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Type | Label | Description |
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Statement | ||
Theorem | pcmpt2 16001* | Dividing two prime count maps yields a number with all dividing primes confined to an interval. (Contributed by Mario Carneiro, 14-Mar-2014.) |
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑𝐴), 1)) & ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0) & ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ (𝜑 → 𝑃 ∈ ℙ) & ⊢ (𝑛 = 𝑃 → 𝐴 = 𝐵) & ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘𝑁)) ⇒ ⊢ (𝜑 → (𝑃 pCnt ((seq1( · , 𝐹)‘𝑀) / (seq1( · , 𝐹)‘𝑁))) = if((𝑃 ≤ 𝑀 ∧ ¬ 𝑃 ≤ 𝑁), 𝐵, 0)) | ||
Theorem | pcmptdvds 16002 | The partial products of the prime power map form a divisibility chain. (Contributed by Mario Carneiro, 12-Mar-2014.) |
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑𝐴), 1)) & ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0) & ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘𝑁)) ⇒ ⊢ (𝜑 → (seq1( · , 𝐹)‘𝑁) ∥ (seq1( · , 𝐹)‘𝑀)) | ||
Theorem | pcprod 16003* | The product of the primes taken to their respective powers reconstructs the original number. (Contributed by Mario Carneiro, 12-Mar-2014.) |
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑(𝑛 pCnt 𝑁)), 1)) ⇒ ⊢ (𝑁 ∈ ℕ → (seq1( · , 𝐹)‘𝑁) = 𝑁) | ||
Theorem | sumhash 16004* | The sum of 1 over a set is the size of the set. (Contributed by Mario Carneiro, 8-Mar-2014.) (Revised by Mario Carneiro, 20-May-2014.) |
⊢ ((𝐵 ∈ Fin ∧ 𝐴 ⊆ 𝐵) → Σ𝑘 ∈ 𝐵 if(𝑘 ∈ 𝐴, 1, 0) = (♯‘𝐴)) | ||
Theorem | fldivp1 16005 | The difference between the floors of adjacent fractions is either 1 or 0. (Contributed by Mario Carneiro, 8-Mar-2014.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → ((⌊‘((𝑀 + 1) / 𝑁)) − (⌊‘(𝑀 / 𝑁))) = if(𝑁 ∥ (𝑀 + 1), 1, 0)) | ||
Theorem | pcfaclem 16006 | Lemma for pcfac 16007. (Contributed by Mario Carneiro, 20-May-2014.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝑀 ∈ (ℤ≥‘𝑁) ∧ 𝑃 ∈ ℙ) → (⌊‘(𝑁 / (𝑃↑𝑀))) = 0) | ||
Theorem | pcfac 16007* | Calculate the prime count of a factorial. (Contributed by Mario Carneiro, 11-Mar-2014.) (Revised by Mario Carneiro, 21-May-2014.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝑀 ∈ (ℤ≥‘𝑁) ∧ 𝑃 ∈ ℙ) → (𝑃 pCnt (!‘𝑁)) = Σ𝑘 ∈ (1...𝑀)(⌊‘(𝑁 / (𝑃↑𝑘)))) | ||
Theorem | pcbc 16008* | Calculate the prime count of a binomial coefficient. (Contributed by Mario Carneiro, 11-Mar-2014.) (Revised by Mario Carneiro, 21-May-2014.) |
⊢ ((𝑁 ∈ ℕ ∧ 𝐾 ∈ (0...𝑁) ∧ 𝑃 ∈ ℙ) → (𝑃 pCnt (𝑁C𝐾)) = Σ𝑘 ∈ (1...𝑁)((⌊‘(𝑁 / (𝑃↑𝑘))) − ((⌊‘((𝑁 − 𝐾) / (𝑃↑𝑘))) + (⌊‘(𝐾 / (𝑃↑𝑘)))))) | ||
