P. 132 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 13101-13200   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	gznegcl 13101	The gaussian integers are closed under negation. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → -𝐴 ∈ ℤ[i])
Theorem	gzcjcl 13102	The gaussian integers are closed under conjugation. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → (∗‘𝐴) ∈ ℤ[i])
Theorem	gzaddcl 13103	The gaussian integers are closed under addition. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 + 𝐵) ∈ ℤ[i])
Theorem	gzmulcl 13104	The gaussian integers are closed under multiplication. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 · 𝐵) ∈ ℤ[i])
Theorem	gzreim 13105	Construct a gaussian integer from real and imaginary parts. (Contributed by Mario Carneiro, 16-Jul-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝐴 + (i · 𝐵)) ∈ ℤ[i])
Theorem	gzsubcl 13106	The gaussian integers are closed under subtraction. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 − 𝐵) ∈ ℤ[i])
Theorem	gzabssqcl 13107	The squared norm of a gaussian integer is an integer. (Contributed by Mario Carneiro, 16-Jul-2014.)
⊢ (𝐴 ∈ ℤ[i] → ((abs‘𝐴)↑2) ∈ ℕ0)
Theorem	4sqlem5 13108	Lemma for 4sq 13136. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → (𝐵 ∈ ℤ ∧ ((𝐴 − 𝐵) / 𝑀) ∈ ℤ))
Theorem	4sqlem6 13109	Lemma for 4sq 13136. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → (-(𝑀 / 2) ≤ 𝐵 ∧ 𝐵 < (𝑀 / 2)))
Theorem	4sqlem7 13110	Lemma for 4sq 13136. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → (𝐵↑2) ≤ (((𝑀↑2) / 2) / 2))
Theorem	4sqlem8 13111	Lemma for 4sq 13136. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → 𝑀 ∥ ((𝐴↑2) − (𝐵↑2)))
Theorem	4sqlem9 13112	Lemma for 4sq 13136. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ ((𝜑 ∧ 𝜓) → (𝐵↑2) = 0)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → (𝑀↑2) ∥ (𝐴↑2))
Theorem	4sqlem10 13113	Lemma for 4sq 13136. (Contributed by Mario Carneiro, 16-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ 𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ ((𝜑 ∧ 𝜓) → ((((𝑀↑2) / 2) / 2) − (𝐵↑2)) = 0)    ⇒   ⊢ ((𝜑 ∧ 𝜓) → (𝑀↑2) ∥ ((𝐴↑2) − (((𝑀↑2) / 2) / 2)))
Theorem	4sqlem1 13114*	Lemma for 4sq 13136. The set 𝑆 is the set of all numbers that are expressible as a sum of four squares. Our goal is to show that 𝑆 = ℕ0; here we show one subset direction. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ 𝑆 ⊆ ℕ0
Theorem	4sqlem2 13115*	Lemma for 4sq 13136. Change bound variables in 𝑆. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ ∃𝑐 ∈ ℤ ∃𝑑 ∈ ℤ 𝐴 = (((𝑎↑2) + (𝑏↑2)) + ((𝑐↑2) + (𝑑↑2))))
Theorem	4sqlem3 13116*	Lemma for 4sq 13136. Sufficient condition to be in 𝑆. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝐶 ∈ ℤ ∧ 𝐷 ∈ ℤ)) → (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))) ∈ 𝑆)
Theorem	4sqlem4a 13117*	Lemma for 4sqlem4 13118. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ ((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (((abs‘𝐴)↑2) + ((abs‘𝐵)↑2)) ∈ 𝑆)
