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Theorem List for Metamath Proof Explorer - 16101-16200   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoreminfpnlem1 16101* Lemma for infpn 16103. The smallest divisor (greater than 1) 𝑀 of 𝑁! + 1 is a prime greater than 𝑁. (Contributed by NM, 5-May-2005.)
𝐾 = ((!‘𝑁) + 1)       ((𝑁 ∈ ℕ ∧ 𝑀 ∈ ℕ) → (((1 < 𝑀 ∧ (𝐾 / 𝑀) ∈ ℕ) ∧ ∀𝑗 ∈ ℕ ((1 < 𝑗 ∧ (𝐾 / 𝑗) ∈ ℕ) → 𝑀𝑗)) → (𝑁 < 𝑀 ∧ ∀𝑗 ∈ ℕ ((𝑀 / 𝑗) ∈ ℕ → (𝑗 = 1 ∨ 𝑗 = 𝑀)))))

Theoreminfpnlem2 16102* Lemma for infpn 16103. For any positive integer 𝑁, there exists a prime number 𝑗 greater than 𝑁. (Contributed by NM, 5-May-2005.)
𝐾 = ((!‘𝑁) + 1)       (𝑁 ∈ ℕ → ∃𝑗 ∈ ℕ (𝑁 < 𝑗 ∧ ∀𝑘 ∈ ℕ ((𝑗 / 𝑘) ∈ ℕ → (𝑘 = 1 ∨ 𝑘 = 𝑗))))

Theoreminfpn 16103* There exist infinitely many prime numbers: for any positive integer 𝑁, there exists a prime number 𝑗 greater than 𝑁. (See infpn2 16104 for the equinumerosity version.) (Contributed by NM, 1-Jun-2006.)
(𝑁 ∈ ℕ → ∃𝑗 ∈ ℕ (𝑁 < 𝑗 ∧ ∀𝑘 ∈ ℕ ((𝑗 / 𝑘) ∈ ℕ → (𝑘 = 1 ∨ 𝑘 = 𝑗))))

Theoreminfpn2 16104* There exist infinitely many prime numbers: the set of all primes 𝑆 is unbounded by infpn 16103, so by unben 16100 it is infinite. This is Metamath 100 proof #11. (Contributed by NM, 5-May-2005.)
𝑆 = {𝑛 ∈ ℕ ∣ (1 < 𝑛 ∧ ∀𝑚 ∈ ℕ ((𝑛 / 𝑚) ∈ ℕ → (𝑚 = 1 ∨ 𝑚 = 𝑛)))}       𝑆 ≈ ℕ

Theoremprmunb 16105* The primes are unbounded. (Contributed by Paul Chapman, 28-Nov-2012.)
(𝑁 ∈ ℕ → ∃𝑝 ∈ ℙ 𝑁 < 𝑝)

Theoremprminf 16106 There are an infinite number of primes. Theorem 1.7 in [ApostolNT] p. 16. (Contributed by Paul Chapman, 28-Nov-2012.)
ℙ ≈ ℕ

6.2.10  Sum of prime reciprocals

Theoremprmreclem1 16107* Lemma for prmrec 16113. Properties of the "square part" function, which extracts the 𝑚 of the decomposition 𝑁 = 𝑟𝑚↑2, with 𝑚 maximal and 𝑟 squarefree. (Contributed by Mario Carneiro, 5-Aug-2014.)
𝑄 = (𝑛 ∈ ℕ ↦ sup({𝑟 ∈ ℕ ∣ (𝑟↑2) ∥ 𝑛}, ℝ, < ))       (𝑁 ∈ ℕ → ((𝑄𝑁) ∈ ℕ ∧ ((𝑄𝑁)↑2) ∥ 𝑁 ∧ (𝐾 ∈ (ℤ‘2) → ¬ (𝐾↑2) ∥ (𝑁 / ((𝑄𝑁)↑2)))))

Theoremprmreclem2 16108* Lemma for prmrec 16113. There are at most 2↑𝐾 squarefree numbers which divide no primes larger than 𝐾. (We could strengthen this to 2↑♯(ℙ ∩ (1...𝐾)) but there's no reason to.) We establish the inequality by showing that the prime counts of the number up to 𝐾 completely determine it because all higher prime counts are zero, and they are all at most 1 because no square divides the number, so there are at most 2↑𝐾 possibilities. (Contributed by Mario Carneiro, 5-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0))    &   (𝜑𝐾 ∈ ℕ)    &   (𝜑𝑁 ∈ ℕ)    &   𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝𝑛}    &   𝑄 = (𝑛 ∈ ℕ ↦ sup({𝑟 ∈ ℕ ∣ (𝑟↑2) ∥ 𝑛}, ℝ, < ))       (𝜑 → (♯‘{𝑥𝑀 ∣ (𝑄𝑥) = 1}) ≤ (2↑𝐾))

Theoremprmreclem3 16109* Lemma for prmrec 16113. The main inequality established here is 𝑀 ≤ ♯{𝑥𝑀 ∣ (𝑄𝑥) = 1} · √𝑁, where {𝑥𝑀 ∣ (𝑄𝑥) = 1} is the set of squarefree numbers in 𝑀. This is demonstrated by the map 𝑦 ↦ ⟨𝑦 / (𝑄𝑦)↑2, (𝑄𝑦)⟩ where 𝑄𝑦 is the largest number whose square divides 𝑦. (Contributed by Mario Carneiro, 5-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0))    &   (𝜑𝐾 ∈ ℕ)    &   (𝜑𝑁 ∈ ℕ)    &   𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝𝑛}    &   𝑄 = (𝑛 ∈ ℕ ↦ sup({𝑟 ∈ ℕ ∣ (𝑟↑2) ∥ 𝑛}, ℝ, < ))       (𝜑 → (♯‘𝑀) ≤ ((2↑𝐾) · (√‘𝑁)))

