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Theorem List for Metamath Proof Explorer - 14901-15000   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremsadid1 14901 The adder sequence function has a left identity, the empty set, which is the representation of the integer zero. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝐴 ⊆ ℕ0 → (𝐴 sadd ∅) = 𝐴)

Theoremsadid2 14902 The adder sequence function has a right identity, the empty set, which is the representation of the integer zero. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝐴 ⊆ ℕ0 → (∅ sadd 𝐴) = 𝐴)

(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑𝐶 ⊆ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (((𝐴 sadd 𝐵) sadd 𝐶) ∩ (0..^𝑁)) = ((𝐴 sadd (𝐵 sadd 𝐶)) ∩ (0..^𝑁)))

Theoremsadeq 14905 Any element of a sequence sum only depends on the values of the argument sequences up to and including that point. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → ((𝐴 sadd 𝐵) ∩ (0..^𝑁)) = (((𝐴 ∩ (0..^𝑁)) sadd (𝐵 ∩ (0..^𝑁))) ∩ (0..^𝑁)))

Theorembitsres 14906 Restrict the bits of a number to an upper integer set. (Contributed by Mario Carneiro, 5-Sep-2016.)
((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → ((bits‘𝐴) ∩ (ℤ𝑁)) = (bits‘((⌊‘(𝐴 / (2↑𝑁))) · (2↑𝑁))))

Theorembitsuz 14907 The bits of a number are all at least 𝑁 iff the number is divisible by 2↑𝑁. (Contributed by Mario Carneiro, 21-Sep-2016.)
((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → ((2↑𝑁) ∥ 𝐴 ↔ (bits‘𝐴) ⊆ (ℤ𝑁)))

Theorembitsshft 14908* Shifting a bit sequence to the left (toward the more significant bits) causes the number to be multiplied by a power of two. (Contributed by Mario Carneiro, 22-Sep-2016.)
((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → {𝑛 ∈ ℕ0 ∣ (𝑛𝑁) ∈ (bits‘𝐴)} = (bits‘(𝐴 · (2↑𝑁))))

Definitiondf-smu 14909* Define the multiplication of two bit sequences, using repeated sequence addition. (Contributed by Mario Carneiro, 9-Sep-2016.)
smul = (𝑥 ∈ 𝒫 ℕ0, 𝑦 ∈ 𝒫 ℕ0 ↦ {𝑘 ∈ ℕ0𝑘 ∈ (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝑥 ∧ (𝑛𝑚) ∈ 𝑦)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1))})

Theoremsmufval 14910* The multiplication of two bit sequences as repeated sequence addition. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑 → (𝐴 smul 𝐵) = {𝑘 ∈ ℕ0𝑘 ∈ (𝑃‘(𝑘 + 1))})

Theoremsmupf 14911* The sequence of partial sums of the sequence multiplication. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑𝑃:ℕ0⟶𝒫 ℕ0)

Theoremsmup0 14912* The initial element of the partial sum sequence. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑 → (𝑃‘0) = ∅)

Theoremsmupp1 14913* The initial element of the partial sum sequence. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝑃‘(𝑁 + 1)) = ((𝑃𝑁) sadd {𝑛 ∈ ℕ0 ∣ (𝑁𝐴 ∧ (𝑛𝑁) ∈ 𝐵)}))

Theoremsmuval 14914* Define the addition of two bit sequences, using df-had 1523 and df-cad 1536 bit operations. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝑁 ∈ (𝐴 smul 𝐵) ↔ 𝑁 ∈ (𝑃‘(𝑁 + 1))))

Theoremsmuval2 14915* The partial sum sequence stabilizes at 𝑁 after the 𝑁 + 1-th element of the sequence; this stable value is the value of the sequence multiplication. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   (𝜑𝑀 ∈ (ℤ‘(𝑁 + 1)))       (𝜑 → (𝑁 ∈ (𝐴 smul 𝐵) ↔ 𝑁 ∈ (𝑃𝑀)))

Theoremsmupvallem 14916* If 𝐴 only has elements less than 𝑁, then all elements of the partial sum sequence past 𝑁 already equal the final value. (Contributed by Mario Carneiro, 20-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   (𝜑𝐴 ⊆ (0..^𝑁))    &   (𝜑𝑀 ∈ (ℤ𝑁))       (𝜑 → (𝑃𝑀) = (𝐴 smul 𝐵))