Theorem	4sqlem4 13118*	Lemma for 4sq 13136. We can express the four-square property more compactly in terms of gaussian integers, because the norms of gaussian integers are exactly sums of two squares. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ ∃𝑢 ∈ ℤ[i] ∃𝑣 ∈ ℤ[i] 𝐴 = (((abs‘𝑢)↑2) + ((abs‘𝑣)↑2)))
Theorem	mul4sqlem 13119*	Lemma for mul4sq 13120: algebraic manipulations. The extra assumptions involving 𝑀 would let us know not just that the product is a sum of squares, but also that it preserves divisibility by 𝑀. (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝐴 ∈ ℤ[i])    &   ⊢ (𝜑 → 𝐵 ∈ ℤ[i])    &   ⊢ (𝜑 → 𝐶 ∈ ℤ[i])    &   ⊢ (𝜑 → 𝐷 ∈ ℤ[i])    &   ⊢ 𝑋 = (((abs‘𝐴)↑2) + ((abs‘𝐵)↑2))    &   ⊢ 𝑌 = (((abs‘𝐶)↑2) + ((abs‘𝐷)↑2))    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → ((𝐴 − 𝐶) / 𝑀) ∈ ℤ[i])    &   ⊢ (𝜑 → ((𝐵 − 𝐷) / 𝑀) ∈ ℤ[i])    &   ⊢ (𝜑 → (𝑋 / 𝑀) ∈ ℕ0)    ⇒   ⊢ (𝜑 → ((𝑋 / 𝑀) · (𝑌 / 𝑀)) ∈ 𝑆)
Theorem	mul4sq 13120*	Euler's four-square identity: The product of two sums of four squares is also a sum of four squares. This is usually quoted as an explicit formula involving eight real variables; we save some time by working with complex numbers (gaussian integers) instead, so that we only have to work with four variables, and also hiding the actual formula for the product in the proof of mul4sqlem 13119. (For the curious, the explicit formula that is used is ( ∣ 𝑎 ∣ ↑2 + ∣ 𝑏 ∣ ↑2)( ∣ 𝑐 ∣ ↑2 + ∣ 𝑑 ∣ ↑2) = ∣ 𝑎∗ · 𝑐 + 𝑏 · 𝑑∗ ∣ ↑2 + ∣ 𝑎∗ · 𝑑 − 𝑏 · 𝑐∗ ∣ ↑2.) (Contributed by Mario Carneiro, 14-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ ((𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆) → (𝐴 · 𝐵) ∈ 𝑆)
Theorem	4sqlemafi 13121*	Lemma for 4sq 13136. 𝐴 is finite. (Contributed by Jim Kingdon, 24-May-2025.)
⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    ⇒   ⊢ (𝜑 → 𝐴 ∈ Fin)
Theorem	4sqlemffi 13122*	Lemma for 4sq 13136. ran 𝐹 is finite. (Contributed by Jim Kingdon, 24-May-2025.)
⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → ran 𝐹 ∈ Fin)
Theorem	4sqleminfi 13123*	Lemma for 4sq 13136. 𝐴 ∩ ran 𝐹 is finite. (Contributed by Jim Kingdon, 24-May-2025.)
⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℕ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → (𝐴 ∩ ran 𝐹) ∈ Fin)
Theorem	4sqexercise1 13124*	Exercise which may help in understanding the proof of 4sqlemsdc 13126. (Contributed by Jim Kingdon, 25-May-2025.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ 𝑛 = (𝑥↑2)}    ⇒   ⊢ (𝐴 ∈ ℕ0 → DECID 𝐴 ∈ 𝑆)
Theorem	4sqexercise2 13125*	Exercise which may help in understanding the proof of 4sqlemsdc 13126. (Contributed by Jim Kingdon, 30-May-2025.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑛 = ((𝑥↑2) + (𝑦↑2))}    ⇒   ⊢ (𝐴 ∈ ℕ0 → DECID 𝐴 ∈ 𝑆)
Theorem	4sqlemsdc 13126*	Lemma for 4sq 13136. The property of being the sum of four squares is decidable. The proof involves showing that (for a particular 𝐴) there are only a finite number of possible ways that it could be the sum of four squares, so checking each of those possibilities in turn decides whether the number is the sum of four squares. If this proof is hard to follow, especially because of its length, the simplified versions at 4sqexercise1 13124 and 4sqexercise2 13125 may help clarify, as they are using very much the same techniques on simplified versions of this lemma. (Contributed by Jim Kingdon, 25-May-2025.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ (𝐴 ∈ ℕ0 → DECID 𝐴 ∈ 𝑆)