Theoremprmreclem4 16110* Lemma for prmrec 16113. Show by induction that the indexed (nondisjoint) union 𝑊𝑘 is at most the size of the prime reciprocal series. The key counting lemma is hashdvds 15967, to show that the number of numbers in 1...𝑁 that divide 𝑘 is at most 𝑁 / 𝑘. (Contributed by Mario Carneiro, 6-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0))    &   (𝜑𝐾 ∈ ℕ)    &   (𝜑𝑁 ∈ ℕ)    &   𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝𝑛}    &   (𝜑 → seq1( + , 𝐹) ∈ dom ⇝ )    &   (𝜑 → Σ𝑘 ∈ (ℤ‘(𝐾 + 1))if(𝑘 ∈ ℙ, (1 / 𝑘), 0) < (1 / 2))    &   𝑊 = (𝑝 ∈ ℕ ↦ {𝑛 ∈ (1...𝑁) ∣ (𝑝 ∈ ℙ ∧ 𝑝𝑛)})       (𝜑 → (𝑁 ∈ (ℤ𝐾) → (♯‘ 𝑘 ∈ ((𝐾 + 1)...𝑁)(𝑊𝑘)) ≤ (𝑁 · Σ𝑘 ∈ ((𝐾 + 1)...𝑁)if(𝑘 ∈ ℙ, (1 / 𝑘), 0))))

Theoremprmreclem5 16111* Lemma for prmrec 16113. Here we show the inequality 𝑁 / 2 < ♯𝑀 by decomposing the set (1...𝑁) into the disjoint union of the set 𝑀 of those numbers that are not divisible by any "large" primes (above 𝐾) and the indexed union over 𝐾 < 𝑘 of the numbers 𝑊𝑘 that divide the prime 𝑘. By prmreclem4 16110 the second of these has size less than 𝑁 times the prime reciprocal series, which is less than 1 / 2 by assumption, we find that the complementary part 𝑀 must be at least 𝑁 / 2 large. (Contributed by Mario Carneiro, 6-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0))    &   (𝜑𝐾 ∈ ℕ)    &   (𝜑𝑁 ∈ ℕ)    &   𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝𝑛}    &   (𝜑 → seq1( + , 𝐹) ∈ dom ⇝ )    &   (𝜑 → Σ𝑘 ∈ (ℤ‘(𝐾 + 1))if(𝑘 ∈ ℙ, (1 / 𝑘), 0) < (1 / 2))    &   𝑊 = (𝑝 ∈ ℕ ↦ {𝑛 ∈ (1...𝑁) ∣ (𝑝 ∈ ℙ ∧ 𝑝𝑛)})       (𝜑 → (𝑁 / 2) < ((2↑𝐾) · (√‘𝑁)))

Theoremprmreclem6 16112* Lemma for prmrec 16113. If the series 𝐹 was convergent, there would be some 𝑘 such that the sum starting from 𝑘 + 1 sums to less than 1 / 2; this is a sufficient hypothesis for prmreclem5 16111 to produce the contradictory bound 𝑁 / 2 < (2↑𝑘)√𝑁, which is false for 𝑁 = 2↑(2𝑘 + 2). (Contributed by Mario Carneiro, 6-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0))        ¬ seq1( + , 𝐹) ∈ dom ⇝

Theoremprmrec 16113* The sum of the reciprocals of the primes diverges. Theorem 1.13 in [ApostolNT] p. 18. This is the "second" proof at http://en.wikipedia.org/wiki/Prime_harmonic_series, attributed to Paul Erdős. This is Metamath 100 proof #81. (Contributed by Mario Carneiro, 6-Aug-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ Σ𝑘 ∈ (ℙ ∩ (1...𝑛))(1 / 𝑘))        ¬ 𝐹 ∈ dom ⇝

6.2.11  Fundamental theorem of arithmetic

Theorem1arithlem1 16114* Lemma for 1arith 16118. (Contributed by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))       (𝑁 ∈ ℕ → (𝑀𝑁) = (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑁)))

Theorem1arithlem2 16115* Lemma for 1arith 16118. (Contributed by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))       ((𝑁 ∈ ℕ ∧ 𝑃 ∈ ℙ) → ((𝑀𝑁)‘𝑃) = (𝑃 pCnt 𝑁))

Theorem1arithlem3 16116* Lemma for 1arith 16118. (Contributed by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))       (𝑁 ∈ ℕ → (𝑀𝑁):ℙ⟶ℕ0)

Theorem1arithlem4 16117* Lemma for 1arith 16118. (Contributed by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    &   𝐺 = (𝑦 ∈ ℕ ↦ if(𝑦 ∈ ℙ, (𝑦↑(𝐹𝑦)), 1))    &   (𝜑𝐹:ℙ⟶ℕ0)    &   (𝜑𝑁 ∈ ℕ)    &   ((𝜑 ∧ (𝑞 ∈ ℙ ∧ 𝑁𝑞)) → (𝐹𝑞) = 0)       (𝜑 → ∃𝑥 ∈ ℕ 𝐹 = (𝑀𝑥))

Theorem1arith 16118* Fundamental theorem of arithmetic, where a prime factorization is represented as a sequence of prime exponents, for which only finitely many primes have nonzero exponent. The function 𝑀 maps the set of positive integers one-to-one onto the set of prime factorizations 𝑅. (Contributed by Paul Chapman, 17-Nov-2012.) (Proof shortened by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    &   𝑅 = {𝑒 ∈ (ℕ0𝑚 ℙ) ∣ (𝑒 “ ℕ) ∈ Fin}       𝑀:ℕ–1-1-onto𝑅

Theorem1arith2 16119* Fundamental theorem of arithmetic, where a prime factorization is represented as a finite monotonic 1-based sequence of primes. Every positive integer has a unique prime factorization. Theorem 1.10 in [ApostolNT] p. 17. This is Metamath 100 proof #80. (Contributed by Paul Chapman, 17-Nov-2012.) (Revised by Mario Carneiro, 30-May-2014.)
𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛)))    &   𝑅 = {𝑒 ∈ (ℕ0𝑚 ℙ) ∣ (𝑒 “ ℕ) ∈ Fin}       𝑧 ∈ ℕ ∃!𝑔𝑅 (𝑀𝑧) = 𝑔

6.2.12  Lagrange's four-square theorem

Syntaxcgz 16120 Extend class notation with the set of gaussian integers.
class ℤ[i]

Definitiondf-gz 16121 Define the set of gaussian integers, which are complex numbers whose real and imaginary parts are integers. (Note that the [i] is actually part of the symbol token and has no independent meaning.) (Contributed by Mario Carneiro, 14-Jul-2014.)
ℤ[i] = {𝑥 ∈ ℂ ∣ ((ℜ‘𝑥) ∈ ℤ ∧ (ℑ‘𝑥) ∈ ℤ)}

Theoremelgz 16122 Elementhood in the gaussian integers. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ))