Theorem | qexpz 16009 | If a power of a rational number is an integer, then the number is an integer. In other words, all n-th roots are irrational unless they are integers (so that the original number is an n-th power). (Contributed by Mario Carneiro, 10-Aug-2015.) |
⊢ ((𝐴 ∈ ℚ ∧ 𝑁 ∈ ℕ ∧ (𝐴↑𝑁) ∈ ℤ) → 𝐴 ∈ ℤ) | ||
Theorem | expnprm 16010 | A second or higher power of a rational number is not a prime number. Or by contraposition, the n-th root of a prime number is irrational. Suggested by Norm Megill. (Contributed by Mario Carneiro, 10-Aug-2015.) |
⊢ ((𝐴 ∈ ℚ ∧ 𝑁 ∈ (ℤ≥‘2)) → ¬ (𝐴↑𝑁) ∈ ℙ) | ||
Theorem | oddprmdvds 16011* | Every positive integer which is not a power of two is divisible by an odd prime number. (Contributed by AV, 6-Aug-2021.) |
⊢ ((𝐾 ∈ ℕ ∧ ¬ ∃𝑛 ∈ ℕ0 𝐾 = (2↑𝑛)) → ∃𝑝 ∈ (ℙ ∖ {2})𝑝 ∥ 𝐾) | ||
Theorem | prmpwdvds 16012 | A relation involving divisibility by a prime power. (Contributed by Mario Carneiro, 2-Mar-2014.) |
⊢ (((𝐾 ∈ ℤ ∧ 𝐷 ∈ ℤ) ∧ (𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) ∧ (𝐷 ∥ (𝐾 · (𝑃↑𝑁)) ∧ ¬ 𝐷 ∥ (𝐾 · (𝑃↑(𝑁 − 1))))) → (𝑃↑𝑁) ∥ 𝐷) | ||
Theorem | pockthlem 16013 | Lemma for pockthg 16014. (Contributed by Mario Carneiro, 2-Mar-2014.) |
⊢ (𝜑 → 𝐴 ∈ ℕ) & ⊢ (𝜑 → 𝐵 ∈ ℕ) & ⊢ (𝜑 → 𝐵 < 𝐴) & ⊢ (𝜑 → 𝑁 = ((𝐴 · 𝐵) + 1)) & ⊢ (𝜑 → 𝑃 ∈ ℙ) & ⊢ (𝜑 → 𝑃 ∥ 𝑁) & ⊢ (𝜑 → 𝑄 ∈ ℙ) & ⊢ (𝜑 → (𝑄 pCnt 𝐴) ∈ ℕ) & ⊢ (𝜑 → 𝐶 ∈ ℤ) & ⊢ (𝜑 → ((𝐶↑(𝑁 − 1)) mod 𝑁) = 1) & ⊢ (𝜑 → (((𝐶↑((𝑁 − 1) / 𝑄)) − 1) gcd 𝑁) = 1) ⇒ ⊢ (𝜑 → (𝑄 pCnt 𝐴) ≤ (𝑄 pCnt (𝑃 − 1))) | ||
Theorem | pockthg 16014* | The generalized Pocklington's theorem. If 𝑁 − 1 = 𝐴 · 𝐵 where 𝐵 < 𝐴, then 𝑁 is prime if and only if for every prime factor 𝑝 of 𝐴, there is an 𝑥 such that 𝑥↑(𝑁 − 1) = 1( mod 𝑁) and gcd (𝑥↑((𝑁 − 1) / 𝑝) − 1, 𝑁) = 1. (Contributed by Mario Carneiro, 2-Mar-2014.) |
⊢ (𝜑 → 𝐴 ∈ ℕ) & ⊢ (𝜑 → 𝐵 ∈ ℕ) & ⊢ (𝜑 → 𝐵 < 𝐴) & ⊢ (𝜑 → 𝑁 = ((𝐴 · 𝐵) + 1)) & ⊢ (𝜑 → ∀𝑝 ∈ ℙ (𝑝 ∥ 𝐴 → ∃𝑥 ∈ ℤ (((𝑥↑(𝑁 − 1)) mod 𝑁) = 1 ∧ (((𝑥↑((𝑁 − 1) / 𝑝)) − 1) gcd 𝑁) = 1))) ⇒ ⊢ (𝜑 → 𝑁 ∈ ℙ) | ||
Theorem | pockthi 16015 | Pocklington's theorem, which gives a sufficient criterion for a number 𝑁 to be prime. This is the preferred method for verifying large primes, being much more efficient to compute than trial division. This form has been optimized for application to specific large primes; see pockthg 16014 for a more general closed-form version. (Contributed by Mario Carneiro, 2-Mar-2014.) |
⊢ 𝑃 ∈ ℙ & ⊢ 𝐺 ∈ ℕ & ⊢ 𝑀 = (𝐺 · 𝑃) & ⊢ 𝑁 = (𝑀 + 1) & ⊢ 𝐷 ∈ ℕ & ⊢ 𝐸 ∈ ℕ & ⊢ 𝐴 ∈ ℕ & ⊢ 𝑀 = (𝐷 · (𝑃↑𝐸)) & ⊢ 𝐷 < (𝑃↑𝐸) & ⊢ ((𝐴↑𝑀) mod 𝑁) = (1 mod 𝑁) & ⊢ (((𝐴↑𝐺) − 1) gcd 𝑁) = 1 ⇒ ⊢ 𝑁 ∈ ℙ | ||
Theorem | unbenlem 16016* | Lemma for unben 16017. (Contributed by NM, 5-May-2005.) (Revised by Mario Carneiro, 15-Sep-2013.) |
⊢ 𝐺 = (rec((𝑥 ∈ V ↦ (𝑥 + 1)), 1) ↾ ω) ⇒ ⊢ ((𝐴 ⊆ ℕ ∧ ∀𝑚 ∈ ℕ ∃𝑛 ∈ 𝐴 𝑚 < 𝑛) → 𝐴 ≈ ω) | ||
Theorem | unben 16017* | An unbounded set of positive integers is infinite. (Contributed by NM, 5-May-2005.) (Revised by Mario Carneiro, 15-Sep-2013.) |
