HomeHome Intuitionistic Logic Explorer
Theorem List (p. 131 of 168)
< Previous  Next >
Bad symbols? Try the
GIF version.

Mirrors  >  Metamath Home Page  >  ILE Home Page  >  Theorem List Contents  >  Recent Proofs       This page: Page List

Theorem List for Intuitionistic Logic Explorer - 13001-13100   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theorem4sqlem15 13001* Lemma for 4sq 13006. (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 13002* Lemma for 4sq 13006. (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 13003* Lemma for 4sq 13006. (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 13004* Lemma for 4sq 13006. 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 13005* Lemma for 4sq 13006. 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 13004. 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 12990 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 13006* 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.)
 
Theoremdec2dvds 13007 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
 
Theoremdec5dvds 13008 Divisibility by five is obvious in base 10. (Contributed by Mario Carneiro, 19-Apr-2015.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ    &   𝐵 < 5        ¬ 5 ∥ 𝐴𝐵
 
Theoremdec5dvds2 13009 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
 
Theoremdec5nprm 13010 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
 
Theoremdec2nprm 13011 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
 
Theoremmodxai 13012 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    &   ((𝐴𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   ((𝐴𝐶) mod 𝑁) = (𝐿 mod 𝑁)    &   (𝐵 + 𝐶) = 𝐸    &   ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐿)       ((𝐴𝐸) mod 𝑁) = (𝑀 mod 𝑁)
 
Theoremmod2xi 13013 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    &   ((𝐴𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   (2 · 𝐵) = 𝐸    &   ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐾)       ((𝐴𝐸) mod 𝑁) = (𝑀 mod 𝑁)
 
Theoremmodxp1i 13014 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    &   ((𝐴𝐵) mod 𝑁) = (𝐾 mod 𝑁)    &   (𝐵 + 1) = 𝐸    &   ((𝐷 · 𝑁) + 𝑀) = (𝐾 · 𝐴)       ((𝐴𝐸) mod 𝑁) = (𝑀 mod 𝑁)
 
Theoremmodsubi 13015 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       ((𝐴𝐵) mod 𝑁) = (𝑀 mod 𝑁)
 
Theoremgcdi 13016 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
 
Theoremgcdmodi 13017 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
 
Theoremnumexp0 13018 Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0       (𝐴↑0) = 1
 
Theoremnumexp1 13019 Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0       (𝐴↑1) = 𝐴
 
Theoremnumexpp1 13020 Calculate an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0    &   𝑀 ∈ ℕ0    &   (𝑀 + 1) = 𝑁    &   ((𝐴𝑀) · 𝐴) = 𝐶       (𝐴𝑁) = 𝐶
 
Theoremnumexp2x 13021 Double an integer power. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0    &   𝑀 ∈ ℕ0    &   (2 · 𝑀) = 𝑁    &   (𝐴𝑀) = 𝐷    &   (𝐷 · 𝐷) = 𝐶       (𝐴𝑁) = 𝐶
 
Theoremdecsplit0b 13022 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)
 
Theoremdecsplit0 13023 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
 
Theoremdecsplit1 13024 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
 
Theoremdecsplit 13025 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
 
Theoremkaratsuba 13026 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 9680. (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
 
Theorem2exp4 13027 Two to the fourth power is 16. (Contributed by Mario Carneiro, 20-Apr-2015.)
(2↑4) = 16
 
Theorem2exp5 13028 Two to the fifth power is 32. (Contributed by AV, 16-Aug-2021.)
(2↑5) = 32
 
Theorem2exp6 13029 Two to the sixth power is 64. (Contributed by Mario Carneiro, 20-Apr-2015.) (Proof shortened by OpenAI, 25-Mar-2020.)
(2↑6) = 64
 
Theorem2exp7 13030 Two to the seventh power is 128. (Contributed by AV, 16-Aug-2021.)
(2↑7) = 128
 
Theorem2exp8 13031 Two to the eighth power is 256. (Contributed by Mario Carneiro, 20-Apr-2015.)
(2↑8) = 256
 
Theorem2exp11 13032 Two to the eleventh power is 2048. (Contributed by AV, 16-Aug-2021.)
(2↑11) = 2048
 
Theorem2exp16 13033 Two to the sixteenth power is 65536. (Contributed by Mario Carneiro, 20-Apr-2015.)
(2↑16) = 65536
 
Theorem3exp3 13034 Three to the third power is 27. (Contributed by Mario Carneiro, 20-Apr-2015.)
(3↑3) = 27
 
Theorem2expltfac 13035 The factorial grows faster than two to the power 𝑁. (Contributed by Mario Carneiro, 15-Sep-2016.)
(𝑁 ∈ (ℤ‘4) → (2↑𝑁) < (!‘𝑁))
 
5.3  Cardinality of real and complex number subsets
 
5.3.1  Countability of integers and rationals
 
Theoremoddennn 13036 There are as many odd positive integers as there are positive integers. (Contributed by Jim Kingdon, 11-May-2022.)
{𝑧 ∈ ℕ ∣ ¬ 2 ∥ 𝑧} ≈ ℕ
 