Theoremgzcn 16123 A gaussian integer is a complex number. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] → 𝐴 ∈ ℂ)

Theoremzgz 16124 An integer is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ → 𝐴 ∈ ℤ[i])

Theoremigz 16125 i is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.)
i ∈ ℤ[i]

Theoremgznegcl 16126 The gaussian integers are closed under negation. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] → -𝐴 ∈ ℤ[i])

Theoremgzcjcl 16127 The gaussian integers are closed under conjugation. (Contributed by Mario Carneiro, 14-Jul-2014.)
(𝐴 ∈ ℤ[i] → (∗‘𝐴) ∈ ℤ[i])

Theoremgzaddcl 16128 The gaussian integers are closed under addition. (Contributed by Mario Carneiro, 14-Jul-2014.)
((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 + 𝐵) ∈ ℤ[i])

Theoremgzmulcl 16129 The gaussian integers are closed under multiplication. (Contributed by Mario Carneiro, 14-Jul-2014.)
((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴 · 𝐵) ∈ ℤ[i])

Theoremgzreim 16130 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])

Theoremgzsubcl 16131 The gaussian integers are closed under subtraction. (Contributed by Mario Carneiro, 14-Jul-2014.)
((𝐴 ∈ ℤ[i] ∧ 𝐵 ∈ ℤ[i]) → (𝐴𝐵) ∈ ℤ[i])

Theoremgzabssqcl 16132 The squared norm of a gaussian integer is an integer. (Contributed by Mario Carneiro, 16-Jul-2014.)
(𝐴 ∈ ℤ[i] → ((abs‘𝐴)↑2) ∈ ℕ0)

Theorem4sqlem5 16133 Lemma for 4sq 16155. (Contributed by Mario Carneiro, 15-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))       (𝜑 → (𝐵 ∈ ℤ ∧ ((𝐴𝐵) / 𝑀) ∈ ℤ))

Theorem4sqlem6 16134 Lemma for 4sq 16155. (Contributed by Mario Carneiro, 15-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))       (𝜑 → (-(𝑀 / 2) ≤ 𝐵𝐵 < (𝑀 / 2)))

Theorem4sqlem7 16135 Lemma for 4sq 16155. (Contributed by Mario Carneiro, 15-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))       (𝜑 → (𝐵↑2) ≤ (((𝑀↑2) / 2) / 2))

Theorem4sqlem8 16136 Lemma for 4sq 16155. (Contributed by Mario Carneiro, 15-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))       (𝜑𝑀 ∥ ((𝐴↑2) − (𝐵↑2)))

Theorem4sqlem9 16137 Lemma for 4sq 16155. (Contributed by Mario Carneiro, 15-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ((𝜑𝜓) → (𝐵↑2) = 0)       ((𝜑𝜓) → (𝑀↑2) ∥ (𝐴↑2))

Theorem4sqlem10 16138 Lemma for 4sq 16155. (Contributed by Mario Carneiro, 16-Jul-2014.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℕ)    &   𝐵 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ((𝜑𝜓) → ((((𝑀↑2) / 2) / 2) − (𝐵↑2)) = 0)       ((𝜑𝜓) → (𝑀↑2) ∥ ((𝐴↑2) − (((𝑀↑2) / 2) / 2)))

Theorem4sqlem1 16139* Lemma for 4sq 16155. 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

Theorem4sqlem2 16140* Lemma for 4sq 16155. 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)))}       (𝐴𝑆 ↔ ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ ∃𝑐 ∈ ℤ ∃𝑑 ∈ ℤ 𝐴 = (((𝑎↑2) + (𝑏↑2)) + ((𝑐↑2) + (𝑑↑2))))

Theorem4sqlem3 16141* Lemma for 4sq 16155. 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)

Theorem4sqlem4a 16142* Lemma for 4sqlem4 16143. (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)

Theorem4sqlem4 16143* Lemma for 4sq 16155. 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)))}       (𝐴𝑆 ↔ ∃𝑢 ∈ ℤ[i] ∃𝑣 ∈ ℤ[i] 𝐴 = (((abs‘𝑢)↑2) + ((abs‘𝑣)↑2)))

Theoremmul4sqlem 16144* Lemma for mul4sq 16145: algebraic manipulations. The extra assumptions involving 𝑀 are for a part of 4sqlem17 16152 which needs to 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)))}    &   (𝜑𝐴 ∈ ℤ[i])    &   (𝜑𝐵 ∈ ℤ[i])    &   (𝜑𝐶 ∈ ℤ[i])    &   (𝜑𝐷 ∈ ℤ[i])    &   𝑋 = (((abs‘𝐴)↑2) + ((abs‘𝐵)↑2))    &   𝑌 = (((abs‘𝐶)↑2) + ((abs‘𝐷)↑2))    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑 → ((𝐴𝐶) / 𝑀) ∈ ℤ[i])    &   (𝜑 → ((𝐵𝐷) / 𝑀) ∈ ℤ[i])    &   (𝜑 → (𝑋 / 𝑀) ∈ ℕ0)       (𝜑 → ((𝑋 / 𝑀) · (𝑌 / 𝑀)) ∈ 𝑆)

Theoremmul4sq 16145* 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 16144. (For the curious, the explicit formula that is used is ( ∣ 𝑎 ∣ ↑2 + ∣ 𝑏 ∣ ↑2)( ∣ 𝑐 ∣ ↑2 + ∣ 𝑑 ∣ ↑2) = 𝑎∗ · 𝑐 + 𝑏 · 𝑑∗ ∣ ↑2 + ∣ 𝑎∗ · 𝑑𝑏 · 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)))}       ((𝐴𝑆𝐵𝑆) → (𝐴 · 𝐵) ∈ 𝑆)

Theorem4sqlem11 16146* Lemma for 4sq 16155. 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. (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)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   𝐹 = (𝑣𝐴 ↦ ((𝑃 − 1) − 𝑣))       (𝜑 → (𝐴 ∩ ran 𝐹) ≠ ∅)

Theorem4sqlem12 16147* Lemma for 4sq 16155. 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)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   𝐴 = {𝑢 ∣ ∃𝑚 ∈ (0...𝑁)𝑢 = ((𝑚↑2) mod 𝑃)}    &   𝐹 = (𝑣𝐴 ↦ ((𝑃 − 1) − 𝑣))       (𝜑 → ∃𝑘 ∈ (1...(𝑃 − 1))∃𝑢 ∈ ℤ[i] (((abs‘𝑢)↑2) + 1) = (𝑘 · 𝑃))