Theoremsmucl 14917 The product of two sequences is a sequence. (Contributed by Mario Carneiro, 19-Sep-2016.)
((𝐴 ⊆ ℕ0𝐵 ⊆ ℕ0) → (𝐴 smul 𝐵) ⊆ ℕ0)

Theoremsmu01lem 14918* Lemma for smu01 14919 and smu02 14920. (Contributed by Mario Carneiro, 19-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   ((𝜑 ∧ (𝑘 ∈ ℕ0𝑛 ∈ ℕ0)) → ¬ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵))       (𝜑 → (𝐴 smul 𝐵) = ∅)

Theoremsmu01 14919 Multiplication of a sequence by 0 on the right. (Contributed by Mario Carneiro, 19-Sep-2016.)
(𝐴 ⊆ ℕ0 → (𝐴 smul ∅) = ∅)

Theoremsmu02 14920 Multiplication of a sequence by 0 on the left. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝐴 ⊆ ℕ0 → (∅ smul 𝐴) = ∅)

Theoremsmupval 14921* Rewrite the elements of the partial sum sequence in terms of sequence multiplication. (Contributed by Mario Carneiro, 20-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝑃𝑁) = ((𝐴 ∩ (0..^𝑁)) smul 𝐵))

Theoremsmup1 14922* Rewrite smupp1 14913 using only smul instead of the internal recursive function 𝑃. (Contributed by Mario Carneiro, 20-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → ((𝐴 ∩ (0..^(𝑁 + 1))) smul 𝐵) = (((𝐴 ∩ (0..^𝑁)) smul 𝐵) sadd {𝑛 ∈ ℕ0 ∣ (𝑁𝐴 ∧ (𝑛𝑁) ∈ 𝐵)}))

Theoremsmueqlem 14923* Any element of a sequence multiplication only depends on the values of the argument sequences up to and including that point. (Contributed by Mario Carneiro, 20-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   𝑄 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ (𝐵 ∩ (0..^𝑁)))})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑 → ((𝐴 smul 𝐵) ∩ (0..^𝑁)) = (((𝐴 ∩ (0..^𝑁)) smul (𝐵 ∩ (0..^𝑁))) ∩ (0..^𝑁)))

Theoremsmueq 14924 Any element of a sequence multiplication only depends on the values of the argument sequences up to and including that point. (Contributed by Mario Carneiro, 20-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → ((𝐴 smul 𝐵) ∩ (0..^𝑁)) = (((𝐴 ∩ (0..^𝑁)) smul (𝐵 ∩ (0..^𝑁))) ∩ (0..^𝑁)))

Theoremsmumullem 14925 Lemma for smumul 14926. (Contributed by Mario Carneiro, 22-Sep-2016.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (((bits‘𝐴) ∩ (0..^𝑁)) smul (bits‘𝐵)) = (bits‘((𝐴 mod (2↑𝑁)) · 𝐵)))

Theoremsmumul 14926 For sequences that correspond to valid integers, the sequence multiplication function produces the sequence for the product. This is effectively a proof of the correctness of the multiplication process, implemented in terms of logic gates for df-sad 14884, whose correctness is verified in sadadd 14900.

Outside this range, the sequences cannot be representing integers, but the smul function still "works". This extended function is best interpreted in terms of the ring structure of the 2-adic integers. (Contributed by Mario Carneiro, 22-Sep-2016.)

((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((bits‘𝐴) smul (bits‘𝐵)) = (bits‘(𝐴 · 𝐵)))

6.1.7  The greatest common divisor operator

Syntaxcgcd 14927 Extend the definition of a class to include the greatest common divisor operator.
class gcd

Definitiondf-gcd 14928* Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 26444). For an alternate definition, based on the definition in [ApostolNT] p. 15, see dfgcd2 14977. (Contributed by Paul Chapman, 21-Mar-2011.)
gcd = (𝑥 ∈ ℤ, 𝑦 ∈ ℤ ↦ if((𝑥 = 0 ∧ 𝑦 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛𝑥𝑛𝑦)}, ℝ, < )))

Theoremgcdval 14929* The value of the gcd operator. (𝑀 gcd 𝑁) is the greatest common divisor of 𝑀 and 𝑁. If 𝑀 and 𝑁 are both 0, the result is defined conventionally as 0. (Contributed by Paul Chapman, 21-Mar-2011.) (Revised by Mario Carneiro, 10-Nov-2013.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = if((𝑀 = 0 ∧ 𝑁 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛𝑀𝑛𝑁)}, ℝ, < )))