⊢ ((𝐴 ⊆ ℕ ∧ ∀𝑚 ∈ ℕ ∃𝑛 ∈ 𝐴 𝑚 < 𝑛) → 𝐴 ≈ ℕ) | ||
Theorem | infpnlem1 16018* | Lemma for infpn 16020. 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) → 𝑀 ≤ 𝑗)) → (𝑁 < 𝑀 ∧ ∀𝑗 ∈ ℕ ((𝑀 / 𝑗) ∈ ℕ → (𝑗 = 1 ∨ 𝑗 = 𝑀))))) | ||
Theorem | infpnlem2 16019* | Lemma for infpn 16020. 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)))) | ||
Theorem | infpn 16020* | There exist infinitely many prime numbers: for any positive integer 𝑁, there exists a prime number 𝑗 greater than 𝑁. (See infpn2 16021 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)))) | ||
Theorem | infpn2 16021* | There exist infinitely many prime numbers: the set of all primes 𝑆 is unbounded by infpn 16020, so by unben 16017 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 | ||
Theorem | prmunb 16022* | The primes are unbounded. (Contributed by Paul Chapman, 28-Nov-2012.) |
⊢ (𝑁 ∈ ℕ → ∃𝑝 ∈ ℙ 𝑁 < 𝑝) | ||
Theorem | prminf 16023 | There are an infinite number of primes. Theorem 1.7 in [ApostolNT] p. 16. (Contributed by Paul Chapman, 28-Nov-2012.) |
⊢ ℙ ≈ ℕ | ||
Theorem | prmreclem1 16024* | Lemma for prmrec 16030. 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 → ((𝑄‘𝑁) ∈ ℕ ∧ ((𝑄‘𝑁)↑2) ∥ 𝑁 ∧ (𝐾 ∈ (ℤ≥‘2) → ¬ (𝐾↑2) ∥ (𝑁 / ((𝑄‘𝑁)↑2))))) | ||
Theorem | prmreclem2 16025* | Lemma for prmrec 16030. 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)) & ⊢ (𝜑 → 𝐾 ∈ ℕ) & ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ 𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝 ∥ 𝑛} & ⊢ 𝑄 = (𝑛 ∈ ℕ ↦ sup({𝑟 ∈ ℕ ∣ (𝑟↑2) ∥ 𝑛}, ℝ, < )) ⇒ ⊢ (𝜑 → (♯‘{𝑥 ∈ 𝑀 ∣ (𝑄‘𝑥) = 1}) ≤ (2↑𝐾)) | ||
Theorem | prmreclem3 16026* | Lemma for prmrec 16030. The main inequality established here is ♯𝑀 ≤ ♯{𝑥 ∈ 𝑀 ∣ (𝑄‘𝑥) = 1} · √𝑁, where {𝑥 ∈ 𝑀 ∣ (𝑄‘𝑥) = 1} is the set of squarefree numbers in 𝑀. This is demonstrated by the map 𝑦 ↦ 〈𝑦 / (𝑄‘𝑦)↑2, (𝑄‘𝑦)〉 where 𝑄‘𝑦 is the largest number whose square divides 𝑦. (Contributed by Mario Carneiro, 5-Aug-2014.) |
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (1 / 𝑛), 0)) & ⊢ (𝜑 → 𝐾 ∈ ℕ) & ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ 𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝 ∥ 𝑛} & ⊢ 𝑄 = (𝑛 ∈ ℕ ↦ sup({𝑟 ∈ ℕ ∣ (𝑟↑2) ∥ 𝑛}, ℝ, < )) ⇒ ⊢ (𝜑 → (♯‘𝑀) ≤ ((2↑𝐾) · (√‘𝑁))) | ||
Theorem | prmreclem4 16027* | Lemma for prmrec 16030. Show by induction that the indexed (nondisjoint) union 𝑊‘𝑘 is at most the size of the prime reciprocal series. The key counting lemma is hashdvds 15884, 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)) & ⊢ (𝜑 → 𝐾 ∈ ℕ) & ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ 𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝 ∥ 𝑛} & ⊢ (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 ∧ 𝑝 ∥ 𝑛)}) ⇒ ⊢ (𝜑 → (𝑁 ∈ (ℤ≥‘𝐾) → (♯‘∪ 𝑘 ∈ ((𝐾 + 1)...𝑁)(𝑊‘𝑘)) ≤ (𝑁 · Σ𝑘 ∈ ((𝐾 + 1)...𝑁)if(𝑘 ∈ ℙ, (1 / 𝑘), 0)))) | ||