Theoremevenennn 13037 There are as many even positive integers as there are positive integers. (Contributed by Jim Kingdon, 12-May-2022.)
{𝑧 ∈ ℕ ∣ 2 ∥ 𝑧} ≈ ℕ
 
Theoremxpnnen 13038 The Cartesian product of the set of positive integers with itself is equinumerous to the set of positive integers. (Contributed by NM, 1-Aug-2004.)
(ℕ × ℕ) ≈ ℕ
 
Theoremxpomen 13039 The Cartesian product of omega (the set of ordinal natural numbers) with itself is equinumerous to omega. Exercise 1 of [Enderton] p. 133. (Contributed by NM, 23-Jul-2004.)
(ω × ω) ≈ ω
 
Theoremxpct 13040 The cartesian product of two sets dominated by ω is dominated by ω. (Contributed by Thierry Arnoux, 24-Sep-2017.)
((𝐴 ≼ ω ∧ 𝐵 ≼ ω) → (𝐴 × 𝐵) ≼ ω)
 
Theoremunennn 13041 The union of two disjoint countably infinite sets is countably infinite. (Contributed by Jim Kingdon, 13-May-2022.)
((𝐴 ≈ ℕ ∧ 𝐵 ≈ ℕ ∧ (𝐴𝐵) = ∅) → (𝐴𝐵) ≈ ℕ)
 
Theoremznnen 13042 The set of integers and the set of positive integers are equinumerous. Corollary 8.1.23 of [AczelRathjen], p. 75. (Contributed by NM, 31-Jul-2004.)
ℤ ≈ ℕ
 
Theoremennnfonelemdc 13043* Lemma for ennnfone 13069. A direct consequence of fidcenumlemrk 7158. (Contributed by Jim Kingdon, 15-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑𝑃 ∈ ω)       (𝜑DECID (𝐹𝑃) ∈ (𝐹𝑃))
 
Theoremennnfonelemk 13044* Lemma for ennnfone 13069. (Contributed by Jim Kingdon, 15-Jul-2023.)
(𝜑𝐹:ω–onto𝐴)    &   (𝜑𝐾 ∈ ω)    &   (𝜑𝑁 ∈ ω)    &   (𝜑 → ∀𝑗 ∈ suc 𝑁(𝐹𝐾) ≠ (𝐹𝑗))       (𝜑𝑁𝐾)
 
Theoremennnfonelemj0 13045* Lemma for ennnfone 13069. Initial state for 𝐽. (Contributed by Jim Kingdon, 20-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       (𝜑 → (𝐽‘0) ∈ {𝑔 ∈ (𝐴pm ω) ∣ dom 𝑔 ∈ ω})
 
Theoremennnfonelemjn 13046* Lemma for ennnfone 13069. Non-initial state for 𝐽. (Contributed by Jim Kingdon, 20-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       ((𝜑𝑓 ∈ (ℤ‘(0 + 1))) → (𝐽𝑓) ∈ ω)
 
Theoremennnfonelemg 13047* Lemma for ennnfone 13069. Closure for 𝐺. (Contributed by Jim Kingdon, 20-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       ((𝜑 ∧ (𝑓 ∈ {𝑔 ∈ (𝐴pm ω) ∣ dom 𝑔 ∈ ω} ∧ 𝑗 ∈ ω)) → (𝑓𝐺𝑗) ∈ {𝑔 ∈ (𝐴pm ω) ∣ dom 𝑔 ∈ ω})
 
Theoremennnfonelemh 13048* Lemma for ennnfone 13069. (Contributed by Jim Kingdon, 8-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       (𝜑𝐻:ℕ0⟶(𝐴pm ω))
 
Theoremennnfonelem0 13049* Lemma for ennnfone 13069. Initial value. (Contributed by Jim Kingdon, 15-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       (𝜑 → (𝐻‘0) = ∅)
 
Theoremennnfonelemp1 13050* Lemma for ennnfone 13069. Value of 𝐻 at a successor. (Contributed by Jim Kingdon, 23-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   (𝜑𝑃 ∈ ℕ0)       (𝜑 → (𝐻‘(𝑃 + 1)) = if((𝐹‘(𝑁𝑃)) ∈ (𝐹 “ (𝑁𝑃)), (𝐻𝑃), ((𝐻𝑃) ∪ {⟨dom (𝐻𝑃), (𝐹‘(𝑁𝑃))⟩})))
 
Theoremennnfonelem1 13051* Lemma for ennnfone 13069. Second value. (Contributed by Jim Kingdon, 19-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)       (𝜑 → (𝐻‘1) = {⟨∅, (𝐹‘∅)⟩})
 
Theoremennnfonelemom 13052* Lemma for ennnfone 13069. 𝐻 yields finite sequences. (Contributed by Jim Kingdon, 19-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   (𝜑𝑃 ∈ ℕ0)       (𝜑 → dom (𝐻𝑃) ∈ ω)
 
Theoremennnfonelemhdmp1 13053* Lemma for ennnfone 13069. Domain at a successor where we need to add an element to the sequence. (Contributed by Jim Kingdon, 23-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   (𝜑𝑃 ∈ ℕ0)    &   (𝜑 → ¬ (𝐹‘(𝑁𝑃)) ∈ (𝐹 “ (𝑁𝑃)))       (𝜑 → dom (𝐻‘(𝑃 + 1)) = suc dom (𝐻𝑃))
 