Theorem4sqlem13 16148* Lemma for 4sq 16155. (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)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )       (𝜑 → (𝑇 ≠ ∅ ∧ 𝑀 < 𝑃))

Theorem4sqlem14 16149* Lemma for 4sq 16155. (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)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )    &   (𝜑𝑀 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ 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))))       (𝜑𝑅 ∈ ℕ0)

Theorem4sqlem15 16150* Lemma for 4sq 16155. (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)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )    &   (𝜑𝑀 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ 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))))       ((𝜑𝑅 = 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)))

Theorem4sqlem16 16151* Lemma for 4sq 16155. (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)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )    &   (𝜑𝑀 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ 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 → (𝑅𝑀 ∧ ((𝑅 = 0 ∨ 𝑅 = 𝑀) → (𝑀↑2) ∥ (𝑀 · 𝑃))))

Theorem4sqlem17 16152* Lemma for 4sq 16155. (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)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )    &   (𝜑𝑀 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ 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

Theorem4sqlem18 16153* Lemma for 4sq 16155. 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)))}    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 = ((2 · 𝑁) + 1))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (0...(2 · 𝑁)) ⊆ 𝑆)    &   𝑇 = {𝑖 ∈ ℕ ∣ (𝑖 · 𝑃) ∈ 𝑆}    &   𝑀 = inf(𝑇, ℝ, < )       (𝜑𝑃𝑆)

Theorem4sqlem19 16154* Lemma for 4sq 16155. 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 16153. If 𝑘 is 0, 1, 2, we show 𝑘𝑆 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 16145 𝑘𝑆. (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)))}       0 = 𝑆

Theorem4sq 16155* 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))))

6.2.13  Van der Waerden's theorem

Syntaxcvdwa 16156 The arithmetic progression function.
class AP

Syntaxcvdwm 16157 The monochromatic arithmetic progression predicate.
class MonoAP

Syntaxcvdwp 16158 The polychromatic arithmetic progression predicate.
class PolyAP

Definitiondf-vdwap 16159* Define the arithmetic progression function, which takes as input a length 𝑘, a start point 𝑎, and a step 𝑑 and outputs the set of points in this progression. (Contributed by Mario Carneiro, 18-Aug-2014.)
AP = (𝑘 ∈ ℕ0 ↦ (𝑎 ∈ ℕ, 𝑑 ∈ ℕ ↦ ran (𝑚 ∈ (0...(𝑘 − 1)) ↦ (𝑎 + (𝑚 · 𝑑)))))

Definitiondf-vdwmc 16160* Define the "contains a monochromatic AP" predicate. (Contributed by Mario Carneiro, 18-Aug-2014.)
MonoAP = {⟨𝑘, 𝑓⟩ ∣ ∃𝑐(ran (AP‘𝑘) ∩ 𝒫 (𝑓 “ {𝑐})) ≠ ∅}

Definitiondf-vdwpc 16161* Define the "contains a polychromatic collection of APs" predicate. See vdwpc 16171 for more information. (Contributed by Mario Carneiro, 18-Aug-2014.)
PolyAP = {⟨⟨𝑚, 𝑘⟩, 𝑓⟩ ∣ ∃𝑎 ∈ ℕ ∃𝑑 ∈ (ℕ ↑𝑚 (1...𝑚))(∀𝑖 ∈ (1...𝑚)((𝑎 + (𝑑𝑖))(AP‘𝑘)(𝑑𝑖)) ⊆ (𝑓 “ {(𝑓‘(𝑎 + (𝑑𝑖)))}) ∧ (♯‘ran (𝑖 ∈ (1...𝑚) ↦ (𝑓‘(𝑎 + (𝑑𝑖))))) = 𝑚)}

Theoremvdwapfval 16162* Define the arithmetic progression function, which takes as input a length 𝑘, a start point 𝑎, and a step 𝑑 and outputs the set of points in this progression. (Contributed by Mario Carneiro, 18-Aug-2014.)
(𝐾 ∈ ℕ0 → (AP‘𝐾) = (𝑎 ∈ ℕ, 𝑑 ∈ ℕ ↦ ran (𝑚 ∈ (0...(𝐾 − 1)) ↦ (𝑎 + (𝑚 · 𝑑)))))

Theoremvdwapf 16163 The arithmetic progression function is a function. (Contributed by Mario Carneiro, 18-Aug-2014.)
(𝐾 ∈ ℕ0 → (AP‘𝐾):(ℕ × ℕ)⟶𝒫 ℕ)

Theoremvdwapval 16164* Value of the arithmetic progression function. (Contributed by Mario Carneiro, 18-Aug-2014.)
((𝐾 ∈ ℕ0𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝑋 ∈ (𝐴(AP‘𝐾)𝐷) ↔ ∃𝑚 ∈ (0...(𝐾 − 1))𝑋 = (𝐴 + (𝑚 · 𝐷))))

Theoremvdwapun 16165 Remove the first element of an arithmetic progression. (Contributed by Mario Carneiro, 11-Sep-2014.)
((𝐾 ∈ ℕ0𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝐴(AP‘(𝐾 + 1))𝐷) = ({𝐴} ∪ ((𝐴 + 𝐷)(AP‘𝐾)𝐷)))

Theoremvdwapid1 16166 The first element of an arithmetic progression. (Contributed by Mario Carneiro, 12-Sep-2014.)
((𝐾 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → 𝐴 ∈ (𝐴(AP‘𝐾)𝐷))

Theoremvdwap0 16167 Value of a length-1 arithmetic progression. (Contributed by Mario Carneiro, 18-Aug-2014.)
((𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝐴(AP‘0)𝐷) = ∅)

Theoremvdwap1 16168 Value of a length-1 arithmetic progression. (Contributed by Mario Carneiro, 18-Aug-2014.)
((𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝐴(AP‘1)𝐷) = {𝐴})

Theoremvdwmc 16169* The predicate " The 𝑅, 𝑁-coloring 𝐹 contains a monochromatic AP of length 𝐾". (Contributed by Mario Carneiro, 18-Aug-2014.)
𝑋 ∈ V    &   (𝜑𝐾 ∈ ℕ0)    &   (𝜑𝐹:𝑋𝑅)       (𝜑 → (𝐾 MonoAP 𝐹 ↔ ∃𝑐𝑎 ∈ ℕ ∃𝑑 ∈ ℕ (𝑎(AP‘𝐾)𝑑) ⊆ (𝐹 “ {𝑐})))