Theorem | prmreclem5 16028* | Lemma for prmrec 16030. 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 𝑊‘𝑘 that divide the prime 𝑘. By prmreclem4 16027 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)) & ⊢ (𝜑 → 𝐾 ∈ ℕ) & ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ 𝑀 = {𝑛 ∈ (1...𝑁) ∣ ∀𝑝 ∈ (ℙ ∖ (1...𝐾)) ¬ 𝑝 ∥ 𝑛} & ⊢ (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 ∧ 𝑝 ∥ 𝑛)}) ⇒ ⊢ (𝜑 → (𝑁 / 2) < ((2↑𝐾) · (√‘𝑁))) | ||
Theorem | prmreclem6 16029* | Lemma for prmrec 16030. 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 16028 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 ⇝ | ||
Theorem | prmrec 16030* | 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 ⇝ | ||
Theorem | 1arithlem1 16031* | Lemma for 1arith 16035. (Contributed by Mario Carneiro, 30-May-2014.) |
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛))) ⇒ ⊢ (𝑁 ∈ ℕ → (𝑀‘𝑁) = (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑁))) | ||
Theorem | 1arithlem2 16032* | Lemma for 1arith 16035. (Contributed by Mario Carneiro, 30-May-2014.) |
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛))) ⇒ ⊢ ((𝑁 ∈ ℕ ∧ 𝑃 ∈ ℙ) → ((𝑀‘𝑁)‘𝑃) = (𝑃 pCnt 𝑁)) | ||
Theorem | 1arithlem3 16033* | Lemma for 1arith 16035. (Contributed by Mario Carneiro, 30-May-2014.) |
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛))) ⇒ ⊢ (𝑁 ∈ ℕ → (𝑀‘𝑁):ℙ⟶ℕ0) | ||
Theorem | 1arithlem4 16034* | Lemma for 1arith 16035. (Contributed by Mario Carneiro, 30-May-2014.) |
⊢ 𝑀 = (𝑛 ∈ ℕ ↦ (𝑝 ∈ ℙ ↦ (𝑝 pCnt 𝑛))) & ⊢ 𝐺 = (𝑦 ∈ ℕ ↦ if(𝑦 ∈ ℙ, (𝑦↑(𝐹‘𝑦)), 1)) & ⊢ (𝜑 → 𝐹:ℙ⟶ℕ0) & ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ ((𝜑 ∧ (𝑞 ∈ ℙ ∧ 𝑁 ≤ 𝑞)) → (𝐹‘𝑞) = 0) ⇒ ⊢ (𝜑 → ∃𝑥 ∈ ℕ 𝐹 = (𝑀‘𝑥)) | ||
Theorem | 1arith 16035* | 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 ∈ (ℕ0 ↑𝑚 ℙ) ∣ (◡𝑒 “ ℕ) ∈ Fin} ⇒ ⊢ 𝑀:ℕ–1-1-onto→𝑅 | ||
Theorem | 1arith2 16036* | 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 ∈ (ℕ0 ↑𝑚 ℙ) ∣ (◡𝑒 “ ℕ) ∈ Fin} ⇒ ⊢ ∀𝑧 ∈ ℕ ∃!𝑔 ∈ 𝑅 (𝑀‘𝑧) = 𝑔 | ||
Syntax | cgz 16037 | Extend class notation with the set of gaussian integers. |
class ℤ[i] | ||
Definition | df-gz 16038 | 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)} | ||
Theorem | elgz 16039 | Elementhood in the gaussian integers. (Contributed by Mario Carneiro, 14-Jul-2014.) |
⊢ (𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ)) | ||
Theorem | gzcn 16040 | A gaussian integer is a complex number. (Contributed by Mario Carneiro, 14-Jul-2014.) |
⊢ (𝐴 ∈ ℤ[i] → 𝐴 ∈ ℂ) | ||
Theorem | zgz 16041 | An integer is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.) |
⊢ (𝐴 ∈ ℤ → 𝐴 ∈ ℤ[i]) | ||