Theoremennnfonelemss 13054* Lemma for ennnfone 13069. We only add elements to 𝐻 as the index increases. (Contributed by Jim Kingdon, 15-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   (𝜑𝑃 ∈ ℕ0)       (𝜑 → (𝐻𝑃) ⊆ (𝐻‘(𝑃 + 1)))
 
Theoremennnfoneleminc 13055* Lemma for ennnfone 13069. We only add elements to 𝐻 as the index increases. (Contributed by Jim Kingdon, 21-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   (𝜑𝑃 ∈ ℕ0)    &   (𝜑𝑄 ∈ ℕ0)    &   (𝜑𝑃𝑄)       (𝜑 → (𝐻𝑃) ⊆ (𝐻𝑄))
 
Theoremennnfonelemkh 13056* Lemma for ennnfone 13069. Because we add zero or one entries for each new index, the length of each sequence is no greater than its index. (Contributed by Jim Kingdon, 19-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   (𝜑𝑃 ∈ ℕ0)       (𝜑 → dom (𝐻𝑃) ⊆ (𝑁𝑃))
 
Theoremennnfonelemhf1o 13057* Lemma for ennnfone 13069. Each of the functions in 𝐻 is one to one and onto an image of 𝐹. (Contributed by Jim Kingdon, 17-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   (𝜑𝑃 ∈ ℕ0)       (𝜑 → (𝐻𝑃):dom (𝐻𝑃)–1-1-onto→(𝐹 “ (𝑁𝑃)))
 
Theoremennnfonelemex 13058* Lemma for ennnfone 13069. Extending the sequence (𝐻𝑃) to include an additional element. (Contributed by Jim Kingdon, 19-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   (𝜑𝑃 ∈ ℕ0)       (𝜑 → ∃𝑖 ∈ ℕ0 dom (𝐻𝑃) ∈ dom (𝐻𝑖))
 
Theoremennnfonelemhom 13059* Lemma for ennnfone 13069. The sequences in 𝐻 increase in length without bound if you go out far enough. (Contributed by Jim Kingdon, 19-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   (𝜑𝑀 ∈ ω)       (𝜑 → ∃𝑖 ∈ ℕ0 𝑀 ∈ dom (𝐻𝑖))
 
Theoremennnfonelemrnh 13060* Lemma for ennnfone 13069. A consequence of ennnfonelemss 13054. (Contributed by Jim Kingdon, 16-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   (𝜑𝑋 ∈ ran 𝐻)    &   (𝜑𝑌 ∈ ran 𝐻)       (𝜑 → (𝑋𝑌𝑌𝑋))
 
Theoremennnfonelemfun 13061* Lemma for ennnfone 13069. 𝐿 is a function. (Contributed by Jim Kingdon, 16-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   𝐿 = 𝑖 ∈ ℕ0 (𝐻𝑖)       (𝜑 → Fun 𝐿)
 
Theoremennnfonelemf1 13062* Lemma for ennnfone 13069. 𝐿 is one-to-one. (Contributed by Jim Kingdon, 16-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   𝐿 = 𝑖 ∈ ℕ0 (𝐻𝑖)       (𝜑𝐿:dom 𝐿1-1𝐴)
 
Theoremennnfonelemrn 13063* Lemma for ennnfone 13069. 𝐿 is onto 𝐴. (Contributed by Jim Kingdon, 16-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   𝐿 = 𝑖 ∈ ℕ0 (𝐻𝑖)       (𝜑 → ran 𝐿 = 𝐴)
 
Theoremennnfonelemdm 13064* Lemma for ennnfone 13069. The function 𝐿 is defined everywhere. (Contributed by Jim Kingdon, 16-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   𝐿 = 𝑖 ∈ ℕ0 (𝐻𝑖)       (𝜑 → dom 𝐿 = ω)
 
Theoremennnfonelemen 13065* Lemma for ennnfone 13069. The result. (Contributed by Jim Kingdon, 16-Jul-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ∀𝑗 ∈ suc 𝑛(𝐹𝑘) ≠ (𝐹𝑗))    &   𝐺 = (𝑥 ∈ (𝐴pm ω), 𝑦 ∈ ω ↦ if((𝐹𝑦) ∈ (𝐹𝑦), 𝑥, (𝑥 ∪ {⟨dom 𝑥, (𝐹𝑦)⟩})))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐽 = (𝑥 ∈ ℕ0 ↦ if(𝑥 = 0, ∅, (𝑁‘(𝑥 − 1))))    &   𝐻 = seq0(𝐺, 𝐽)    &   𝐿 = 𝑖 ∈ ℕ0 (𝐻𝑖)       (𝜑𝐴 ≈ ℕ)
 
Theoremennnfonelemnn0 13066* Lemma for ennnfone 13069. A version of ennnfonelemen 13065 expressed in terms of 0 instead of ω. (Contributed by Jim Kingdon, 27-Oct-2022.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ℕ0onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ℕ0𝑘 ∈ ℕ0𝑗 ∈ (0...𝑛)(𝐹𝑘) ≠ (𝐹𝑗))    &   𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)       (𝜑𝐴 ≈ ℕ)
 