Theoremgcd0val 14930 The value, by convention, of the gcd operator when both operands are 0. (Contributed by Paul Chapman, 21-Mar-2011.)
(0 gcd 0) = 0

Theoremgcdn0val 14931* The value of the gcd operator when at least one operand is nonzero. (Contributed by Paul Chapman, 21-Mar-2011.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → (𝑀 gcd 𝑁) = sup({𝑛 ∈ ℤ ∣ (𝑛𝑀𝑛𝑁)}, ℝ, < ))

Theoremgcdcllem1 14932* Lemma for gcdn0cl 14935, gcddvds 14936 and dvdslegcd 14937. (Contributed by Paul Chapman, 21-Mar-2011.)
𝑆 = {𝑧 ∈ ℤ ∣ ∀𝑛𝐴 𝑧𝑛}       ((𝐴 ⊆ ℤ ∧ ∃𝑛𝐴 𝑛 ≠ 0) → (𝑆 ≠ ∅ ∧ ∃𝑥 ∈ ℤ ∀𝑦𝑆 𝑦𝑥))

Theoremgcdcllem2 14933* Lemma for gcdn0cl 14935, gcddvds 14936 and dvdslegcd 14937. (Contributed by Paul Chapman, 21-Mar-2011.)
𝑆 = {𝑧 ∈ ℤ ∣ ∀𝑛 ∈ {𝑀, 𝑁}𝑧𝑛}    &   𝑅 = {𝑧 ∈ ℤ ∣ (𝑧𝑀𝑧𝑁)}       ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑅 = 𝑆)

Theoremgcdcllem3 14934* Lemma for gcdn0cl 14935, gcddvds 14936 and dvdslegcd 14937. (Contributed by Paul Chapman, 21-Mar-2011.)
𝑆 = {𝑧 ∈ ℤ ∣ ∀𝑛 ∈ {𝑀, 𝑁}𝑧𝑛}    &   𝑅 = {𝑧 ∈ ℤ ∣ (𝑧𝑀𝑧𝑁)}       (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → (sup(𝑅, ℝ, < ) ∈ ℕ ∧ (sup(𝑅, ℝ, < ) ∥ 𝑀 ∧ sup(𝑅, ℝ, < ) ∥ 𝑁) ∧ ((𝐾 ∈ ℤ ∧ 𝐾𝑀𝐾𝑁) → 𝐾 ≤ sup(𝑅, ℝ, < ))))

Theoremgcdn0cl 14935 Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → (𝑀 gcd 𝑁) ∈ ℕ)

Theoremgcddvds 14936 The gcd of two integers divides each of them. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 gcd 𝑁) ∥ 𝑀 ∧ (𝑀 gcd 𝑁) ∥ 𝑁))

Theoremdvdslegcd 14937 An integer which divides both operands of the gcd operator is bounded by it. (Contributed by Paul Chapman, 21-Mar-2011.)
(((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → ((𝐾𝑀𝐾𝑁) → 𝐾 ≤ (𝑀 gcd 𝑁)))

Theoremnndvdslegcd 14938 A positive integer which divides both positive operands of the gcd operator is bounded by it. (Contributed by AV, 9-Aug-2020.)
((𝐾 ∈ ℕ ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐾𝑀𝐾𝑁) → 𝐾 ≤ (𝑀 gcd 𝑁)))

Theoremgcdcl 14939 Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)

Theoremgcdnncl 14940 Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)

Theoremgcdcld 14941 Closure of the gcd operator. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝑀 gcd 𝑁) ∈ ℕ0)

Theoremgcd2n0cl 14942 Closure of the gcd operator if the second operand is not 0. (Contributed by AV, 10-Jul-2021.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → (𝑀 gcd 𝑁) ∈ ℕ)

Theoremzeqzmulgcd 14943* An integer is the product of an integer and the gcd of it and another integer. (Contributed by AV, 11-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑛 ∈ ℤ 𝐴 = (𝑛 · (𝐴 gcd 𝐵)))

Theoremdivgcdz 14944 An integer divided by the gcd of it and a nonzero integer is an integer. (Contributed by AV, 11-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐵 ≠ 0) → (𝐴 / (𝐴 gcd 𝐵)) ∈ ℤ)