Theorem | igz 16042 | i is a gaussian integer. (Contributed by Mario Carneiro, 14-Jul-2014.) |
⊢ i ∈ ℤ[i] | ||
Theorem | gznegcl 16043 | The gaussian integers are closed under negation. (Contributed by Mario Carneiro, 14-Jul-2014.) |
⊢ (𝐴 ∈ ℤ[i] → -𝐴 ∈ ℤ[i]) | ||
Theorem | gzcjcl 16044 | The gaussian integers are closed under conjugation. (Contributed by Mario Carneiro, 14-Jul-2014.) |
⊢ (𝐴 ∈ ℤ[i] → (∗‘𝐴) ∈ ℤ[i]) | ||
Theorem | gzaddcl 16045 | 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 16046 | 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 16047 | 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 16048 | 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 16049 | 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 16050 | Lemma for 4sq 16072. (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 16051 | Lemma for 4sq 16072. (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 16052 | Lemma for 4sq 16072. (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 16053 | Lemma for 4sq 16072. (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 16054 | Lemma for 4sq 16072. (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 16055 | Lemma for 4sq 16072. (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 16056* | Lemma for 4sq 16072. 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 16057* | Lemma for 4sq 16072. 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 16058* | Lemma for 4sq 16072. 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 16059* | Lemma for 4sqlem4 16060. (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 16060* | Lemma for 4sq 16072. 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 16061* | Lemma for mul4sq 16062: algebraic manipulations. The extra assumptions involving 𝑀 are for a part of 4sqlem17 16069 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)))} & ⊢ (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 16062* | 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 16061. (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 | 4sqlem11 16063* | Lemma for 4sq 16072. 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)))} & ⊢ (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 16064* | Lemma for 4sq 16072. 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 | 4sqlem13 16065* | Lemma for 4sq 16072. (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 → (𝑇 ≠ ∅ ∧ 𝑀 < 𝑃)) | ||