Theoremvdwmc2 16170* Expand out the definition of an arithmetic progression. (Contributed by Mario Carneiro, 18-Aug-2014.)
𝑋 ∈ V    &   (𝜑𝐾 ∈ ℕ0)    &   (𝜑𝐹:𝑋𝑅)    &   (𝜑𝐴𝑋)       (𝜑 → (𝐾 MonoAP 𝐹 ↔ ∃𝑐𝑅𝑎 ∈ ℕ ∃𝑑 ∈ ℕ ∀𝑚 ∈ (0...(𝐾 − 1))(𝑎 + (𝑚 · 𝑑)) ∈ (𝐹 “ {𝑐})))

Theoremvdwpc 16171* The predicate " The coloring 𝐹 contains a polychromatic 𝑀-tuple of AP's of length 𝐾". A polychromatic 𝑀-tuple of AP's is a set of AP's with the same base point but different step lengths, such that each individual AP is monochromatic, but the AP's all have mutually distinct colors. (The common basepoint is not required to have the same color as any of the AP's.) (Contributed by Mario Carneiro, 18-Aug-2014.)
𝑋 ∈ V    &   (𝜑𝐾 ∈ ℕ0)    &   (𝜑𝐹:𝑋𝑅)    &   (𝜑𝑀 ∈ ℕ)    &   𝐽 = (1...𝑀)       (𝜑 → (⟨𝑀, 𝐾⟩ PolyAP 𝐹 ↔ ∃𝑎 ∈ ℕ ∃𝑑 ∈ (ℕ ↑𝑚 𝐽)(∀𝑖𝐽 ((𝑎 + (𝑑𝑖))(AP‘𝐾)(𝑑𝑖)) ⊆ (𝐹 “ {(𝐹‘(𝑎 + (𝑑𝑖)))}) ∧ (♯‘ran (𝑖𝐽 ↦ (𝐹‘(𝑎 + (𝑑𝑖))))) = 𝑀)))

Theoremvdwlem1 16172* Lemma for vdw 16185. (Contributed by Mario Carneiro, 12-Sep-2014.)
(𝜑𝑅 ∈ Fin)    &   (𝜑𝐾 ∈ ℕ)    &   (𝜑𝑊 ∈ ℕ)    &   (𝜑𝐹:(1...𝑊)⟶𝑅)    &   (𝜑𝐴 ∈ ℕ)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝐷:(1...𝑀)⟶ℕ)    &   (𝜑 → ∀𝑖 ∈ (1...𝑀)((𝐴 + (𝐷𝑖))(AP‘𝐾)(𝐷𝑖)) ⊆ (𝐹 “ {(𝐹‘(𝐴 + (𝐷𝑖)))}))    &   (𝜑𝐼 ∈ (1...𝑀))    &   (𝜑 → (𝐹𝐴) = (𝐹‘(𝐴 + (𝐷𝐼))))       (𝜑 → (𝐾 + 1) MonoAP 𝐹)

Theoremvdwlem2 16173* Lemma for vdw 16185. (Contributed by Mario Carneiro, 12-Sep-2014.)
(𝜑𝑅 ∈ Fin)    &   (𝜑𝐾 ∈ ℕ0)    &   (𝜑𝑊 ∈ ℕ)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝐹:(1...𝑀)⟶𝑅)    &   (𝜑𝑀 ∈ (ℤ‘(𝑊 + 𝑁)))    &   𝐺 = (𝑥 ∈ (1...𝑊) ↦ (𝐹‘(𝑥 + 𝑁)))       (𝜑 → (𝐾 MonoAP 𝐺𝐾 MonoAP 𝐹))

Theoremvdwlem3 16174 Lemma for vdw 16185. (Contributed by Mario Carneiro, 13-Sep-2014.)
(𝜑𝑉 ∈ ℕ)    &   (𝜑𝑊 ∈ ℕ)    &   (𝜑𝐴 ∈ (1...𝑉))    &   (𝜑𝐵 ∈ (1...𝑊))       (𝜑 → (𝐵 + (𝑊 · ((𝐴 − 1) + 𝑉))) ∈ (1...(𝑊 · (2 · 𝑉))))

Theoremvdwlem4 16175* Lemma for vdw 16185. (Contributed by Mario Carneiro, 12-Sep-2014.)
(𝜑𝑉 ∈ ℕ)    &   (𝜑𝑊 ∈ ℕ)    &   (𝜑𝑅 ∈ Fin)    &   (𝜑𝐻:(1...(𝑊 · (2 · 𝑉)))⟶𝑅)    &   𝐹 = (𝑥 ∈ (1...𝑉) ↦ (𝑦 ∈ (1...𝑊) ↦ (𝐻‘(𝑦 + (𝑊 · ((𝑥 − 1) + 𝑉))))))       (𝜑𝐹:(1...𝑉)⟶(𝑅𝑚 (1...𝑊)))

Theoremvdwlem5 16176* Lemma for vdw 16185. (Contributed by Mario Carneiro, 12-Sep-2014.)
(𝜑𝑉 ∈ ℕ)    &   (𝜑𝑊 ∈ ℕ)    &   (𝜑𝑅 ∈ Fin)    &   (𝜑𝐻:(1...(𝑊 · (2 · 𝑉)))⟶𝑅)    &   𝐹 = (𝑥 ∈ (1...𝑉) ↦ (𝑦 ∈ (1...𝑊) ↦ (𝐻‘(𝑦 + (𝑊 · ((𝑥 − 1) + 𝑉))))))    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝐺:(1...𝑊)⟶𝑅)    &   (𝜑𝐾 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℕ)    &   (𝜑𝐷 ∈ ℕ)    &   (𝜑 → (𝐴(AP‘𝐾)𝐷) ⊆ (𝐹 “ {𝐺}))    &   (𝜑𝐵 ∈ ℕ)    &   (𝜑𝐸:(1...𝑀)⟶ℕ)    &   (𝜑 → ∀𝑖 ∈ (1...𝑀)((𝐵 + (𝐸𝑖))(AP‘𝐾)(𝐸𝑖)) ⊆ (𝐺 “ {(𝐺‘(𝐵 + (𝐸𝑖)))}))    &   𝐽 = (𝑖 ∈ (1...𝑀) ↦ (𝐺‘(𝐵 + (𝐸𝑖))))    &   (𝜑 → (♯‘ran 𝐽) = 𝑀)    &   𝑇 = (𝐵 + (𝑊 · ((𝐴 + (𝑉𝐷)) − 1)))    &   𝑃 = (𝑗 ∈ (1...(𝑀 + 1)) ↦ (if(𝑗 = (𝑀 + 1), 0, (𝐸𝑗)) + (𝑊 · 𝐷)))       (𝜑𝑇 ∈ ℕ)