Theorem	4sqlem11 13127*	Lemma for 4sq 13136. Use the pigeonhole principle to show that the sets {𝑚↑2 ∣ 𝑚 ∈ (0...𝑁)} and {-1 − 𝑛↑2 ∣ 𝑛 ∈ (0...𝑁)} have a common element, mod 𝑃. Note that although the conclusion is stated in terms of 𝐴 ∩ ran 𝐹 being nonempty, it is also inhabited by 4sqleminfi 13123 and fin0 7155. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → (𝐴 ∩ ran 𝐹) ≠ ∅)
Theorem	4sqlem12 13128*	Lemma for 4sq 13136. For any odd prime 𝑃, there is a 𝑘 < 𝑃 such that 𝑘𝑃 − 1 is a sum of two squares. (Contributed by Mario Carneiro, 15-Jul-2014.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ 𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   ⊢ 𝐹 = (𝑣 ∈ 𝐴 ↦ ((𝑃 − 1) − 𝑣))    ⇒   ⊢ (𝜑 → ∃𝑘 ∈ (1...(𝑃 − 1))∃𝑢 ∈ ℤ[i] (((abs‘𝑢)↑2) + 1) = (𝑘 · 𝑃))
Theorem	4sqlem13m 13129*	Lemma for 4sq 13136. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    ⇒   ⊢ (𝜑 → (∃𝑗 𝑗 ∈ 𝑇 ∧ 𝑀 < 𝑃))
Theorem	4sqlem14 13130*	Lemma for 4sq 13136. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ (𝜑 → 𝑅 ∈ ℕ0)
Theorem	4sqlem15 13131*	Lemma for 4sq 13136. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ ((𝜑 ∧ 𝑅 = 𝑀) → ((((((𝑀↑2) / 2) / 2) − (𝐸↑2)) = 0 ∧ ((((𝑀↑2) / 2) / 2) − (𝐹↑2)) = 0) ∧ (((((𝑀↑2) / 2) / 2) − (𝐺↑2)) = 0 ∧ ((((𝑀↑2) / 2) / 2) − (𝐻↑2)) = 0)))
Theorem	4sqlem16 13132*	Lemma for 4sq 13136. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ (𝜑 → (𝑅 ≤ 𝑀 ∧ ((𝑅 = 0 ∨ 𝑅 = 𝑀) → (𝑀↑2) ∥ (𝑀 · 𝑃))))
Theorem	4sqlem17 13133*	Lemma for 4sq 13136. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ 𝐸 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐹 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐺 = (((𝐶 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐻 = (((𝐷 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝑅 = ((((𝐸↑2) + (𝐹↑2)) + ((𝐺↑2) + (𝐻↑2))) / 𝑀)    &   ⊢ (𝜑 → (𝑀 · 𝑃) = (((𝐴↑2) + (𝐵↑2)) + ((𝐶↑2) + (𝐷↑2))))    ⇒   ⊢ ¬ 𝜑
Theorem	4sqlem18 13134*	Lemma for 4sq 13136. Inductive step, odd prime case. (Contributed by Mario Carneiro, 16-Jul-2014.) (Revised by AV, 14-Sep-2020.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 = ((2 · 𝑁) + 1))    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   ⊢ 𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   ⊢ 𝑀 = inf(𝑇, ℝ, < )    ⇒   ⊢ (𝜑 → 𝑃 ∈ 𝑆)
Theorem	4sqlem19 13135*	Lemma for 4sq 13136. The proof is by strong induction - we show that if all the integers less than 𝑘 are in 𝑆, then 𝑘 is as well. In this part of the proof we do the induction argument and dispense with all the cases except the odd prime case, which is sent to 4sqlem18 13134. If 𝑘 is 0, 1, 2, we show 𝑘 ∈ 𝑆 directly; otherwise if 𝑘 is composite, 𝑘 is the product of two numbers less than it (and hence in 𝑆 by assumption), so by mul4sq 13120 𝑘 ∈ 𝑆. (Contributed by Mario Carneiro, 14-Jul-2014.) (Revised by Mario Carneiro, 20-Jun-2015.)
