry Arnoux, 6-Feb-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    ⇒   ⊢ (𝜑 → ran 𝑋 ⊆ ℝ)
Theorem	rrvf2 34605	A real-valued random variable is a function. (Contributed by Thierry Arnoux, 25-Jan-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    ⇒   ⊢ (𝜑 → 𝑋:dom 𝑋⟶ℝ)
Theorem	rrvdmss 34606	The domain of a random variable. This is useful to shorten proofs. (Contributed by Thierry Arnoux, 25-Jan-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    ⇒   ⊢ (𝜑 → ∪ dom 𝑃 ⊆ dom 𝑋)
Theorem	rrvfinvima 34607*	For a real-value random variable 𝑋, any open interval in ℝ is the image of a measurable set. (Contributed by Thierry Arnoux, 25-Jan-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    ⇒   ⊢ (𝜑 → ∀𝑦 ∈ 𝔅ℝ (◡𝑋 “ 𝑦) ∈ dom 𝑃)
Theorem	0rrv 34608*	The constant function equal to zero is a random variable. (Contributed by Thierry Arnoux, 16-Jan-2017.) (Revised by Thierry Arnoux, 30-Jan-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    ⇒   ⊢ (𝜑 → (𝑥 ∈ ∪ dom 𝑃 ↦ 0) ∈ (rRndVar‘𝑃))
Theorem	rrvadd 34609	The sum of two random variables is a random variable. (Contributed by Thierry Arnoux, 4-Jun-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝑌 ∈ (rRndVar‘𝑃))    ⇒   ⊢ (𝜑 → (𝑋 ∘f + 𝑌) ∈ (rRndVar‘𝑃))
Theorem	rrvmulc 34610	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.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐶 ∈ ℝ)    ⇒   ⊢ (𝜑 → (𝑋 ∘f/c · 𝐶) ∈ (rRndVar‘𝑃))
Theorem	rrvsum 34611	An indexed sum of random variables is a random variable. (Contributed by Thierry Arnoux, 22-May-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋:ℕ⟶(rRndVar‘𝑃))    &   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → 𝑆 = (seq1( ∘f + , 𝑋)‘𝑁))    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → 𝑆 ∈ (rRndVar‘𝑃))
Theorem	boolesineq 34612*	Boole's inequality (union bound). For any finite or countable collection of events, the probability of their union is at most the sum of their probabilities. (Suggested by DeepSeek R1.) (Contributed by Ender Ting, 30-Apr-2025.)
⊢ ((𝑃 ∈ Prob ∧ 𝐴:ℕ⟶dom 𝑃) → (𝑃‘∪ 𝑛 ∈ ℕ (𝐴‘𝑛)) ≤ Σ*𝑛 ∈ ℕ(𝑃‘(𝐴‘𝑛)))
21.3.23.4  Preimage set mapping operator
Syntax	corvc 34613	Extend class notation to include the preimage set mapping operator.
class ∘RV/𝑐𝑅
Definition	df-orvc 34614*	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‘(𝑋∘RV/𝑐 I 𝐴)), and keep a similar notation. Because with this notation (𝑋∘RV/𝑐 I 𝐴) is a set, we can also apply it to conditional probabilities, like in (𝑃‘(𝑋∘RV/𝑐 I 𝐴) ∣ (𝑌∘RV/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 (𝑋∘RV/𝑐 I 𝐴) cf. ideq 5801- elementhood: {𝑋 ∈ 𝐴} as (𝑋∘RV/𝑐 E 𝐴) cf. epel 5527- less-than: {𝑋 ≤ 𝐴} as (𝑋∘RV/𝑐 ≤ 𝐴) 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 “ {𝑦 ∣ 𝑦𝑅𝑎}))
Theorem	orvcval 34615*	Value of the preimage mapping operator applied on a given random variable and constant. (Contributed by Thierry Arnoux, 19-Jan-2017.)
⊢ (𝜑 → Fun 𝑋)    &   ⊢ (𝜑 → 𝑋 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑊)    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐𝑅𝐴) = (◡𝑋 “ {𝑦 ∣ 𝑦𝑅𝐴}))
Theorem	orvcval2 34616*	Another way to express the value of the preimage mapping operator. (Contributed by Thierry Arnoux, 19-Jan-2017.)