Theoremvdwlem6 16177* Lemma for vdw 16185. (Contributed by Mario Carneiro, 13-Sep-2014.)
(𝜑𝑉 ∈ ℕ)    &   (𝜑𝑊 ∈ ℕ)    &   (𝜑𝑅 ∈ Fin)    &   (𝜑𝐻:(1...(𝑊 · (2 · 𝑉)))⟶𝑅)    &   𝐹 = (𝑥 ∈ (1...𝑉) ↦ (𝑦 ∈ (1...𝑊) ↦ (𝐻‘(𝑦 + (𝑊 · ((𝑥 − 1) + 𝑉))))))    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝐺:(1...𝑊)⟶𝑅)    &   (𝜑𝐾 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℕ)    &   (𝜑𝐷 ∈ ℕ)    &   (𝜑 → (𝐴(AP‘𝐾)𝐷) ⊆ (𝐹 “ {𝐺}))    &   (𝜑𝐵 ∈ ℕ)    &   (𝜑𝐸:(1...𝑀)⟶ℕ)    &   (𝜑 → ∀𝑖 ∈ (1...𝑀)((𝐵 + (𝐸𝑖))(AP‘𝐾)(𝐸𝑖)) ⊆ (𝐺 “ {(𝐺‘(𝐵 + (𝐸𝑖)))}))    &   𝐽 = (𝑖 ∈ (1...𝑀) ↦ (𝐺‘(𝐵 + (𝐸𝑖))))    &   (𝜑 → (♯‘ran 𝐽) = 𝑀)    &   𝑇 = (𝐵 + (𝑊 · ((𝐴 + (𝑉𝐷)) − 1)))    &   𝑃 = (𝑗 ∈ (1...(𝑀 + 1)) ↦ (if(𝑗 = (𝑀 + 1), 0, (𝐸𝑗)) + (𝑊 · 𝐷)))       (𝜑 → (⟨(𝑀 + 1), 𝐾⟩ PolyAP 𝐻 ∨ (𝐾 + 1) MonoAP 𝐺))

Theoremvdwlem7 16178* Lemma for vdw 16185. (Contributed by Mario Carneiro, 12-Sep-2014.)
(𝜑𝑉 ∈ ℕ)    &   (𝜑𝑊 ∈ ℕ)    &   (𝜑𝑅 ∈ Fin)    &   (𝜑𝐻:(1...(𝑊 · (2 · 𝑉)))⟶𝑅)    &   𝐹 = (𝑥 ∈ (1...𝑉) ↦ (𝑦 ∈ (1...𝑊) ↦ (𝐻‘(𝑦 + (𝑊 · ((𝑥 − 1) + 𝑉))))))    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝐺:(1...𝑊)⟶𝑅)    &   (𝜑𝐾 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℕ)    &   (𝜑𝐷 ∈ ℕ)    &   (𝜑 → (𝐴(AP‘𝐾)𝐷) ⊆ (𝐹 “ {𝐺}))       (𝜑 → (⟨𝑀, 𝐾⟩ PolyAP 𝐺 → (⟨(𝑀 + 1), 𝐾⟩ PolyAP 𝐻 ∨ (𝐾 + 1) MonoAP 𝐺)))

Theoremvdwlem8 16179* Lemma for vdw 16185. (Contributed by Mario Carneiro, 18-Aug-2014.)
(𝜑𝑅 ∈ Fin)    &   (𝜑𝐾 ∈ (ℤ‘2))    &   (𝜑𝑊 ∈ ℕ)    &   (𝜑𝐹:(1...(2 · 𝑊))⟶𝑅)    &   𝐶 ∈ V    &   (𝜑𝐴 ∈ ℕ)    &   (𝜑𝐷 ∈ ℕ)    &   (𝜑 → (𝐴(AP‘𝐾)𝐷) ⊆ (𝐺 “ {𝐶}))    &   𝐺 = (𝑥 ∈ (1...𝑊) ↦ (𝐹‘(𝑥 + 𝑊)))       (𝜑 → ⟨1, 𝐾⟩ PolyAP 𝐹)

Theoremvdwlem9 16180* Lemma for vdw 16185. (Contributed by Mario Carneiro, 12-Sep-2014.)
(𝜑𝑅 ∈ Fin)    &   (𝜑𝐾 ∈ (ℤ‘2))    &   (𝜑 → ∀𝑠 ∈ Fin ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑠𝑚 (1...𝑛))𝐾 MonoAP 𝑓)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑊 ∈ ℕ)    &   (𝜑 → ∀𝑔 ∈ (𝑅𝑚 (1...𝑊))(⟨𝑀, 𝐾⟩ PolyAP 𝑔 ∨ (𝐾 + 1) MonoAP 𝑔))    &   (𝜑𝑉 ∈ ℕ)    &   (𝜑 → ∀𝑓 ∈ ((𝑅𝑚 (1...𝑊)) ↑𝑚 (1...𝑉))𝐾 MonoAP 𝑓)    &   (𝜑𝐻:(1...(𝑊 · (2 · 𝑉)))⟶𝑅)    &   𝐹 = (𝑥 ∈ (1...𝑉) ↦ (𝑦 ∈ (1...𝑊) ↦ (𝐻‘(𝑦 + (𝑊 · ((𝑥 − 1) + 𝑉))))))       (𝜑 → (⟨(𝑀 + 1), 𝐾⟩ PolyAP 𝐻 ∨ (𝐾 + 1) MonoAP 𝐻))

Theoremvdwlem10 16181* Lemma for vdw 16185. Set up secondary induction on 𝑀. (Contributed by Mario Carneiro, 18-Aug-2014.)
(𝜑𝑅 ∈ Fin)    &   (𝜑𝐾 ∈ (ℤ‘2))    &   (𝜑 → ∀𝑠 ∈ Fin ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑠𝑚 (1...𝑛))𝐾 MonoAP 𝑓)    &   (𝜑𝑀 ∈ ℕ)       (𝜑 → ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑅𝑚 (1...𝑛))(⟨𝑀, 𝐾⟩ PolyAP 𝑓 ∨ (𝐾 + 1) MonoAP 𝑓))