Theoremgcdf 14945 Domain and codomain of the gcd operator. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 16-Nov-2013.)
gcd :(ℤ × ℤ)⟶ℕ0

Theoremgcdcom 14946 The gcd operator is commutative. Theorem 1.4(a) in [ApostolNT] p. 16. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑁 gcd 𝑀))

Theoremdivgcdnn 14947 A positive integer divided by the gcd of it and another integer is a positive integer. (Contributed by AV, 10-Jul-2021.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → (𝐴 / (𝐴 gcd 𝐵)) ∈ ℕ)

Theoremdivgcdnnr 14948 A positive integer divided by the gcd of it and another integer is a positive integer. (Contributed by AV, 10-Jul-2021.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → (𝐴 / (𝐵 gcd 𝐴)) ∈ ℕ)

Theoremgcdeq0 14949 The gcd of two integers is zero iff they are both zero. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 gcd 𝑁) = 0 ↔ (𝑀 = 0 ∧ 𝑁 = 0)))

Theoremgcdn0gt0 14950 The gcd of two integers is positive (nonzero) iff they are not both zero. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (¬ (𝑀 = 0 ∧ 𝑁 = 0) ↔ 0 < (𝑀 gcd 𝑁)))

Theoremgcd0id 14951 The gcd of 0 and an integer is the integer's absolute value. (Contributed by Paul Chapman, 21-Mar-2011.)
(𝑁 ∈ ℤ → (0 gcd 𝑁) = (abs‘𝑁))

Theoremgcdid0 14952 The gcd of an integer and 0 is the integer's absolute value. Theorem 1.4(d)2 in [ApostolNT] p. 16. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑁 ∈ ℤ → (𝑁 gcd 0) = (abs‘𝑁))

Theoremnn0gcdid0 14953 The gcd of a nonnegative integer with 0 is itself. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑁 ∈ ℕ0 → (𝑁 gcd 0) = 𝑁)

Theoremgcdneg 14954 Negating one operand of the gcd operator does not alter the result. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd -𝑁) = (𝑀 gcd 𝑁))

Theoremneggcd 14955 Negating one operand of the gcd operator does not alter the result. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (-𝑀 gcd 𝑁) = (𝑀 gcd 𝑁))

𝐾 ∈ ℤ    &   𝑀 ∈ ℤ    &   𝑁 ∈ ℤ       (𝑀 gcd 𝑁) = (𝑀 gcd ((𝐾 · 𝑀) + 𝑁))

Theoremgcdaddm 14957 Adding a multiple of one operand of the gcd operator to the other does not alter the result. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑀 gcd (𝑁 + (𝐾 · 𝑀))))

Theoremgcdadd 14958 The GCD of two numbers is the same as the GCD of the left and their sum. (Contributed by Scott Fenton, 20-Apr-2014.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑀 gcd (𝑁 + 𝑀)))

Theoremgcdid 14959 The gcd of a number and itself is its absolute value. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑁 ∈ ℤ → (𝑁 gcd 𝑁) = (abs‘𝑁))

Theoremgcd1 14960 The gcd of a number with 1 is 1. Theorem 1.4(d)1 in [ApostolNT] p. 16. (Contributed by Mario Carneiro, 19-Feb-2014.)
(𝑀 ∈ ℤ → (𝑀 gcd 1) = 1)

Theoremgcdabs 14961 The gcd of two integers is the same as that of their absolute values. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((abs‘𝑀) gcd (abs‘𝑁)) = (𝑀 gcd 𝑁))

Theoremgcdabs1 14962 gcd of the absolute value of the first operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → ((abs‘𝑁) gcd 𝑀) = (𝑁 gcd 𝑀))

Theoremgcdabs2 14963 gcd of the absolute value of the second operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → (𝑁 gcd (abs‘𝑀)) = (𝑁 gcd 𝑀))

Theoremmodgcd 14964 The gcd remains unchanged if one operand is replaced with its remainder modulo the other. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → ((𝑀 mod 𝑁) gcd 𝑁) = (𝑀 gcd 𝑁))

Theorem1gcd 14965 The GCD of one and an integer is one. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(𝑀 ∈ ℤ → (1 gcd 𝑀) = 1)