Theoremennnfonelemr 13067* Lemma for ennnfone 13069. The interesting direction, expressed in deduction form. (Contributed by Jim Kingdon, 27-Oct-2022.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑𝐹:ℕ0onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ℕ0𝑘 ∈ ℕ0𝑗 ∈ (0...𝑛)(𝐹𝑘) ≠ (𝐹𝑗))       (𝜑𝐴 ≈ ℕ)
 
Theoremennnfonelemim 13068* Lemma for ennnfone 13069. The trivial direction. (Contributed by Jim Kingdon, 27-Oct-2022.)
(𝐴 ≈ ℕ → (∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓(𝑓:ℕ0onto𝐴 ∧ ∀𝑛 ∈ ℕ0𝑘 ∈ ℕ0𝑗 ∈ (0...𝑛)(𝑓𝑘) ≠ (𝑓𝑗))))
 
Theoremennnfone 13069* A condition for a set being countably infinite. Corollary 8.1.13 of [AczelRathjen], p. 73. Roughly speaking, the condition says that 𝐴 is countable (that's the 𝑓:ℕ0onto𝐴 part, as seen in theorems like ctm 7313), infinite (that's the part about being able to find an element of 𝐴 distinct from any mapping of a natural number via 𝑓), and has decidable equality. (Contributed by Jim Kingdon, 27-Oct-2022.)
(𝐴 ≈ ℕ ↔ (∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓(𝑓:ℕ0onto𝐴 ∧ ∀𝑛 ∈ ℕ0𝑘 ∈ ℕ0𝑗 ∈ (0...𝑛)(𝑓𝑘) ≠ (𝑓𝑗))))
 
Theoremexmidunben 13070* If any unbounded set of positive integers is equinumerous to , then the Limited Principle of Omniscience (LPO) implies excluded middle. (Contributed by Jim Kingdon, 29-Jul-2023.)
((∀𝑥((𝑥 ⊆ ℕ ∧ ∀𝑚 ∈ ℕ ∃𝑛𝑥 𝑚 < 𝑛) → 𝑥 ≈ ℕ) ∧ ω ∈ Omni) → EXMID)
 
Theoremctinfomlemom 13071* Lemma for ctinfom 13072. Converting between ω and 0. (Contributed by Jim Kingdon, 10-Aug-2023.)
𝑁 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    &   𝐺 = (𝐹𝑁)    &   (𝜑𝐹:ω–onto𝐴)    &   (𝜑 → ∀𝑛 ∈ ω ∃𝑘 ∈ ω ¬ (𝐹𝑘) ∈ (𝐹𝑛))       (𝜑 → (𝐺:ℕ0onto𝐴 ∧ ∀𝑚 ∈ ℕ0𝑗 ∈ ℕ0𝑖 ∈ (0...𝑚)(𝐺𝑗) ≠ (𝐺𝑖)))
 
Theoremctinfom 13072* A condition for a set being countably infinite. Restates ennnfone 13069 in terms of ω and function image. Like ennnfone 13069 the condition can be summarized as 𝐴 being countable, infinite, and having decidable equality. (Contributed by Jim Kingdon, 7-Aug-2023.)
(𝐴 ≈ ℕ ↔ (∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓(𝑓:ω–onto𝐴 ∧ ∀𝑛 ∈ ω ∃𝑘 ∈ ω ¬ (𝑓𝑘) ∈ (𝑓𝑛))))
 
Theoreminffinp1 13073* An infinite set contains an element not contained in a given finite subset. (Contributed by Jim Kingdon, 7-Aug-2023.)
(𝜑 → ∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦)    &   (𝜑 → ω ≼ 𝐴)    &   (𝜑𝐵𝐴)    &   (𝜑𝐵 ∈ Fin)       (𝜑 → ∃𝑥𝐴 ¬ 𝑥𝐵)
 
Theoremctinf 13074* A set is countably infinite if and only if it has decidable equality, is countable, and is infinite. (Contributed by Jim Kingdon, 7-Aug-2023.)
(𝐴 ≈ ℕ ↔ (∀𝑥𝐴𝑦𝐴 DECID 𝑥 = 𝑦 ∧ ∃𝑓 𝑓:ω–onto𝐴 ∧ ω ≼ 𝐴))
 
Theoremqnnen 13075 The rational numbers are countably infinite. Corollary 8.1.23 of [AczelRathjen], p. 75. This is Metamath 100 proof #3. (Contributed by Jim Kingdon, 11-Aug-2023.)
ℚ ≈ ℕ
 
Theoremenctlem 13076* Lemma for enct 13077. One direction of the biconditional. (Contributed by Jim Kingdon, 23-Dec-2023.)
(𝐴𝐵 → (∃𝑓 𝑓:ω–onto→(𝐴 ⊔ 1o) → ∃𝑔 𝑔:ω–onto→(𝐵 ⊔ 1o)))
 