Theorem | 4sqlem14 16066* | Lemma for 4sq 16072. (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 16067* | Lemma for 4sq 16072. (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 16068* | Lemma for 4sq 16072. (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 16069* | Lemma for 4sq 16072. (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 16070* | Lemma for 4sq 16072. 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 16071* | Lemma for 4sq 16072. 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 16070. 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 16062 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 16072* | 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)))) | ||
Syntax | cvdwa 16073 | The arithmetic progression function. |
class AP | ||
Syntax | cvdwm 16074 | The monochromatic arithmetic progression predicate. |
class MonoAP | ||
Syntax | cvdwp 16075 | The polychromatic arithmetic progression predicate. |
class PolyAP | ||
Definition | df-vdwap 16076* | 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))))) | ||
Definition | df-vdwmc 16077* | Define the "contains a monochromatic AP" predicate. (Contributed by Mario Carneiro, 18-Aug-2014.) |
⊢ MonoAP = {〈𝑘, 𝑓〉 ∣ ∃𝑐(ran (AP‘𝑘) ∩ 𝒫 (◡𝑓 “ {𝑐})) ≠ ∅} | ||
Definition | df-vdwpc 16078* | Define the "contains a polychromatic collection of APs" predicate. See vdwpc 16088 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 + (𝑑‘𝑖))(AP‘𝑘)(𝑑‘𝑖)) ⊆ (◡𝑓 “ {(𝑓‘(𝑎 + (𝑑‘𝑖)))}) ∧ (♯‘ran (𝑖 ∈ (1...𝑚) ↦ (𝑓‘(𝑎 + (𝑑‘𝑖))))) = 𝑚)} | ||
Theorem | vdwapfval 16079* | 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))))) | ||
Theorem | vdwapf 16080 | The arithmetic progression function is a function. (Contributed by Mario Carneiro, 18-Aug-2014.) |
⊢ (𝐾 ∈ ℕ0 → (AP‘𝐾):(ℕ × ℕ)⟶𝒫 ℕ) | ||
Theorem | vdwapval 16081* | Value of the arithmetic progression function. (Contributed by Mario Carneiro, 18-Aug-2014.) |
⊢ ((𝐾 ∈ ℕ0 ∧ 𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝑋 ∈ (𝐴(AP‘𝐾)𝐷) ↔ ∃𝑚 ∈ (0...(𝐾 − 1))𝑋 = (𝐴 + (𝑚 · 𝐷)))) | ||
Theorem | vdwapun 16082 | Remove the first element of an arithmetic progression. (Contributed by Mario Carneiro, 11-Sep-2014.) |
⊢ ((𝐾 ∈ ℕ0 ∧ 𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝐴(AP‘(𝐾 + 1))𝐷) = ({𝐴} ∪ ((𝐴 + 𝐷)(AP‘𝐾)𝐷))) | ||
Theorem | vdwapid1 16083 | 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)) | ||
Theorem | vdwap0 16084 | Value of a length-1 arithmetic progression. (Contributed by Mario Carneiro, 18-Aug-2014.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝐴(AP‘0)𝐷) = ∅) | ||
Theorem | vdwap1 16085 | Value of a length-1 arithmetic progression. (Contributed by Mario Carneiro, 18-Aug-2014.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐷 ∈ ℕ) → (𝐴(AP‘1)𝐷) = {𝐴}) | ||
Theorem | vdwmc 16086* | The predicate " The 〈𝑅, 𝑁〉-coloring 𝐹 contains a monochromatic AP of length 𝐾". (Contributed by Mario Carneiro, 18-Aug-2014.) |
⊢ 𝑋 ∈ V & ⊢ (𝜑 → 𝐾 ∈ ℕ0) & ⊢ (𝜑 → 𝐹:𝑋⟶𝑅) ⇒ ⊢ (𝜑 → (𝐾 MonoAP 𝐹 ↔ ∃𝑐∃𝑎 ∈ ℕ ∃𝑑 ∈ ℕ (𝑎(AP‘𝐾)𝑑) ⊆ (◡𝐹 “ {𝑐}))) | ||