Theorem6gcd4e2 14966 The greatest common divisor of six and four is two. To calculate this gcd, a simple form of Euclid's algorithm is used: (6 gcd 4) = ((4 + 2) gcd 4) = (2 gcd 4) and (2 gcd 4) = (2 gcd (2 + 2)) = (2 gcd 2) = 2. (Contributed by AV, 27-Aug-2020.)
(6 gcd 4) = 2

6.1.8  Bézout's identity

Theorembezoutlem1 14967* Lemma for bezout 14974. (Contributed by Mario Carneiro, 15-Mar-2014.)
𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)       (𝜑 → (𝐴 ≠ 0 → (abs‘𝐴) ∈ 𝑀))

Theorembezoutlem2OLD 14968* Lemma for bezout 14974. (Contributed by Mario Carneiro, 15-Mar-2014.) Obsolete version of bezoutlem2 14971 as of 30-Sep-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   𝐺 = sup(𝑀, ℝ, < )    &   (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))       (𝜑𝐺𝑀)

Theorembezoutlem3OLD 14969* Lemma for bezout 14974. (Contributed by Mario Carneiro, 22-Feb-2014.) Obsolete version of bezoutlem3 14972 as of 30-Sep-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   𝐺 = sup(𝑀, ℝ, < )    &   (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))       (𝜑 → (𝐶𝑀𝐺𝐶))

Theorembezoutlem4OLD 14970* Lemma for bezout 14974. (Contributed by Mario Carneiro, 22-Feb-2014.) Obsolete version of bezoutlem4 14973 as of 30-Sep-2020. (New usage is discouraged.) (Proof modification is discouraged.)
𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   𝐺 = sup(𝑀, ℝ, < )    &   (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))       (𝜑 → (𝐴 gcd 𝐵) ∈ 𝑀)

Theorembezoutlem2 14971* Lemma for bezout 14974. (Contributed by Mario Carneiro, 15-Mar-2014.) ( Revised by AV, 30-Sep-2020.)
𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   𝐺 = inf(𝑀, ℝ, < )    &   (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))       (𝜑𝐺𝑀)

Theorembezoutlem3 14972* Lemma for bezout 14974. (Contributed by Mario Carneiro, 22-Feb-2014.) ( Revised by AV, 30-Sep-2020.)
𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   𝐺 = inf(𝑀, ℝ, < )    &   (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))       (𝜑 → (𝐶𝑀𝐺𝐶))

Theorembezoutlem4 14973* Lemma for bezout 14974. (Contributed by Mario Carneiro, 22-Feb-2014.)
𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   𝐺 = inf(𝑀, ℝ, < )    &   (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))       (𝜑 → (𝐴 gcd 𝐵) ∈ 𝑀)

Theorembezout 14974* Bézout's identity: For any integers 𝐴 and 𝐵, there are integers 𝑥, 𝑦 such that (𝐴 gcd 𝐵) = 𝐴 · 𝑥 + 𝐵 · 𝑦. This is Metamath 100 proof #60. (Contributed by Mario Carneiro, 22-Feb-2014.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ (𝐴 gcd 𝐵) = ((𝐴 · 𝑥) + (𝐵 · 𝑦)))

Theoremdvdsgcd 14975 An integer which divides each of two others also divides their gcd. (Contributed by Paul Chapman, 22-Jun-2011.) (Revised by Mario Carneiro, 30-May-2014.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾𝑀𝐾𝑁) → 𝐾 ∥ (𝑀 gcd 𝑁)))

Theoremdvdsgcdb 14976 Biconditional form of dvdsgcd 14975. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾𝑀𝐾𝑁) ↔ 𝐾 ∥ (𝑀 gcd 𝑁)))

Theoremdfgcd2 14977* Alternate definition of the gcd operator, see definition in [ApostolNT] p. 15. (Contributed by AV, 8-Aug-2021.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐷 = (𝑀 gcd 𝑁) ↔ (0 ≤ 𝐷 ∧ (𝐷𝑀𝐷𝑁) ∧ ∀𝑒 ∈ ℤ ((𝑒𝑀𝑒𝑁) → 𝑒𝐷))))

Theoremgcdass 14978 Associative law for gcd operator. Theorem 1.4(b) in [ApostolNT] p. 16. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑃 ∈ ℤ) → ((𝑁 gcd 𝑀) gcd 𝑃) = (𝑁 gcd (𝑀 gcd 𝑃)))