Theoremenct 13077* Countability is invariant relative to equinumerosity. (Contributed by Jim Kingdon, 23-Dec-2023.)
(𝐴𝐵 → (∃𝑓 𝑓:ω–onto→(𝐴 ⊔ 1o) ↔ ∃𝑔 𝑔:ω–onto→(𝐵 ⊔ 1o)))
 
Theoremctiunctlemu1st 13078* Lemma for ctiunct 13084. (Contributed by Jim Kingdon, 28-Oct-2023.)
(𝜑𝑆 ⊆ ω)    &   (𝜑 → ∀𝑛 ∈ ω DECID 𝑛𝑆)    &   (𝜑𝐹:𝑆onto𝐴)    &   ((𝜑𝑥𝐴) → 𝑇 ⊆ ω)    &   ((𝜑𝑥𝐴) → ∀𝑛 ∈ ω DECID 𝑛𝑇)    &   ((𝜑𝑥𝐴) → 𝐺:𝑇onto𝐵)    &   (𝜑𝐽:ω–1-1-onto→(ω × ω))    &   𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽𝑧)) ∈ (𝐹‘(1st ‘(𝐽𝑧))) / 𝑥𝑇)}    &   (𝜑𝑁𝑈)       (𝜑 → (1st ‘(𝐽𝑁)) ∈ 𝑆)
 
Theoremctiunctlemu2nd 13079* Lemma for ctiunct 13084. (Contributed by Jim Kingdon, 28-Oct-2023.)
(𝜑𝑆 ⊆ ω)    &   (𝜑 → ∀𝑛 ∈ ω DECID 𝑛𝑆)    &   (𝜑𝐹:𝑆onto𝐴)    &   ((𝜑𝑥𝐴) → 𝑇 ⊆ ω)    &   ((𝜑𝑥𝐴) → ∀𝑛 ∈ ω DECID 𝑛𝑇)    &   ((𝜑𝑥𝐴) → 𝐺:𝑇onto𝐵)    &   (𝜑𝐽:ω–1-1-onto→(ω × ω))    &   𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽𝑧)) ∈ (𝐹‘(1st ‘(𝐽𝑧))) / 𝑥𝑇)}    &   (𝜑𝑁𝑈)       (𝜑 → (2nd ‘(𝐽𝑁)) ∈ (𝐹‘(1st ‘(𝐽𝑁))) / 𝑥𝑇)
 
Theoremctiunctlemuom 13080 Lemma for ctiunct 13084. (Contributed by Jim Kingdon, 28-Oct-2023.)
(𝜑𝑆 ⊆ ω)    &   (𝜑 → ∀𝑛 ∈ ω DECID 𝑛𝑆)    &   (𝜑𝐹:𝑆onto𝐴)    &   ((𝜑𝑥𝐴) → 𝑇 ⊆ ω)    &   ((𝜑𝑥𝐴) → ∀𝑛 ∈ ω DECID 𝑛𝑇)    &   ((𝜑𝑥𝐴) → 𝐺:𝑇onto𝐵)    &   (𝜑𝐽:ω–1-1-onto→(ω × ω))    &   𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽𝑧)) ∈ (𝐹‘(1st ‘(𝐽𝑧))) / 𝑥𝑇)}       (𝜑𝑈 ⊆ ω)
 
Theoremctiunctlemudc 13081* Lemma for ctiunct 13084. (Contributed by Jim Kingdon, 28-Oct-2023.)
(𝜑𝑆 ⊆ ω)    &   (𝜑 → ∀𝑛 ∈ ω DECID 𝑛𝑆)    &   (𝜑𝐹:𝑆onto𝐴)    &   ((𝜑𝑥𝐴) → 𝑇 ⊆ ω)    &   ((𝜑𝑥𝐴) → ∀𝑛 ∈ ω DECID 𝑛𝑇)    &   ((𝜑𝑥𝐴) → 𝐺:𝑇onto𝐵)    &   (𝜑𝐽:ω–1-1-onto→(ω × ω))    &   𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽𝑧)) ∈ (𝐹‘(1st ‘(𝐽𝑧))) / 𝑥𝑇)}       (𝜑 → ∀𝑛 ∈ ω DECID 𝑛𝑈)
 
Theoremctiunctlemf 13082* Lemma for ctiunct 13084. (Contributed by Jim Kingdon, 28-Oct-2023.)
(𝜑𝑆 ⊆ ω)    &   (𝜑 → ∀𝑛 ∈ ω DECID 𝑛𝑆)    &   (𝜑𝐹:𝑆onto𝐴)    &   ((𝜑𝑥𝐴) → 𝑇 ⊆ ω)    &   ((𝜑𝑥𝐴) → ∀𝑛 ∈ ω DECID 𝑛𝑇)    &   ((𝜑𝑥𝐴) → 𝐺:𝑇onto𝐵)    &   (𝜑𝐽:ω–1-1-onto→(ω × ω))    &   𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽𝑧)) ∈ (𝐹‘(1st ‘(𝐽𝑧))) / 𝑥𝑇)}    &   𝐻 = (𝑛𝑈 ↦ ((𝐹‘(1st ‘(𝐽𝑛))) / 𝑥𝐺‘(2nd ‘(𝐽𝑛))))       (𝜑𝐻:𝑈 𝑥𝐴 𝐵)
 