Theoremvdwlem11 16182* Lemma for vdw 16185. (Contributed by Mario Carneiro, 18-Aug-2014.)
(𝜑𝑅 ∈ Fin)    &   (𝜑𝐾 ∈ (ℤ‘2))    &   (𝜑 → ∀𝑠 ∈ Fin ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑠𝑚 (1...𝑛))𝐾 MonoAP 𝑓)       (𝜑 → ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑅𝑚 (1...𝑛))(𝐾 + 1) MonoAP 𝑓)

Theoremvdwlem12 16183 Lemma for vdw 16185. 𝐾 = 2 base case of induction. (Contributed by Mario Carneiro, 18-Aug-2014.)
(𝜑𝑅 ∈ Fin)    &   (𝜑𝐹:(1...((♯‘𝑅) + 1))⟶𝑅)    &   (𝜑 → ¬ 2 MonoAP 𝐹)        ¬ 𝜑

Theoremvdwlem13 16184* Lemma for vdw 16185. Main induction on 𝐾; 𝐾 = 0, 𝐾 = 1 base cases. (Contributed by Mario Carneiro, 18-Aug-2014.)
(𝜑𝑅 ∈ Fin)    &   (𝜑𝐾 ∈ ℕ0)       (𝜑 → ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑅𝑚 (1...𝑛))𝐾 MonoAP 𝑓)

Theoremvdw 16185* Van der Waerden's theorem. For any finite coloring 𝑅 and integer 𝐾, there is an 𝑁 such that every coloring function from 1...𝑁 to 𝑅 contains a monochromatic arithmetic progression (which written out in full means that there is a color 𝑐 and base, increment values 𝑎, 𝑑 such that all the numbers 𝑎, 𝑎 + 𝑑, ..., 𝑎 + (𝑘 − 1)𝑑 lie in the preimage of {𝑐}, i.e. they are all in 1...𝑁 and 𝑓 evaluated at each one yields 𝑐). (Contributed by Mario Carneiro, 13-Sep-2014.)
((𝑅 ∈ Fin ∧ 𝐾 ∈ ℕ0) → ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑅𝑚 (1...𝑛))∃𝑐𝑅𝑎 ∈ ℕ ∃𝑑 ∈ ℕ ∀𝑚 ∈ (0...(𝐾 − 1))(𝑎 + (𝑚 · 𝑑)) ∈ (𝑓 “ {𝑐}))

Theoremvdwnnlem1 16186* Corollary of vdw 16185, and lemma for vdwnn 16189. If 𝐹 is a coloring of the integers, then there are arbitrarily long monochromatic APs in 𝐹. (Contributed by Mario Carneiro, 13-Sep-2014.)
((𝑅 ∈ Fin ∧ 𝐹:ℕ⟶𝑅𝐾 ∈ ℕ0) → ∃𝑐𝑅𝑎 ∈ ℕ ∃𝑑 ∈ ℕ ∀𝑚 ∈ (0...(𝐾 − 1))(𝑎 + (𝑚 · 𝑑)) ∈ (𝐹 “ {𝑐}))

Theoremvdwnnlem2 16187* Lemma for vdwnn 16189. The set of all "bad" 𝑘 for the theorem is upwards-closed, because a long AP implies a short AP. (Contributed by Mario Carneiro, 13-Sep-2014.)
(𝜑𝑅 ∈ Fin)    &   (𝜑𝐹:ℕ⟶𝑅)    &   𝑆 = {𝑘 ∈ ℕ ∣ ¬ ∃𝑎 ∈ ℕ ∃𝑑 ∈ ℕ ∀𝑚 ∈ (0...(𝑘 − 1))(𝑎 + (𝑚 · 𝑑)) ∈ (𝐹 “ {𝑐})}       ((𝜑𝐵 ∈ (ℤ𝐴)) → (𝐴𝑆𝐵𝑆))

Theoremvdwnnlem3 16188* Lemma for vdwnn 16189. (Contributed by Mario Carneiro, 13-Sep-2014.) (Proof shortened by AV, 27-Sep-2020.)
(𝜑𝑅 ∈ Fin)    &   (𝜑𝐹:ℕ⟶𝑅)    &   𝑆 = {𝑘 ∈ ℕ ∣ ¬ ∃𝑎 ∈ ℕ ∃𝑑 ∈ ℕ ∀𝑚 ∈ (0...(𝑘 − 1))(𝑎 + (𝑚 · 𝑑)) ∈ (𝐹 “ {𝑐})}    &   (𝜑 → ∀𝑐𝑅 𝑆 ≠ ∅)        ¬ 𝜑

Theoremvdwnn 16189* Van der Waerden's theorem, infinitary version. For any finite coloring 𝐹 of the positive integers, there is a color 𝑐 that contains arbitrarily long arithmetic progressions. (Contributed by Mario Carneiro, 13-Sep-2014.)
((𝑅 ∈ Fin ∧ 𝐹:ℕ⟶𝑅) → ∃𝑐𝑅𝑘 ∈ ℕ ∃𝑎 ∈ ℕ ∃𝑑 ∈ ℕ ∀𝑚 ∈ (0...(𝑘 − 1))(𝑎 + (𝑚 · 𝑑)) ∈ (𝐹 “ {𝑐}))

6.2.14  Ramsey's theorem

Syntaxcram 16190 Extend class notation with the Ramsey number function.
class Ramsey

Theoremramtlecl 16191* The set 𝑇 of numbers with the Ramsey number property is upward-closed. (Contributed by Mario Carneiro, 21-Apr-2015.)
𝑇 = {𝑛 ∈ ℕ0 ∣ ∀𝑠(𝑛 ≤ (♯‘𝑠) → 𝜑)}       (𝑀𝑇 → (ℤ𝑀) ⊆ 𝑇)