⊢ (𝜑 → Fun 𝑋)    &   ⊢ (𝜑 → 𝑋 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑊)    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐𝑅𝐴) = {𝑧 ∈ dom 𝑋 ∣ (𝑋‘𝑧)𝑅𝐴})
Theorem	elorvc 34617*	Elementhood of a preimage. (Contributed by Thierry Arnoux, 21-Jan-2017.)
⊢ (𝜑 → Fun 𝑋)    &   ⊢ (𝜑 → 𝑋 ∈ 𝑉)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑊)    ⇒   ⊢ ((𝜑 ∧ 𝑧 ∈ dom 𝑋) → (𝑧 ∈ (𝑋∘RV/𝑐𝑅𝐴) ↔ (𝑋‘𝑧)𝑅𝐴))
Theorem	orvcval4 34618*	The value of the preimage mapping operator can be restricted to preimages in the base set of the topology. Cf. orvcval 34615. (Contributed by Thierry Arnoux, 21-Jan-2017.)
⊢ (𝜑 → 𝑆 ∈ ∪ ran sigAlgebra)    &   ⊢ (𝜑 → 𝐽 ∈ Top)    &   ⊢ (𝜑 → 𝑋 ∈ (𝑆MblFnM(sigaGen‘𝐽)))    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐𝑅𝐴) = (◡𝑋 “ {𝑦 ∈ ∪ 𝐽 ∣ 𝑦𝑅𝐴}))
Theorem	orvcoel 34619*	If the relation produces open sets, preimage maps by a measurable function are measurable sets. (Contributed by Thierry Arnoux, 21-Jan-2017.)
⊢ (𝜑 → 𝑆 ∈ ∪ ran sigAlgebra)    &   ⊢ (𝜑 → 𝐽 ∈ Top)    &   ⊢ (𝜑 → 𝑋 ∈ (𝑆MblFnM(sigaGen‘𝐽)))    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → {𝑦 ∈ ∪ 𝐽 ∣ 𝑦𝑅𝐴} ∈ 𝐽)    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐𝑅𝐴) ∈ 𝑆)
Theorem	orvccel 34620*	If the relation produces closed sets, preimage maps by a measurable function are measurable sets. (Contributed by Thierry Arnoux, 21-Jan-2017.)
⊢ (𝜑 → 𝑆 ∈ ∪ ran sigAlgebra)    &   ⊢ (𝜑 → 𝐽 ∈ Top)    &   ⊢ (𝜑 → 𝑋 ∈ (𝑆MblFnM(sigaGen‘𝐽)))    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → {𝑦 ∈ ∪ 𝐽 ∣ 𝑦𝑅𝐴} ∈ (Clsd‘𝐽))    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐𝑅𝐴) ∈ 𝑆)
Theorem	elorrvc 34621*	Elementhood of a preimage for a real-valued random variable. (Contributed by Thierry Arnoux, 21-Jan-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    ⇒   ⊢ ((𝜑 ∧ 𝑧 ∈ ∪ dom 𝑃) → (𝑧 ∈ (𝑋∘RV/𝑐𝑅𝐴) ↔ (𝑋‘𝑧)𝑅𝐴))
Theorem	orrvcval4 34622*	The value of the preimage mapping operator can be restricted to preimages of subsets of ℝ. (Contributed by Thierry Arnoux, 21-Jan-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐𝑅𝐴) = (◡𝑋 “ {𝑦 ∈ ℝ ∣ 𝑦𝑅𝐴}))
Theorem	orrvcoel 34623*	If the relation produces open sets, preimage maps of a random variable are measurable sets. (Contributed by Thierry Arnoux, 21-Jan-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → {𝑦 ∈ ℝ ∣ 𝑦𝑅𝐴} ∈ (topGen‘ran (,)))    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐𝑅𝐴) ∈ dom 𝑃)
Theorem	orrvccel 34624*	If the relation produces closed sets, preimage maps are measurable sets. (Contributed by Thierry Arnoux, 21-Jan-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    &   ⊢ (𝜑 → {𝑦 ∈ ℝ ∣ 𝑦𝑅𝐴} ∈ (Clsd‘(topGen‘ran (,))))    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐𝑅𝐴) ∈ dom 𝑃)