Theoremctiunctlemfo 13083* Lemma for ctiunct 13084. (Contributed by Jim Kingdon, 28-Oct-2023.)
(𝜑𝑆 ⊆ ω)    &   (𝜑 → ∀𝑛 ∈ ω DECID 𝑛𝑆)    &   (𝜑𝐹:𝑆onto𝐴)    &   ((𝜑𝑥𝐴) → 𝑇 ⊆ ω)    &   ((𝜑𝑥𝐴) → ∀𝑛 ∈ ω DECID 𝑛𝑇)    &   ((𝜑𝑥𝐴) → 𝐺:𝑇onto𝐵)    &   (𝜑𝐽:ω–1-1-onto→(ω × ω))    &   𝑈 = {𝑧 ∈ ω ∣ ((1st ‘(𝐽𝑧)) ∈ 𝑆 ∧ (2nd ‘(𝐽𝑧)) ∈ (𝐹‘(1st ‘(𝐽𝑧))) / 𝑥𝑇)}    &   𝐻 = (𝑛𝑈 ↦ ((𝐹‘(1st ‘(𝐽𝑛))) / 𝑥𝐺‘(2nd ‘(𝐽𝑛))))    &   𝑥𝐻    &   𝑥𝑈       (𝜑𝐻:𝑈onto 𝑥𝐴 𝐵)
 
Theoremctiunct 13084* A sequence of enumerations gives an enumeration of the union. We refer to "sequence of enumerations" rather than "countably many countable sets" because the hypothesis provides more than countability for each 𝐵(𝑥): it refers to 𝐵(𝑥) together with the 𝐺(𝑥) which enumerates it. Theorem 8.1.19 of [AczelRathjen], p. 74.

For "countably many countable sets" the key hypothesis would be (𝜑𝑥𝐴) → ∃𝑔𝑔:ω–onto→(𝐵 ⊔ 1o). This is almost omiunct 13088 (which uses countable choice) although that is for a countably infinite collection not any countable collection.

Compare with the case of two sets instead of countably many, as seen at unct 13086, which says that the union of two countable sets is countable .

The proof proceeds by mapping a natural number to a pair of natural numbers (by xpomen 13039) and using the first number to map to an element 𝑥 of 𝐴 and the second number to map to an element of B(x) . In this way we are able to map to every element of 𝑥𝐴𝐵. Although it would be possible to work directly with countability expressed as 𝐹:ω–onto→(𝐴 ⊔ 1o), we instead use functions from subsets of the natural numbers via ctssdccl 7315 and ctssdc 7317.

(Contributed by Jim Kingdon, 31-Oct-2023.)

(𝜑𝐹:ω–onto→(𝐴 ⊔ 1o))    &   ((𝜑𝑥𝐴) → 𝐺:ω–onto→(𝐵 ⊔ 1o))       (𝜑 → ∃ :ω–onto→( 𝑥𝐴 𝐵 ⊔ 1o))
 
Theoremctiunctal 13085* Variation of ctiunct 13084 which allows 𝑥 to be present in 𝜑. (Contributed by Jim Kingdon, 5-May-2024.)
(𝜑𝐹:ω–onto→(𝐴 ⊔ 1o))    &   (𝜑 → ∀𝑥𝐴 𝐺:ω–onto→(𝐵 ⊔ 1o))       (𝜑 → ∃ :ω–onto→( 𝑥𝐴 𝐵 ⊔ 1o))
 
Theoremunct 13086* The union of two countable sets is countable. Corollary 8.1.20 of [AczelRathjen], p. 75. (Contributed by Jim Kingdon, 1-Nov-2023.)
((∃𝑓 𝑓:ω–onto→(𝐴 ⊔ 1o) ∧ ∃𝑔 𝑔:ω–onto→(𝐵 ⊔ 1o)) → ∃ :ω–onto→((𝐴𝐵) ⊔ 1o))
 
Theoremomctfn 13087* Using countable choice to find a sequence of enumerations for a collection of countable sets. Lemma 8.1.27 of [AczelRathjen], p. 77. (Contributed by Jim Kingdon, 19-Apr-2024.)
(𝜑CCHOICE)    &   ((𝜑𝑥 ∈ ω) → ∃𝑔 𝑔:ω–onto→(𝐵 ⊔ 1o))       (𝜑 → ∃𝑓(𝑓 Fn ω ∧ ∀𝑥 ∈ ω (𝑓𝑥):ω–onto→(𝐵 ⊔ 1o)))
 
Theoremomiunct 13088* The union of a countably infinite collection of countable sets is countable. Theorem 8.1.28 of [AczelRathjen], p. 78. Compare with ctiunct 13084 which has a stronger hypothesis but does not require countable choice. (Contributed by Jim Kingdon, 5-May-2024.)
(𝜑CCHOICE)    &   ((𝜑𝑥 ∈ ω) → ∃𝑔 𝑔:ω–onto→(𝐵 ⊔ 1o))       (𝜑 → ∃ :ω–onto→( 𝑥 ∈ ω 𝐵 ⊔ 1o))
 