Theoremmulgcd 14979 Distribute multiplication by a nonnegative integer over gcd. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Mario Carneiro, 30-May-2014.)
((𝐾 ∈ ℕ0𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) gcd (𝐾 · 𝑁)) = (𝐾 · (𝑀 gcd 𝑁)))

Theoremabsmulgcd 14980 Distribute absolute value of multiplication over gcd. Theorem 1.4(c) in [ApostolNT] p. 16. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) gcd (𝐾 · 𝑁)) = (abs‘(𝐾 · (𝑀 gcd 𝑁))))

Theoremmulgcdr 14981 Reverse distribution law for the gcd operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ0) → ((𝐴 · 𝐶) gcd (𝐵 · 𝐶)) = ((𝐴 gcd 𝐵) · 𝐶))

Theoremgcddiv 14982 Division law for GCD. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ) ∧ (𝐶𝐴𝐶𝐵)) → ((𝐴 gcd 𝐵) / 𝐶) = ((𝐴 / 𝐶) gcd (𝐵 / 𝐶)))

Theoremgcdmultiple 14983 The GCD of a multiple of a number is the number itself. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)

Theoremgcdmultiplez 14984 Extend gcdmultiple 14983 so 𝑁 can be an integer. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)

Theoremgcdzeq 14985 A positive integer 𝐴 is equal to its gcd with an integer 𝐵 if and only if 𝐴 divides 𝐵. Generalization of gcdeq 14986. (Contributed by AV, 1-Jul-2020.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵) = 𝐴𝐴𝐵))

Theoremgcdeq 14986 𝐴 is equal to its gcd with 𝐵 if and only if 𝐴 divides 𝐵. (Contributed by Mario Carneiro, 23-Feb-2014.) (Proof shortened by AV, 8-Aug-2021.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → ((𝐴 gcd 𝐵) = 𝐴𝐴𝐵))

Theoremdvdssqim 14987 Unidirectional form of dvdssq 14994. (Contributed by Scott Fenton, 19-Apr-2014.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁 → (𝑀↑2) ∥ (𝑁↑2)))

Theoremdvdsmulgcd 14988 A divisibility equivalent for odmulg 17701. (Contributed by Stefan O'Rear, 6-Sep-2015.)
((𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) → (𝐴 ∥ (𝐵 · 𝐶) ↔ 𝐴 ∥ (𝐵 · (𝐶 gcd 𝐴))))

Theoremrpmulgcd 14989 If 𝐾 and 𝑀 are relatively prime, then the GCD of 𝐾 and 𝑀 · 𝑁 is the GCD of 𝐾 and 𝑁. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐾 ∈ ℕ ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ (𝐾 gcd 𝑀) = 1) → (𝐾 gcd (𝑀 · 𝑁)) = (𝐾 gcd 𝑁))

Theoremrplpwr 14990 If 𝐴 and 𝐵 are relatively prime, then so are 𝐴𝑁 and 𝐵. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → ((𝐴𝑁) gcd 𝐵) = 1))

Theoremrppwr 14991 If 𝐴 and 𝐵 are relatively prime, then so are 𝐴𝑁 and 𝐵𝑁. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → ((𝐴𝑁) gcd (𝐵𝑁)) = 1))

Theoremsqgcd 14992 Square distributes over GCD. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝑀 gcd 𝑁)↑2) = ((𝑀↑2) gcd (𝑁↑2)))

Theoremdvdssqlem 14993 Lemma for dvdssq 14994. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))

Theoremdvdssq 14994 Two numbers are divisible iff their squares are. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))

Theorembezoutr 14995 Partial converse to bezout 14974. Existence of a linear combination does not set the GCD, but it does upper bound it. (Contributed by Stefan O'Rear, 23-Sep-2014.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ)) → (𝐴 gcd 𝐵) ∥ ((𝐴 · 𝑋) + (𝐵 · 𝑌)))

Theorembezoutr1 14996 Converse of bezout 14974 for the gcd = 1 case, sufficient condition for relative primality. (Contributed by Stefan O'Rear, 23-Sep-2014.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ)) → (((𝐴 · 𝑋) + (𝐵 · 𝑌)) = 1 → (𝐴 gcd 𝐵) = 1))