Theorem	orvcgteel 34625	Preimage maps produced by the "greater than or equal to" relation are measurable sets. (Contributed by Thierry Arnoux, 5-Feb-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐◡ ≤ 𝐴) ∈ dom 𝑃)
21.3.23.5  Distribution Functions
Theorem	orvcelval 34626	Preimage maps produced by the membership relation. (Contributed by Thierry Arnoux, 6-Feb-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐴 ∈ 𝔅ℝ)    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐 E 𝐴) = (◡𝑋 “ 𝐴))
Theorem	orvcelel 34627	Preimage maps produced by the membership relation are measurable sets. (Contributed by Thierry Arnoux, 5-Feb-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐴 ∈ 𝔅ℝ)    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐 E 𝐴) ∈ dom 𝑃)
Theorem	dstrvval 34628*	The value of the distribution of a random variable. (Contributed by Thierry Arnoux, 9-Feb-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐷 = (𝑎 ∈ 𝔅ℝ ↦ (𝑃‘(𝑋∘RV/𝑐 E 𝑎))))    &   ⊢ (𝜑 → 𝐴 ∈ 𝔅ℝ)    ⇒   ⊢ (𝜑 → (𝐷‘𝐴) = (𝑃‘(◡𝑋 “ 𝐴)))
Theorem	dstrvprob 34629*	The distribution of a random variable is a probability law. (TODO: could be shortened using dstrvval 34628). (Contributed by Thierry Arnoux, 10-Feb-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐷 = (𝑎 ∈ 𝔅ℝ ↦ (𝑃‘(𝑋∘RV/𝑐 E 𝑎))))    ⇒   ⊢ (𝜑 → 𝐷 ∈ Prob)
21.3.23.6  Cumulative Distribution Functions
Theorem	orvclteel 34630	Preimage maps produced by the "less than or equal to" relation are measurable sets. (Contributed by Thierry Arnoux, 4-Feb-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐 ≤ 𝐴) ∈ dom 𝑃)
Theorem	dstfrvel 34631	Elementhood of preimage maps produced by the "less than or equal to" relation. (Contributed by Thierry Arnoux, 13-Feb-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ∪ dom 𝑃)    &   ⊢ (𝜑 → (𝑋‘𝐵) ≤ 𝐴)    ⇒   ⊢ (𝜑 → 𝐵 ∈ (𝑋∘RV/𝑐 ≤ 𝐴))
Theorem	dstfrvunirn 34632*	The limit of all preimage maps by the "less than or equal to" relation is the universe. (Contributed by Thierry Arnoux, 12-Feb-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    ⇒   ⊢ (𝜑 → ∪ ran (𝑛 ∈ ℕ ↦ (𝑋∘RV/𝑐 ≤ 𝑛)) = ∪ dom 𝑃)
Theorem	orvclteinc 34633	Preimage maps produced by the "less than or equal to" relation are increasing. (Contributed by Thierry Arnoux, 11-Feb-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ)    &   ⊢ (𝜑 → 𝐴 ≤ 𝐵)    ⇒   ⊢ (𝜑 → (𝑋∘RV/𝑐 ≤ 𝐴) ⊆ (𝑋∘RV/𝑐 ≤ 𝐵))
Theorem	dstfrvinc 34634*	A cumulative distribution function is nondecreasing. (Contributed by Thierry Arnoux, 11-Feb-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐹 = (𝑥 ∈ ℝ ↦ (𝑃‘(𝑋∘RV/𝑐 ≤ 𝑥))))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ)    &   ⊢ (𝜑 → 𝐴 ≤ 𝐵)    ⇒   ⊢ (𝜑 → (𝐹‘𝐴) ≤ (𝐹‘𝐵))
Theorem	dstfrvclim1 34635*	The limit of the cumulative distribution function is one. (Contributed by Thierry Arnoux, 12-Feb-2017.) (Revised by Thierry Arnoux, 11-Jul-2017.)