⊢ 𝑆 = {𝑛 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ∃𝑧 ∈ ℤ ∃𝑤 ∈ ℤ 𝑛 = (((𝑥↑2) + (𝑦↑2)) + ((𝑧↑2) + (𝑤↑2)))}    ⇒   ⊢ ℕ0 = 𝑆
Theorem	4sq 13136*	Lagrange's four-square theorem, or Bachet's conjecture: every nonnegative integer is expressible as a sum of four squares. This is Metamath 100 proof #19. (Contributed by Mario Carneiro, 16-Jul-2014.)
⊢ (𝐴 ∈ ℕ0 ↔ ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ ∃𝑐 ∈ ℤ ∃𝑑 ∈ ℤ 𝐴 = (((𝑎↑2) + (𝑏↑2)) + ((𝑐↑2) + (𝑑↑2))))
5.2.13  Decimal arithmetic (cont.)
Theorem	dec2dvds 13137	Divisibility by two is obvious in base 10. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ (𝐵 · 2) = 𝐶    &   ⊢ 𝐷 = (𝐶 + 1)    ⇒   ⊢ ¬ 2 ∥ ;𝐴𝐷
Theorem	dec5dvds 13138	Divisibility by five is obvious in base 10. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ    &   ⊢ 𝐵 < 5    ⇒   ⊢ ¬ 5 ∥ ;𝐴𝐵
Theorem	dec5dvds2 13139	Divisibility by five is obvious in base 10. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ    &   ⊢ 𝐵 < 5    &   ⊢ (5 + 𝐵) = 𝐶    ⇒   ⊢ ¬ 5 ∥ ;𝐴𝐶
Theorem	dec5nprm 13140	A decimal number greater than 10 and ending with five is not a prime number. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ    ⇒   ⊢ ¬ ;𝐴5 ∈ ℙ
Theorem	dec2nprm 13141	A decimal number greater than 10 and ending with an even digit is not a prime number. (Contributed by Mario Carneiro, 19-Apr-2015.)
⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ (𝐵 · 2) = 𝐶    ⇒   ⊢ ¬ ;𝐴𝐶 ∈ ℙ
Theorem	modxai 13142	Add exponents in a power mod calculation. (Contributed by Mario Carneiro, 21-Feb-2014.) (Revised by Mario Carneiro, 5-Feb-2015.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ 𝐶 ∈ ℕ0    &   ⊢ 𝐿 ∈ ℕ0    &   ⊢ ((𝐴↑𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ ((𝐴↑𝐶) mod 𝑁) = (𝐿 mod 𝑁)    &   ⊢ (𝐵 + 𝐶) = 𝐸    &   ⊢ ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐿)    ⇒   ⊢ ((𝐴↑𝐸) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	mod2xi 13143	Double exponents in a power mod calculation. (Contributed by Mario Carneiro, 21-Feb-2014.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ ((𝐴↑𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ (2 · 𝐵) = 𝐸    &   ⊢ ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐾)    ⇒   ⊢ ((𝐴↑𝐸) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	modxp1i 13144	Add one to an exponent in a power mod calculation. (Contributed by Mario Carneiro, 21-Feb-2014.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ ((𝐴↑𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ (𝐵 + 1) = 𝐸    &   ⊢ ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐴)    ⇒   ⊢ ((𝐴↑𝐸) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	modsubi 13145	Subtract from within a mod calculation. (Contributed by Mario Carneiro, 18-Feb-2014.)
⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐴 ∈ ℕ    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝐴 mod 𝑁) = (𝐾 mod 𝑁)    &   ⊢ (𝑀 + 𝐵) = 𝐾    ⇒   ⊢ ((𝐴 − 𝐵) mod 𝑁) = (𝑀 mod 𝑁)
Theorem	gcdi 13146	Calculate a GCD via Euclid's algorithm. (Contributed by Mario Carneiro, 19-Feb-2014.)
⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑅 ∈ ℕ0    &   ⊢ 𝑁 ∈ ℕ0    &   ⊢ (𝑁 gcd 𝑅) = 𝐺    &   ⊢ ((𝐾 · 𝑁) + 𝑅) = 𝑀    ⇒   ⊢ (𝑀 gcd 𝑁) = 𝐺
Theorem	gcdmodi 13147	Calculate a GCD via Euclid's algorithm. Theorem 5.6 in [ApostolNT] p. 109. (Contributed by Mario Carneiro, 19-Feb-2014.)
⊢ 𝐾 ∈ ℕ0    &   ⊢ 𝑅 ∈ ℕ0    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ (𝐾 mod 𝑁) = (𝑅 mod 𝑁)    &   ⊢ (𝑁 gcd 𝑅) = 𝐺    ⇒   ⊢ (𝐾 gcd 𝑁) = 𝐺
Theorem	numexp0 13148	Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ (𝐴↑0) = 1
Theorem	numexp1 13149	Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ (𝐴↑1) = 𝐴
Theorem	numexpp1 13150	Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝑀 + 1) = 𝑁    &   ⊢ ((𝐴↑𝑀) · 𝐴) = 𝐶    ⇒   ⊢ (𝐴↑𝑁) = 𝐶
Theorem	numexp2x 13151	Double an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (2 · 𝑀) = 𝑁    &   ⊢ (𝐴↑𝑀) = 𝐷    &   ⊢ (𝐷 · 𝐷) = 𝐶    ⇒   ⊢ (𝐴↑𝑁) = 𝐶
Theorem	decsplit0b 13152	Split a decimal number into two parts. Base case: 𝑁 = 0. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ ((𝐴 · (;10↑0)) + 𝐵) = (𝐴 + 𝐵)
Theorem	decsplit0 13153	Split a decimal number into two parts. Base case: 𝑁 = 0. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ ((𝐴 · (;10↑0)) + 0) = 𝐴
Theorem	decsplit1 13154	Split a decimal number into two parts. Base case: 𝑁 = 1. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    ⇒   ⊢ ((𝐴 · (;10↑1)) + 𝐵) = ;𝐴𝐵
Theorem	decsplit 13155	Split a decimal number into two parts. Inductive step. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝑀 + 1) = 𝑁    &   ⊢ ((𝐴 · (;10↑𝑀)) + 𝐵) = 𝐶    ⇒   ⊢ ((𝐴 · (;10↑𝑁)) + ;𝐵𝐷) = ;𝐶𝐷
Theorem	karatsuba 13156	The Karatsuba multiplication algorithm. If 𝑋 and 𝑌 are decomposed into two groups of digits of length 𝑀 (only the lower group is known to be this size but the algorithm is most efficient when the partition is chosen near the middle of the digit string), then 𝑋𝑌 can be written in three groups of digits, where each group needs only one multiplication. Thus, we can halve both inputs with only three multiplications on the smaller operands, yielding an asymptotic improvement of n^(log2 3) instead of n^2 for the "naive" algorithm decmul1c 9794. (Contributed by Mario Carneiro, 16-Jul-2015.) (Revised by AV, 9-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ0    &   ⊢ 𝐶 ∈ ℕ0    &   ⊢ 𝐷 ∈ ℕ0    &   ⊢ 𝑆 ∈ ℕ0    &   ⊢ 𝑀 ∈ ℕ0    &   ⊢ (𝐴 · 𝐶) = 𝑅    &   ⊢ (𝐵 · 𝐷) = 𝑇    &   ⊢ ((𝐴 + 𝐵) · (𝐶 + 𝐷)) = ((𝑅 + 𝑆) + 𝑇)    &   ⊢ ((𝐴 · (;10↑𝑀)) + 𝐵) = 𝑋    &   ⊢ ((𝐶 · (;10↑𝑀)) + 𝐷) = 𝑌    &   ⊢ ((𝑅 · (;10↑𝑀)) + 𝑆) = 𝑊    &   ⊢ ((𝑊 · (;10↑𝑀)) + 𝑇) = 𝑍    ⇒   ⊢ (𝑋 · 𝑌) = 𝑍
Theorem	2exp4 13157	Two to the fourth power is 16. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑4) = ;16
Theorem	2exp5 13158	Two to the fifth power is 32. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑5) = ;32
Theorem	2exp6 13159	Two to the sixth power is 64. (Contributed by Mario Carneiro, 20-Apr-2015.) (Proof shortened by OpenAI, 25-Mar-2020.)