Theoremssomct 13089* A decidable subset of ω is countable. (Contributed by Jim Kingdon, 19-Sep-2024.)
((𝐴 ⊆ ω ∧ ∀𝑥 ∈ ω DECID 𝑥𝐴) → ∃𝑓 𝑓:ω–onto→(𝐴 ⊔ 1o))
 
Theoremssnnctlemct 13090* Lemma for ssnnct 13091. The result. (Contributed by Jim Kingdon, 29-Sep-2024.)
𝐺 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 1)       ((𝐴 ⊆ ℕ ∧ ∀𝑥 ∈ ℕ DECID 𝑥𝐴) → ∃𝑓 𝑓:ω–onto→(𝐴 ⊔ 1o))
 
Theoremssnnct 13091* A decidable subset of is countable. (Contributed by Jim Kingdon, 29-Sep-2024.)
((𝐴 ⊆ ℕ ∧ ∀𝑥 ∈ ℕ DECID 𝑥𝐴) → ∃𝑓 𝑓:ω–onto→(𝐴 ⊔ 1o))
 
Theoremnninfdclemcl 13092* Lemma for nninfdc 13097. (Contributed by Jim Kingdon, 25-Sep-2024.)
(𝜑𝐴 ⊆ ℕ)    &   (𝜑 → ∀𝑥 ∈ ℕ DECID 𝑥𝐴)    &   (𝜑 → ∀𝑚 ∈ ℕ ∃𝑛𝐴 𝑚 < 𝑛)    &   (𝜑𝑃𝐴)    &   (𝜑𝑄𝐴)       (𝜑 → (𝑃(𝑦 ∈ ℕ, 𝑧 ∈ ℕ ↦ inf((𝐴 ∩ (ℤ‘(𝑦 + 1))), ℝ, < ))𝑄) ∈ 𝐴)
 
Theoremnninfdclemf 13093* Lemma for nninfdc 13097. A function from the natural numbers into 𝐴. (Contributed by Jim Kingdon, 23-Sep-2024.)
(𝜑𝐴 ⊆ ℕ)    &   (𝜑 → ∀𝑥 ∈ ℕ DECID 𝑥𝐴)    &   (𝜑 → ∀𝑚 ∈ ℕ ∃𝑛𝐴 𝑚 < 𝑛)    &   (𝜑 → (𝐽𝐴 ∧ 1 < 𝐽))    &   𝐹 = seq1((𝑦 ∈ ℕ, 𝑧 ∈ ℕ ↦ inf((𝐴 ∩ (ℤ‘(𝑦 + 1))), ℝ, < )), (𝑖 ∈ ℕ ↦ 𝐽))       (𝜑𝐹:ℕ⟶𝐴)
 
Theoremnninfdclemp1 13094* Lemma for nninfdc 13097. Each element of the sequence 𝐹 is greater than the previous element. (Contributed by Jim Kingdon, 26-Sep-2024.)
(𝜑𝐴 ⊆ ℕ)    &   (𝜑 → ∀𝑥 ∈ ℕ DECID 𝑥𝐴)    &   (𝜑 → ∀𝑚 ∈ ℕ ∃𝑛𝐴 𝑚 < 𝑛)    &   (𝜑 → (𝐽𝐴 ∧ 1 < 𝐽))    &   𝐹 = seq1((𝑦 ∈ ℕ, 𝑧 ∈ ℕ ↦ inf((𝐴 ∩ (ℤ‘(𝑦 + 1))), ℝ, < )), (𝑖 ∈ ℕ ↦ 𝐽))    &   (𝜑𝑈 ∈ ℕ)       (𝜑 → (𝐹𝑈) < (𝐹‘(𝑈 + 1)))
 
Theoremnninfdclemlt 13095* Lemma for nninfdc 13097. The function from nninfdclemf 13093 is strictly monotonic. (Contributed by Jim Kingdon, 24-Sep-2024.)
(𝜑𝐴 ⊆ ℕ)    &   (𝜑 → ∀𝑥 ∈ ℕ DECID 𝑥𝐴)    &   (𝜑 → ∀𝑚 ∈ ℕ ∃𝑛𝐴 𝑚 < 𝑛)    &   (𝜑 → (𝐽𝐴 ∧ 1 < 𝐽))    &   𝐹 = seq1((𝑦 ∈ ℕ, 𝑧 ∈ ℕ ↦ inf((𝐴 ∩ (ℤ‘(𝑦 + 1))), ℝ, < )), (𝑖 ∈ ℕ ↦ 𝐽))    &   (𝜑𝑈 ∈ ℕ)    &   (𝜑𝑉 ∈ ℕ)    &   (𝜑𝑈 < 𝑉)       (𝜑 → (𝐹𝑈) < (𝐹𝑉))
 