Theorem | vdwmc2 16087* | Expand out the definition of an arithmetic progression. (Contributed by Mario Carneiro, 18-Aug-2014.) |
⊢ 𝑋 ∈ V & ⊢ (𝜑 → 𝐾 ∈ ℕ0) & ⊢ (𝜑 → 𝐹:𝑋⟶𝑅) & ⊢ (𝜑 → 𝐴 ∈ 𝑋) ⇒ ⊢ (𝜑 → (𝐾 MonoAP 𝐹 ↔ ∃𝑐 ∈ 𝑅 ∃𝑎 ∈ ℕ ∃𝑑 ∈ ℕ ∀𝑚 ∈ (0...(𝐾 − 1))(𝑎 + (𝑚 · 𝑑)) ∈ (◡𝐹 “ {𝑐}))) | ||
Theorem | vdwpc 16088* | 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 & ⊢ (𝜑 → 𝐾 ∈ ℕ0) & ⊢ (𝜑 → 𝐹:𝑋⟶𝑅) & ⊢ (𝜑 → 𝑀 ∈ ℕ) & ⊢ 𝐽 = (1...𝑀) ⇒ ⊢ (𝜑 → (〈𝑀, 𝐾〉 PolyAP 𝐹 ↔ ∃𝑎 ∈ ℕ ∃𝑑 ∈ (ℕ ↑𝑚 𝐽)(∀𝑖 ∈ 𝐽 ((𝑎 + (𝑑‘𝑖))(AP‘𝐾)(𝑑‘𝑖)) ⊆ (◡𝐹 “ {(𝐹‘(𝑎 + (𝑑‘𝑖)))}) ∧ (♯‘ran (𝑖 ∈ 𝐽 ↦ (𝐹‘(𝑎 + (𝑑‘𝑖))))) = 𝑀))) | ||
Theorem | vdwlem1 16089* | Lemma for vdw 16102. (Contributed by Mario Carneiro, 12-Sep-2014.) |
⊢ (𝜑 → 𝑅 ∈ Fin) & ⊢ (𝜑 → 𝐾 ∈ ℕ) & ⊢ (𝜑 → 𝑊 ∈ ℕ) & ⊢ (𝜑 → 𝐹:(1...𝑊)⟶𝑅) & ⊢ (𝜑 → 𝐴 ∈ ℕ) & ⊢ (𝜑 → 𝑀 ∈ ℕ) & ⊢ (𝜑 → 𝐷:(1...𝑀)⟶ℕ) & ⊢ (𝜑 → ∀𝑖 ∈ (1...𝑀)((𝐴 + (𝐷‘𝑖))(AP‘𝐾)(𝐷‘𝑖)) ⊆ (◡𝐹 “ {(𝐹‘(𝐴 + (𝐷‘𝑖)))})) & ⊢ (𝜑 → 𝐼 ∈ (1...𝑀)) & ⊢ (𝜑 → (𝐹‘𝐴) = (𝐹‘(𝐴 + (𝐷‘𝐼)))) ⇒ ⊢ (𝜑 → (𝐾 + 1) MonoAP 𝐹) | ||
Theorem | vdwlem2 16090* | Lemma for vdw 16102. (Contributed by Mario Carneiro, 12-Sep-2014.) |
⊢ (𝜑 → 𝑅 ∈ Fin) & ⊢ (𝜑 → 𝐾 ∈ ℕ0) & ⊢ (𝜑 → 𝑊 ∈ ℕ) & ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ (𝜑 → 𝐹:(1...𝑀)⟶𝑅) & ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘(𝑊 + 𝑁))) & ⊢ 𝐺 = (𝑥 ∈ (1...𝑊) ↦ (𝐹‘(𝑥 + 𝑁))) ⇒ ⊢ (𝜑 → (𝐾 MonoAP 𝐺 → 𝐾 MonoAP 𝐹)) | ||
Theorem | vdwlem3 16091 | Lemma for vdw 16102. (Contributed by Mario Carneiro, 13-Sep-2014.) |
⊢ (𝜑 → 𝑉 ∈ ℕ) & ⊢ (𝜑 → 𝑊 ∈ ℕ) & ⊢ (𝜑 → 𝐴 ∈ (1...𝑉)) & ⊢ (𝜑 → 𝐵 ∈ (1...𝑊)) ⇒ ⊢ (𝜑 → (𝐵 + (𝑊 · ((𝐴 − 1) + 𝑉))) ∈ (1...(𝑊 · (2 · 𝑉)))) | ||
Theorem | vdwlem4 16092* | Lemma for vdw 16102. (Contributed by Mario Carneiro, 12-Sep-2014.) |
⊢ (𝜑 → 𝑉 ∈ ℕ) & ⊢ (𝜑 → 𝑊 ∈ ℕ) & ⊢ (𝜑 → 𝑅 ∈ Fin) & ⊢ (𝜑 → 𝐻:(1...(𝑊 · (2 · 𝑉)))⟶𝑅) & ⊢ 𝐹 = (𝑥 ∈ (1...𝑉) ↦ (𝑦 ∈ (1...𝑊) ↦ (𝐻‘(𝑦 + (𝑊 · ((𝑥 − 1) + 𝑉)))))) ⇒ ⊢ (𝜑 → 𝐹:(1...𝑉)⟶(𝑅 ↑𝑚 (1...𝑊))) | ||
Theorem | vdwlem5 16093* | Lemma for vdw 16102. (Contributed by Mario Carneiro, 12-Sep-2014.) |
⊢ (𝜑 → 𝑉 ∈ ℕ) & ⊢ (𝜑 → 𝑊 ∈ ℕ) & ⊢ (𝜑 → 𝑅 ∈ Fin) & ⊢ (𝜑 → 𝐻:(1...(𝑊 · (2 · 𝑉)))⟶𝑅) & ⊢ 𝐹 = (𝑥 ∈ (1...𝑉) ↦ (𝑦 ∈ (1...𝑊) ↦ (𝐻‘(𝑦 + (𝑊 · ((𝑥 − 1) + 𝑉)))))) & ⊢ (𝜑 → 𝑀 ∈ ℕ) & ⊢ (𝜑 → 𝐺:(1...𝑊)⟶𝑅) & ⊢ (𝜑 → 𝐾 ∈ (ℤ≥‘2)) & ⊢ (𝜑 → 𝐴 ∈ ℕ) & ⊢ (𝜑 → 𝐷 ∈ ℕ) & ⊢ (𝜑 → (𝐴(AP‘𝐾)𝐷) ⊆ (◡𝐹 “ {𝐺})) & ⊢ (𝜑 → 𝐵 ∈ ℕ) & ⊢ (𝜑 → 𝐸:(1...𝑀)⟶ℕ) & ⊢ (𝜑 → ∀𝑖 ∈ (1...𝑀)((𝐵 + (𝐸‘𝑖))(AP‘𝐾)(𝐸‘𝑖)) ⊆ (◡𝐺 “ {(𝐺‘(𝐵 + (𝐸‘𝑖)))})) & ⊢ 𝐽 = (𝑖 ∈ (1...𝑀) ↦ (𝐺‘(𝐵 + (𝐸‘𝑖)))) & ⊢ (𝜑 → (♯‘ran 𝐽) = 𝑀) & ⊢ 𝑇 = (𝐵 + (𝑊 · ((𝐴 + (𝑉 − 𝐷)) − 1))) & ⊢ 𝑃 = (𝑗 ∈ (1...(𝑀 + 1)) ↦ (if(𝑗 = (𝑀 + 1), 0, (𝐸‘𝑗)) + (𝑊 · 𝐷))) ⇒ ⊢ (𝜑 → 𝑇 ∈ ℕ) | ||
Theorem | vdwlem6 16094* | Lemma for vdw 16102. (Contributed by Mario Carneiro, 13-Sep-2014.) |