⊢ (2↑6) = ;64
Theorem	2exp7 13160	Two to the seventh power is 128. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑7) = ;;128
Theorem	2exp8 13161	Two to the eighth power is 256. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑8) = ;;256
Theorem	2exp11 13162	Two to the eleventh power is 2048. (Contributed by AV, 16-Aug-2021.)
⊢ (2↑;11) = ;;;2048
Theorem	2exp16 13163	Two to the sixteenth power is 65536. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (2↑;16) = ;;;;65536
Theorem	3exp3 13164	Three to the third power is 27. (Contributed by Mario Carneiro, 20-Apr-2015.)
⊢ (3↑3) = ;27
Theorem	2expltfac 13165	The factorial grows faster than two to the power 𝑁. (Contributed by Mario Carneiro, 15-Sep-2016.)
⊢ (𝑁 ∈ (ℤ≥‘4) → (2↑𝑁) < (!‘𝑁))
5.2.14  Bertrand's Ballot Problem
Theorem	ballotfilemofi 13166*	𝑂 is finite. (Contributed by Jim Kingdon, 20-May-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    ⇒   ⊢ 𝑂 ∈ Fin
Theorem	ballotfilem1 13167*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    ⇒   ⊢ (♯‘𝑂) = ((𝑀 + 𝑁)C𝑀)
Theorem	ballotfilemonn 13168*	The size of the universe is at least one. (Contributed by Jim Kingdon, 4-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    ⇒   ⊢ (♯‘𝑂) ∈ ℕ
Theorem	ballotfilemelo 13169*	Elementhood in 𝑂. (Contributed by Thierry Arnoux, 17-Apr-2017.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    ⇒   ⊢ (𝐶 ∈ 𝑂 ↔ (𝐶 ⊆ (1...(𝑀 + 𝑁)) ∧ 𝐶 ∈ Fin ∧ (♯‘𝐶) = 𝑀))
Theorem	ballotfilemcdc 13170*	Lemma for ballotfi . It is decidable whether a given integer is an element of a particular element of 𝑂. (Contributed by Jim Kingdon, 7-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐾 ∈ ℤ)    ⇒   ⊢ (𝜑 → DECID 𝐾 ∈ 𝐶)
Theorem	ballotfilemcinfi 13171*	Lemma for ballotfi . The portion of a counting representing votes for A up to a specified integer is finite. (Contributed by Jim Kingdon, 8-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((1...𝐽) ∩ 𝐶) ∈ Fin)
Theorem	ballotfilemdifcfi 13172*	Lemma for ballotfi . The portion of a counting representing votes for B up to a specified integer is finite. (Contributed by Jim Kingdon, 8-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((1...𝐽) ∖ 𝐶) ∈ Fin)
Theorem	ballotfilemcinfz 13173*	Lemma for ballotfi . The portion of a counting representing votes for A within a specified integer range is finite. (Contributed by Jim Kingdon, 15-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((𝐽...𝐾) ∩ 𝐶) ∈ Fin)
Theorem	ballotfilemdifcfz 13174*	Lemma for ballotfi . The portion of a counting representing votes for B within a specified integer range is finite. (Contributed by Jim Kingdon, 15-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ (𝜑 → 𝐶 ∈ 𝑂)    &   ⊢ (𝜑 → 𝐽 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((𝐽...𝐾) ∖ 𝐶) ∈ Fin)
Theorem	ballotfilem2 13175*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    ⇒   ⊢ (𝑃‘{𝑐 ∈ 𝑂 ∣ ¬ 1 ∈ 𝑐}) = (𝑁 / (𝑀 + 𝑁))
Theorem	ballotfilemfval 13176*	The value of 𝐹. (Contributed by Thierry Arnoux, 23-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemfelz 13177*	(𝐹‘𝐶) has values in ℤ. (Contributed by Thierry Arnoux, 23-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemfp1 13178*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemfc0 13179*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemfcc 13180*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemfmpn 13181*	(𝐹‘𝐶) finishes counting at (𝑀 − 𝑁). (Contributed by Thierry Arnoux, 25-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    ⇒   ⊢ (𝐶 ∈ 𝑂 → ((𝐹‘𝐶)‘(𝑀 + 𝑁)) = (𝑀 − 𝑁))