Theoremnninfdclemf1 13096* Lemma for nninfdc 13097. The function from nninfdclemf 13093 is one-to-one. (Contributed by Jim Kingdon, 23-Sep-2024.)
(𝜑𝐴 ⊆ ℕ)    &   (𝜑 → ∀𝑥 ∈ ℕ DECID 𝑥𝐴)    &   (𝜑 → ∀𝑚 ∈ ℕ ∃𝑛𝐴 𝑚 < 𝑛)    &   (𝜑 → (𝐽𝐴 ∧ 1 < 𝐽))    &   𝐹 = seq1((𝑦 ∈ ℕ, 𝑧 ∈ ℕ ↦ inf((𝐴 ∩ (ℤ‘(𝑦 + 1))), ℝ, < )), (𝑖 ∈ ℕ ↦ 𝐽))       (𝜑𝐹:ℕ–1-1𝐴)
 
Theoremnninfdc 13097* An unbounded decidable set of positive integers is infinite. (Contributed by Jim Kingdon, 23-Sep-2024.)
((𝐴 ⊆ ℕ ∧ ∀𝑥 ∈ ℕ DECID 𝑥𝐴 ∧ ∀𝑚 ∈ ℕ ∃𝑛𝐴 𝑚 < 𝑛) → ω ≼ 𝐴)
 
Theoremunbendc 13098* An unbounded decidable set of positive integers is infinite. (Contributed by NM, 5-May-2005.) (Revised by Jim Kingdon, 30-Sep-2024.)
((𝐴 ⊆ ℕ ∧ ∀𝑥 ∈ ℕ DECID 𝑥𝐴 ∧ ∀𝑚 ∈ ℕ ∃𝑛𝐴 𝑚 < 𝑛) → 𝐴 ≈ ℕ)
 
Theoremprminf 13099 There are an infinite number of primes. Theorem 1.7 in [ApostolNT] p. 16. (Contributed by Paul Chapman, 28-Nov-2012.)
ℙ ≈ ℕ
 
Theoreminfpn2 13100* There exist infinitely many prime numbers: the set of all primes 𝑆 is unbounded by infpn 12957, so by unbendc 13098 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
    < Previous  Next >

Page List
Jump to page: Contents  1 1-100 2 101-200 3 201-300 4 301-400 5 401-500 6 501-600 7 601-700 8 701-800 9 801-900 10 901-1000 11 1001-1100 12 1101-1200 13 1201-1300 14 1301-1400 15 1401-1500 16 1501-1600 17 1601-1700 18 1701-1800 19 1801-1900 20 1901-2000 21 2001-2100 22 2101-2200 23 2201-2300 24 2301-2400 25 2401-2500 26 2501-2600 27 2601-2700 28 2701-2800 29 2801-2900 30 2901-3000 31 3001-3100 32 3101-3200 33 3201-3300 34 3301-3400 35 3401-3500 36 3501-3600 37 3601-3700 38 3701-3800 39 3801-3900 40 3901-4000 41 4001-4100 42 4101-4200 43 4201-4300 44 4301-4400 45 4401-4500 46 4501-4600 47 4601-4700 48 4701-4800 49 4801-4900 50 4901-5000 51 5001-5100 52 5101-5200 53 5201-5300 54 5301-5400 55 5401-5500 56 5501-5600 57 5601-5700 58 5701-5800 59 5801-5900 60 5901-6000 61 6001-6100 62 6101-6200 63 6201-6300 64 6301-6400 65 6401-6500 66 6501-6600 67 6601-6700 68 6701-6800 69 6801-6900 70 6901-7000 71 7001-7100 72 7101-7200 73 7201-7300 74 7301-7400 75 7401-7500 76 7501-7600 77 7601-7700 78 7701-7800 79 7801-7900 80 7901-8000 81 8001-8100 82 8101-8200 83 8201-8300 84 8301-8400 85 8401-8500 86 8501-8600 87 8601-8700 88 8701-8800 89 8801-8900 90 8901-9000 91 9001-9100 92 9101-9200 93 9201-9300 94 9301-9400 95 9401-9500 96 9501-9600 97 9601-9700 98 9701-9800 99 9801-9900 100 9901-10000 101 10001-10100 102 10101-10200 103 10201-10300 104 10301-10400 105 10401-10500 106 10501-10600 107 10601-10700 108 10701-10800 109 10801-10900 110 10901-11000 111 11001-11100 112 11101-11200 113 11201-11300 114 11301-11400 115 11401-11500 116 11501-11600 117 11601-11700 118 11701-11800 119 11801-11900 120 11901-12000 121 12001-12100 122 12101-12200 123 12201-12300 124 12301-12400 125 12401-12500 126 12501-12600 127 12601-12700 128 12701-12800 129 12801-12900 130 12901-13000 131 13001-13100 132 13101-13200 133 13201-13300 134 13301-13400 135 13401-13500 136 13501-13600 137 13601-13700 138 13701-13800 139 13801-13900 140 13901-14000 141 14001-14100 142 14101-14200 143 14201-14300 144 14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14800 149 14801-14900 150 14901-15000 151 15001-15100 152 15101-15200 153 15201-15300 154 15301-15400 155 15401-15500 156 15501-15600 157 15601-15700 158 15701-15800 159 15801-15900 160 15901-16000 161 16001-16100 162 16101-16200 163 16201-16300 164 16301-16400 165 16401-16500 166 16501-16600 167 16601-16700 168 16701-16766
  Copyright terms: Public domain < Previous  Next >