⊢ (𝜑 → 𝑉 ∈ ℕ) & ⊢ (𝜑 → 𝑊 ∈ ℕ) & ⊢ (𝜑 → 𝑅 ∈ Fin) & ⊢ (𝜑 → 𝐻:(1...(𝑊 · (2 · 𝑉)))⟶𝑅) & ⊢ 𝐹 = (𝑥 ∈ (1...𝑉) ↦ (𝑦 ∈ (1...𝑊) ↦ (𝐻‘(𝑦 + (𝑊 · ((𝑥 − 1) + 𝑉)))))) & ⊢ (𝜑 → 𝑀 ∈ ℕ) & ⊢ (𝜑 → 𝐺:(1...𝑊)⟶𝑅) & ⊢ (𝜑 → 𝐾 ∈ (ℤ≥‘2)) & ⊢ (𝜑 → 𝐴 ∈ ℕ) & ⊢ (𝜑 → 𝐷 ∈ ℕ) & ⊢ (𝜑 → (𝐴(AP‘𝐾)𝐷) ⊆ (◡𝐹 “ {𝐺})) & ⊢ (𝜑 → 𝐵 ∈ ℕ) & ⊢ (𝜑 → 𝐸:(1...𝑀)⟶ℕ) & ⊢ (𝜑 → ∀𝑖 ∈ (1...𝑀)((𝐵 + (𝐸‘𝑖))(AP‘𝐾)(𝐸‘𝑖)) ⊆ (◡𝐺 “ {(𝐺‘(𝐵 + (𝐸‘𝑖)))})) & ⊢ 𝐽 = (𝑖 ∈ (1...𝑀) ↦ (𝐺‘(𝐵 + (𝐸‘𝑖)))) & ⊢ (𝜑 → (♯‘ran 𝐽) = 𝑀) & ⊢ 𝑇 = (𝐵 + (𝑊 · ((𝐴 + (𝑉 − 𝐷)) − 1))) & ⊢ 𝑃 = (𝑗 ∈ (1...(𝑀 + 1)) ↦ (if(𝑗 = (𝑀 + 1), 0, (𝐸‘𝑗)) + (𝑊 · 𝐷))) ⇒ ⊢ (𝜑 → (〈(𝑀 + 1), 𝐾〉 PolyAP 𝐻 ∨ (𝐾 + 1) MonoAP 𝐺)) | ||
Theorem | vdwlem7 16095* | Lemma for vdw 16102. (Contributed by Mario Carneiro, 12-Sep-2014.) |
⊢ (𝜑 → 𝑉 ∈ ℕ) & ⊢ (𝜑 → 𝑊 ∈ ℕ) & ⊢ (𝜑 → 𝑅 ∈ Fin) & ⊢ (𝜑 → 𝐻:(1...(𝑊 · (2 · 𝑉)))⟶𝑅) & ⊢ 𝐹 = (𝑥 ∈ (1...𝑉) ↦ (𝑦 ∈ (1...𝑊) ↦ (𝐻‘(𝑦 + (𝑊 · ((𝑥 − 1) + 𝑉)))))) & ⊢ (𝜑 → 𝑀 ∈ ℕ) & ⊢ (𝜑 → 𝐺:(1...𝑊)⟶𝑅) & ⊢ (𝜑 → 𝐾 ∈ (ℤ≥‘2)) & ⊢ (𝜑 → 𝐴 ∈ ℕ) & ⊢ (𝜑 → 𝐷 ∈ ℕ) & ⊢ (𝜑 → (𝐴(AP‘𝐾)𝐷) ⊆ (◡𝐹 “ {𝐺})) ⇒ ⊢ (𝜑 → (〈𝑀, 𝐾〉 PolyAP 𝐺 → (〈(𝑀 + 1), 𝐾〉 PolyAP 𝐻 ∨ (𝐾 + 1) MonoAP 𝐺))) | ||
Theorem | vdwlem8 16096* | Lemma for vdw 16102. (Contributed by Mario Carneiro, 18-Aug-2014.) |
⊢ (𝜑 → 𝑅 ∈ Fin) & ⊢ (𝜑 → 𝐾 ∈ (ℤ≥‘2)) & ⊢ (𝜑 → 𝑊 ∈ ℕ) & ⊢ (𝜑 → 𝐹:(1...(2 · 𝑊))⟶𝑅) & ⊢ 𝐶 ∈ V & ⊢ (𝜑 → 𝐴 ∈ ℕ) & ⊢ (𝜑 → 𝐷 ∈ ℕ) & ⊢ (𝜑 → (𝐴(AP‘𝐾)𝐷) ⊆ (◡𝐺 “ {𝐶})) & ⊢ 𝐺 = (𝑥 ∈ (1...𝑊) ↦ (𝐹‘(𝑥 + 𝑊))) ⇒ ⊢ (𝜑 → 〈1, 𝐾〉 PolyAP 𝐹) | ||
Theorem | vdwlem9 16097* | Lemma for vdw 16102. (Contributed by Mario Carneiro, 12-Sep-2014.) |
⊢ (𝜑 → 𝑅 ∈ Fin) & ⊢ (𝜑 → 𝐾 ∈ (ℤ≥‘2)) & ⊢ (𝜑 → ∀𝑠 ∈ Fin ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑠 ↑𝑚 (1...𝑛))𝐾 MonoAP 𝑓) & ⊢ (𝜑 → 𝑀 ∈ ℕ) & ⊢ (𝜑 → 𝑊 ∈ ℕ) & ⊢ (𝜑 → ∀𝑔 ∈ (𝑅 ↑𝑚 (1...𝑊))(〈𝑀, 𝐾〉 PolyAP 𝑔 ∨ (𝐾 + 1) MonoAP 𝑔)) & ⊢ (𝜑 → 𝑉 ∈ ℕ) & ⊢ (𝜑 → ∀𝑓 ∈ ((𝑅 ↑𝑚 (1...𝑊)) ↑𝑚 (1...𝑉))𝐾 MonoAP 𝑓) & ⊢ (𝜑 → 𝐻:(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)) | ||
Theorem | vdwlem10 16098* | Lemma for vdw 16102. Set up secondary induction on 𝑀. (Contributed by Mario Carneiro, 18-Aug-2014.) |
⊢ (𝜑 → 𝑅 ∈ Fin) & ⊢ (𝜑 → 𝐾 ∈ (ℤ≥‘2)) & ⊢ (𝜑 → ∀𝑠 ∈ Fin ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑠 ↑𝑚 (1...𝑛))𝐾 MonoAP 𝑓) & ⊢ (𝜑 → 𝑀 ∈ ℕ) ⇒ ⊢ (𝜑 → ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑅 ↑𝑚 (1...𝑛))(〈𝑀, 𝐾〉 PolyAP 𝑓 ∨ (𝐾 + 1) MonoAP 𝑓)) | ||
Theorem | vdwlem11 16099* | Lemma for vdw 16102. (Contributed by Mario Carneiro, 18-Aug-2014.) |
⊢ (𝜑 → 𝑅 ∈ Fin) & ⊢ (𝜑 → 𝐾 ∈ (ℤ≥‘2)) & ⊢ (𝜑 → ∀𝑠 ∈ Fin ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑠 ↑𝑚 (1...𝑛))𝐾 MonoAP 𝑓) ⇒ ⊢ (𝜑 → ∃𝑛 ∈ ℕ ∀𝑓 ∈ (𝑅 ↑𝑚 (1...𝑛))(𝐾 + 1) MonoAP 𝑓) | ||
Theorem | vdwlem12 16100 | Lemma for vdw 16102. 𝐾 = 2 base case of induction. (Contributed by Mario Carneiro, 18-Aug-2014.) |
⊢ (𝜑 → 𝑅 ∈ Fin) & ⊢ (𝜑 → 𝐹:(1...((♯‘𝑅) + 1))⟶𝑅) & ⊢ (𝜑 → ¬ 2 MonoAP 𝐹) ⇒ ⊢ ¬ 𝜑 |
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