Theorem	ballotfilemfval0 13182*	(𝐹‘𝐶) always starts counting at 0 . (Contributed by Thierry Arnoux, 25-Nov-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    ⇒   ⊢ (𝐶 ∈ 𝑂 → ((𝐹‘𝐶)‘0) = 0)
Theorem	ballotfileme 13183*	Elements of 𝐸. (Contributed by Thierry Arnoux, 14-Dec-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemefi 13184*	𝐸 is finite. (Contributed by Jim Kingdon, 17-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘𝑥) / (♯‘𝑂)))    &   ⊢ 𝐹 = (𝑐 ∈ 𝑂 ↦ (𝑖 ∈ ℤ ↦ ((♯‘((1...𝑖) ∩ 𝑐)) − (♯‘((1...𝑖) ∖ 𝑐)))))    &   ⊢ 𝐸 = {𝑐 ∈ 𝑂 ∣ ∀𝑖 ∈ (1...(𝑀 + 𝑁))0 < ((𝐹‘𝑐)‘𝑖)}    ⇒   ⊢ 𝐸 ∈ Fin
Theorem	ballotfilemafi 13185*	The set of countings where A got the first vote, but does not stay strictly ahead throughout, is finite. (Contributed by Jim Kingdon, 17-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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) ∣ 1 ∈ 𝑐} ∈ Fin
Theorem	ballotfilembfi 13186*	The set of countings where B got the first vote is finite. (Contributed by Jim Kingdon, 17-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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) ∣ ¬ 1 ∈ 𝑐} ∈ Fin
Theorem	ballotfilemodife 13187*	Elements of (𝑂 ∖ 𝐸). (Contributed by Thierry Arnoux, 7-Dec-2016.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilem4 13188*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilem5 13189*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemi 13190*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemiex 13191*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemi1 13192*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemii 13193*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemscl 13194*	The set of zeroes of 𝐹 has an infimum. (Contributed by Jim Kingdon, 12-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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 = {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝐶)‘𝑘) = 0}    ⇒   ⊢ (𝐶 ∈ (𝑂 ∖ 𝐸) → inf(𝑆, ℝ, < ) ∈ 𝑆)
Theorem	ballotfilemsle 13195*	The infimum of the set of zeroes of 𝐹 is a lower bound. (Contributed by Jim Kingdon, 12-Jun-2026.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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 = {𝑘 ∈ (1...(𝑀 + 𝑁)) ∣ ((𝐹‘𝐶)‘𝑘) = 0}    ⇒   ⊢ ((𝐶 ∈ (𝑂 ∖ 𝐸) ∧ 𝑋 ∈ 𝑆) → inf(𝑆, ℝ, < ) ≤ 𝑋)
Theorem	ballotfilemimin 13196*	(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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemic 13197*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilem1c 13198*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemsval 13199*	Value of 𝑆. (Contributed by Thierry Arnoux, 12-Apr-2017.)
⊢ 𝑀 ∈ ℕ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝑂 = {𝑐 ∈ (𝒫 (1...(𝑀 + 𝑁)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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	ballotfilemsv 13200*	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)) ∩ Fin) ∣ (♯‘𝑐) = 𝑀}    &   ⊢ 𝑃 = (𝑥 ∈ (𝒫 𝑂 ∩ Fin) ↦ ((♯‘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))