Definitiondf-ram 16192* Define the Ramsey number function. The input is a number 𝑚 for the size of the edges of the hypergraph, and a tuple 𝑟 from the finite color set to lower bounds for each color. The Ramsey number (𝑀 Ramsey 𝑅) is the smallest number such that for any set 𝑆 with (𝑀 Ramsey 𝑅) ≤ ♯𝑆 and any coloring 𝐹 of the set of 𝑀-element subsets of 𝑆 (with color set dom 𝑅), there is a color 𝑐 ∈ dom 𝑅 and a subset 𝑥𝑆 such that 𝑅(𝑐) ≤ ♯𝑥 and all the hyperedges of 𝑥 (that is, subsets of 𝑥 of size 𝑀) have color 𝑐. (Contributed by Mario Carneiro, 20-Apr-2015.) (Revised by AV, 14-Sep-2020.)
Ramsey = (𝑚 ∈ ℕ0, 𝑟 ∈ V ↦ inf({𝑛 ∈ ℕ0 ∣ ∀𝑠(𝑛 ≤ (♯‘𝑠) → ∀𝑓 ∈ (dom 𝑟𝑚 {𝑦 ∈ 𝒫 𝑠 ∣ (♯‘𝑦) = 𝑚})∃𝑐 ∈ dom 𝑟𝑥 ∈ 𝒫 𝑠((𝑟𝑐) ≤ (♯‘𝑥) ∧ ∀𝑦 ∈ 𝒫 𝑥((♯‘𝑦) = 𝑚 → (𝑓𝑦) = 𝑐)))}, ℝ*, < ))

Theoremhashbcval 16193* Value of the "binomial set", the set of all 𝑁-element subsets of 𝐴. (Contributed by Mario Carneiro, 20-Apr-2015.)
𝐶 = (𝑎 ∈ V, 𝑖 ∈ ℕ0 ↦ {𝑏 ∈ 𝒫 𝑎 ∣ (♯‘𝑏) = 𝑖})       ((𝐴𝑉𝑁 ∈ ℕ0) → (𝐴𝐶𝑁) = {𝑥 ∈ 𝒫 𝐴 ∣ (♯‘𝑥) = 𝑁})

Theoremhashbccl 16194* The binomial set is a finite set. (Contributed by Mario Carneiro, 20-Apr-2015.)
𝐶 = (𝑎 ∈ V, 𝑖 ∈ ℕ0 ↦ {𝑏 ∈ 𝒫 𝑎 ∣ (♯‘𝑏) = 𝑖})       ((𝐴 ∈ Fin ∧ 𝑁 ∈ ℕ0) → (𝐴𝐶𝑁) ∈ Fin)

Theoremhashbcss 16195* Subset relation for the binomial set. (Contributed by Mario Carneiro, 20-Apr-2015.)
𝐶 = (𝑎 ∈ V, 𝑖 ∈ ℕ0 ↦ {𝑏 ∈ 𝒫 𝑎 ∣ (♯‘𝑏) = 𝑖})       ((𝐴𝑉𝐵𝐴𝑁 ∈ ℕ0) → (𝐵𝐶𝑁) ⊆ (𝐴𝐶𝑁))

Theoremhashbc0 16196* The set of subsets of size zero is the singleton of the empty set. (Contributed by Mario Carneiro, 22-Apr-2015.)
𝐶 = (𝑎 ∈ V, 𝑖 ∈ ℕ0 ↦ {𝑏 ∈ 𝒫 𝑎 ∣ (♯‘𝑏) = 𝑖})       (𝐴𝑉 → (𝐴𝐶0) = {∅})

Theoremhashbc2 16197* The size of the binomial set is the binomial coefficient. (Contributed by Mario Carneiro, 20-Apr-2015.)
𝐶 = (𝑎 ∈ V, 𝑖 ∈ ℕ0 ↦ {𝑏 ∈ 𝒫 𝑎 ∣ (♯‘𝑏) = 𝑖})       ((𝐴 ∈ Fin ∧ 𝑁 ∈ ℕ0) → (♯‘(𝐴𝐶𝑁)) = ((♯‘𝐴)C𝑁))

Theorem0hashbc 16198* There are no subsets of the empty set with size greater than zero. (Contributed by Mario Carneiro, 22-Apr-2015.)
𝐶 = (𝑎 ∈ V, 𝑖 ∈ ℕ0 ↦ {𝑏 ∈ 𝒫 𝑎 ∣ (♯‘𝑏) = 𝑖})       (𝑁 ∈ ℕ → (∅𝐶𝑁) = ∅)

Theoremramval 16199* The value of the Ramsey number function. (Contributed by Mario Carneiro, 21-Apr-2015.) (Revised by AV, 14-Sep-2020.)
𝐶 = (𝑎 ∈ V, 𝑖 ∈ ℕ0 ↦ {𝑏 ∈ 𝒫 𝑎 ∣ (♯‘𝑏) = 𝑖})    &   𝑇 = {𝑛 ∈ ℕ0 ∣ ∀𝑠(𝑛 ≤ (♯‘𝑠) → ∀𝑓 ∈ (𝑅𝑚 (𝑠𝐶𝑀))∃𝑐𝑅𝑥 ∈ 𝒫 𝑠((𝐹𝑐) ≤ (♯‘𝑥) ∧ (𝑥𝐶𝑀) ⊆ (𝑓 “ {𝑐})))}       ((𝑀 ∈ ℕ0𝑅𝑉𝐹:𝑅⟶ℕ0) → (𝑀 Ramsey 𝐹) = inf(𝑇, ℝ*, < ))

Theoremramcl2lem 16200* Lemma for extended real closure of the Ramsey number function. (Contributed by Mario Carneiro, 20-Apr-2015.) (Revised by AV, 14-Sep-2020.)
𝐶 = (𝑎 ∈ V, 𝑖 ∈ ℕ0 ↦ {𝑏 ∈ 𝒫 𝑎 ∣ (♯‘𝑏) = 𝑖})    &   𝑇 = {𝑛 ∈ ℕ0 ∣ ∀𝑠(𝑛 ≤ (♯‘𝑠) → ∀𝑓 ∈ (𝑅𝑚 (𝑠𝐶𝑀))∃𝑐𝑅𝑥 ∈ 𝒫 𝑠((𝐹𝑐) ≤ (♯‘𝑥) ∧ (𝑥𝐶𝑀) ⊆ (𝑓 “ {𝑐})))}       ((𝑀 ∈ ℕ0𝑅𝑉𝐹:𝑅⟶ℕ0) → (𝑀 Ramsey 𝐹) = if(𝑇 = ∅, +∞, inf(𝑇, ℝ, < )))

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