6.1.9  Algorithms

Theoremnn0seqcvgd 14997* A strictly-decreasing nonnegative integer sequence with initial term 𝑁 reaches zero by the 𝑁 th term. Deduction version. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝜑𝐹:ℕ0⟶ℕ0)    &   (𝜑𝑁 = (𝐹‘0))    &   ((𝜑𝑘 ∈ ℕ0) → ((𝐹‘(𝑘 + 1)) ≠ 0 → (𝐹‘(𝑘 + 1)) < (𝐹𝑘)))       (𝜑 → (𝐹𝑁) = 0)

Theoremseq1st 14998 A sequence whose iteration function ignores the second argument is only affected by the first point of the initial value function. (Contributed by Mario Carneiro, 11-Feb-2015.)
𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))       ((𝑀 ∈ ℤ ∧ 𝐴𝑉) → 𝑅 = seq𝑀((𝐹 ∘ 1st ), {⟨𝑀, 𝐴⟩}))

Theoremalgr0 14999 The value of the algorithm iterator 𝑅 at 0 is the initial state 𝐴. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.)
𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)       (𝜑 → (𝑅𝑀) = 𝐴)

Theoremalgrf 15000 An algorithm is a step function 𝐹:𝑆𝑆 on a state space 𝑆. An algorithm acts on an initial state 𝐴𝑆 by iteratively applying 𝐹 to give 𝐴, (𝐹𝐴), (𝐹‘(𝐹𝐴)) and so on. An algorithm is said to halt if a fixed point of 𝐹 is reached after a finite number of iterations.

The algorithm iterator 𝑅:ℕ0𝑆 "runs" the algorithm 𝐹 so that (𝑅𝑘) is the state after 𝑘 iterations of 𝐹 on the initial state 𝐴.

Domain and codomain of the algorithm iterator 𝑅. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.)

𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)       (𝜑𝑅:𝑍𝑆)

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268 26701-26800 269 26801-26900 270 26901-27000 271 27001-27100 272 27101-27200 273 27201-27300 274 27301-27400 275 27401-27500 276 27501-27600 277 27601-27700 278 27701-27800 279 27801-27900 280 27901-28000 281 28001-28100 282 28101-28200 283 28201-28300 284 28301-28400 285 28401-28500 286 28501-28600 287 28601-28700 288 28701-28800 289 28801-28900 290 28901-29000 291 29001-29100 292 29101-29200 293 29201-29300 294 29301-29400 295 29401-29500 296 29501-29600 297 29601-29700 298 29701-29800 299 29801-29900 300 29901-30000 301 30001-30100 302 30101-30200 303 30201-30300 304 30301-30400 305 30401-30500 306 30501-30600 307 30601-30700 308 30701-30800 309 30801-30900 310 30901-31000 311 31001-31100 312 31101-31200 313 31201-31300 314 31301-31400 315 31401-31500 316 31501-31600 317 31601-31700 318 31701-31800 319 31801-31900 320 31901-32000 321 32001-32100 322 32101-32200 323 32201-32300 324 32301-32400 325 32401-32500 326 32501-32600 327 32601-32700 328 32701-32800 329 32801-32900 330 32901-33000 331 33001-33100 332 33101-33200 333 33201-33300 334 33301-33400 335 33401-33500 336 33501-33600 337 33601-33700 338 33701-33800 339 33801-33900 340 33901-34000 341 34001-34100 342 34101-34200 343 34201-34300 344 34301-34400 345 34401-34500 346 34501-34600 347 34601-34700 348 34701-34800 349 34801-34900 350 34901-35000 351 35001-35100 352 35101-35200 353 35201-35300 354 35301-35400 355 35401-35500 356 35501-35600 357 35601-35700 358 35701-35800 359 35801-35900 360 35901-36000 361 36001-36100 362 36101-36200 363 36201-36300 364 36301-36400 365 36401-36500 366 36501-36600 367 36601-36700 368 36701-36800 369 36801-36900 370 36901-37000 371 37001-37100 372 37101-37200 373 37201-37300 374 37301-37400 375 37401-37500 376 37501-37600 377 37601-37700 378 37701-37800 379 37801-37900 380 37901-38000 381 38001-38100 382 38101-38200 383 38201-38300 384 38301-38400 385 38401-38500 386 38501-38600 387 38601-38700 388 38701-38800 389 38801-38900 390 38901-39000 391 39001-39100 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