⊢ (𝜑 → 𝑃 ∈ Prob)    &   ⊢ (𝜑 → 𝑋 ∈ (rRndVar‘𝑃))    &   ⊢ (𝜑 → 𝐹 = (𝑥 ∈ ℝ ↦ (𝑃‘(𝑋∘RV/𝑐 ≤ 𝑥))))    ⇒   ⊢ (𝜑 → 𝐹 ⇝ 1)
21.3.23.7  Probabilities - example
Theorem	coinfliplem 34636	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    &   ⊢ 𝐻 ≠ 𝑇    &   ⊢ 𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   ⊢ 𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}    ⇒   ⊢ 𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c /𝑒 2)
Theorem	coinflipprob 34637	The 𝑃 we defined for coin-flip is a probability law. (Contributed by Thierry Arnoux, 15-Jan-2017.)
⊢ 𝐻 ∈ V    &   ⊢ 𝑇 ∈ V    &   ⊢ 𝐻 ≠ 𝑇    &   ⊢ 𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   ⊢ 𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}    ⇒   ⊢ 𝑃 ∈ Prob
Theorem	coinflipspace 34638	The space of our coin-flip probability. (Contributed by Thierry Arnoux, 15-Jan-2017.)
⊢ 𝐻 ∈ V    &   ⊢ 𝑇 ∈ V    &   ⊢ 𝐻 ≠ 𝑇    &   ⊢ 𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   ⊢ 𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}    ⇒   ⊢ dom 𝑃 = 𝒫 {𝐻, 𝑇}
Theorem	coinflipuniv 34639	The universe of our coin-flip probability is {𝐻, 𝑇}. (Contributed by Thierry Arnoux, 15-Jan-2017.)
⊢ 𝐻 ∈ V    &   ⊢ 𝑇 ∈ V    &   ⊢ 𝐻 ≠ 𝑇    &   ⊢ 𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   ⊢ 𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}    ⇒   ⊢ ∪ dom 𝑃 = {𝐻, 𝑇}
Theorem	coinfliprv 34640	The 𝑋 we defined for coin-flip is a random variable. (Contributed by Thierry Arnoux, 12-Jan-2017.)
⊢ 𝐻 ∈ V    &   ⊢ 𝑇 ∈ V    &   ⊢ 𝐻 ≠ 𝑇    &   ⊢ 𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   ⊢ 𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}    ⇒   ⊢ 𝑋 ∈ (rRndVar‘𝑃)
Theorem	coinflippv 34641	The probability of heads is one-half. (Contributed by Thierry Arnoux, 15-Jan-2017.)
⊢ 𝐻 ∈ V    &   ⊢ 𝑇 ∈ V    &   ⊢ 𝐻 ≠ 𝑇    &   ⊢ 𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   ⊢ 𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}    ⇒   ⊢ (𝑃‘{𝐻}) = (1 / 2)
Theorem	coinflippvt 34642	The probability of tails is one-half. (Contributed by Thierry Arnoux, 5-Feb-2017.)
⊢ 𝐻 ∈ V    &   ⊢ 𝑇 ∈ V    &   ⊢ 𝐻 ≠ 𝑇    &   ⊢ 𝑃 = ((♯ ↾ 𝒫 {𝐻, 𝑇}) ∘f/c / 2)    &   ⊢ 𝑋 = {⟨𝐻, 1⟩, ⟨𝑇, 0⟩}    ⇒   ⊢ (𝑃‘{𝑇}) = (1 / 2)
21.3.23.8  Bertrand's Ballot Problem
Theorem	ballotlemoex 34643*	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
Theorem	ballotlem1 34644*	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)
Theorem	ballotlemelo 34645*	Elementhood in 𝑂. (Contributed by Thierry Arnoux, 17-Apr-2017.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ 𝒫 (1...(𝑀 + 𝑁)) ∣ (♯‘𝑐) = 𝑀}    ⇒   ⊢ (𝐶 ∈ 𝑂 ↔ (𝐶 ⊆ (1...(𝑀 + 𝑁)) ∧ (♯‘𝐶) = 𝑀))
Theorem	ballotlem2 34646*	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‘{𝑐 ∈ 𝑂 ∣ ¬ 1 ∈ 𝑐}) = (𝑁 / (𝑀 + 𝑁))
Theorem	ballotlemfval 34647*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((𝐹‘𝐶)‘𝐽) = ((♯‘((1...𝐽) ∩ 𝐶)) − (♯‘((1...𝐽) ∖ 𝐶))))
Theorem	ballotlemfelz 34648*	(𝐹‘𝐶) has values in ℤ. (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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((𝐹‘𝐶)‘𝐽) ∈ ℤ)
Theorem	ballotlemfp1 34649*	If the 𝐽 th ballot is for A, (𝐹‘𝐶) goes up 1. If the 𝐽 th ballot is for B, (𝐹‘𝐶) 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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℕ)    ⇒   ⊢ (𝜑 → ((¬ 𝐽 ∈ 𝐶 → ((𝐹‘𝐶)‘𝐽) = (((𝐹‘𝐶)‘(𝐽 − 1)) − 1)) ∧ (𝐽 ∈ 𝐶 → ((𝐹‘𝐶)‘𝐽) = (((𝐹‘𝐶)‘(𝐽 − 1)) + 1))))
Theorem	ballotlemfc0 34650*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℕ)    &   ⊢ (𝜑 → ∃𝑖 ∈ (1...𝐽)((𝐹‘𝐶)‘𝑖) ≤ 0)    &   ⊢ (𝜑 → 0 < ((𝐹‘𝐶)‘𝐽))    ⇒   ⊢ (𝜑 → ∃𝑘 ∈ (1...𝐽)((𝐹‘𝐶)‘𝑘) = 0)
Theorem	ballotlemfcc 34651*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℕ)    &   ⊢ (𝜑 → ∃𝑖 ∈ (1...𝐽)0 ≤ ((𝐹‘𝐶)‘𝑖))    &   ⊢ (𝜑 → ((𝐹‘𝐶)‘𝐽) < 0)    ⇒   ⊢ (𝜑 → ∃𝑘 ∈ (1...𝐽)((𝐹‘𝐶)‘𝑘) = 0)
Theorem	ballotlemfmpn 34652*	(𝐹‘𝐶) finishes counting at (𝑀 − 𝑁). (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)))))    ⇒   ⊢ (𝐶 ∈ 𝑂 → ((𝐹‘𝐶)‘(𝑀 + 𝑁)) = (𝑀 − 𝑁))
Theorem	ballotlemfval0 34653*	(𝐹‘𝐶) 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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    ⇒   ⊢ (𝐶 ∈ 𝑂 → ((𝐹‘𝐶)‘0) = 0)
Theorem	ballotleme 34654*	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 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    ⇒   ⊢ (𝐶 ∈ 𝐸 ↔ (𝐶 ∈ 𝑂 ∧ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝐶)‘𝑖)))
Theorem	ballotlemodife 34655*	Elements of (𝑂 ∖ 𝐸). (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)}    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) ↔ (𝐶 ∈ 𝑂 ∧ ∃𝑖 ∈ (1...(𝑀 + 𝑁))((𝐹‘𝐶)‘𝑖) ≤ 0))
Theorem	ballotlem4 34656*	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 34657*	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 34658*	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 34659*	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 34660*	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 34661*	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 34662*	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 34663*	(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 34664*	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 34665*	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 34666*	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 34667*	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 34668*	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 34669*	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 34670*	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 34671*	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 34672*	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 34673*	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 34674*	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 34675*	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 34676*	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 34677*	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 34678*	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 34679*	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 34680*	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 34681*	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 34682*	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 34683*	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 34684*	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 34685*	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 34686*	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 34687*	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 34688*	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 34689*	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 34690*	Lemma for ballotlemrinv 34691. (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 34691*	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 34692*	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 34693*	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 34694*	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 34695*	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	fzssfzo 34696	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 34697*	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 34698*	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 34699	Concatenation of words follow the rule mulgnn0dir 19034 (although applying mulgnn0dir 19034 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 34700	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)))