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Theorem List for Metamath Proof Explorer - 15601-15700   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theorembitsres 15601 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 15602 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 15603* 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 15604* 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 15605* 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 15606* 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 15607* 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 15608* 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 15609* Define the addition of two bit sequences, using df-had 1652 and df-cad 1665 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 15610* 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 15611* 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 15612 The product of two sequences is a sequence. (Contributed by Mario Carneiro, 19-Sep-2016.)
((𝐴 ⊆ ℕ0𝐵 ⊆ ℕ0) → (𝐴 smul 𝐵) ⊆ ℕ0)
 
Theoremsmu01lem 15613* Lemma for smu01 15614 and smu02 15615. (Contributed by Mario Carneiro, 19-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   ((𝜑 ∧ (𝑘 ∈ ℕ0𝑛 ∈ ℕ0)) → ¬ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵))       (𝜑 → (𝐴 smul 𝐵) = ∅)
 
Theoremsmu01 15614 Multiplication of a sequence by 0 on the right. (Contributed by Mario Carneiro, 19-Sep-2016.)
(𝐴 ⊆ ℕ0 → (𝐴 smul ∅) = ∅)
 
Theoremsmu02 15615 Multiplication of a sequence by 0 on the left. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝐴 ⊆ ℕ0 → (∅ smul 𝐴) = ∅)
 
Theoremsmupval 15616* 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 15617* Rewrite smupp1 15608 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 15618* 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 15619 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 15620 Lemma for smumul 15621. (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 15621 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 15579, whose correctness is verified in sadadd 15595.

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 15622 Extend the definition of a class to include the greatest common divisor operator.
class gcd
 
Definitiondf-gcd 15623* Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 27889). For an alternate definition, based on the definition in [ApostolNT] p. 15, see dfgcd2 15669. (Contributed by Paul Chapman, 21-Mar-2011.)
gcd = (𝑥 ∈ ℤ, 𝑦 ∈ ℤ ↦ if((𝑥 = 0 ∧ 𝑦 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛𝑥𝑛𝑦)}, ℝ, < )))
 
Theoremgcdval 15624* 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 15625 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 15626* 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 15627* Lemma for gcdn0cl 15630, gcddvds 15631 and dvdslegcd 15632. (Contributed by Paul Chapman, 21-Mar-2011.)
𝑆 = {𝑧 ∈ ℤ ∣ ∀𝑛𝐴 𝑧𝑛}       ((𝐴 ⊆ ℤ ∧ ∃𝑛𝐴 𝑛 ≠ 0) → (𝑆 ≠ ∅ ∧ ∃𝑥 ∈ ℤ ∀𝑦𝑆 𝑦𝑥))
 
Theoremgcdcllem2 15628* Lemma for gcdn0cl 15630, gcddvds 15631 and dvdslegcd 15632. (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 15629* Lemma for gcdn0cl 15630, gcddvds 15631 and dvdslegcd 15632. (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 15630 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 15631 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 15632 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 15633 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 15634 Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)
 
Theoremgcdnncl 15635 Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)
 
Theoremgcdcld 15636 Closure of the gcd operator. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝑀 gcd 𝑁) ∈ ℕ0)
 
Theoremgcd2n0cl 15637 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 15638* 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 15639 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 15640 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 15641 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 15642 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 15643 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 15644 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 15645 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 15646 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 15647 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 15648 The gcd of a nonnegative integer with 0 is itself. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑁 ∈ ℕ0 → (𝑁 gcd 0) = 𝑁)
 
Theoremgcdneg 15649 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 15650 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))
 
Theoremgcdaddmlem 15651 Lemma for gcdaddm 15652. (Contributed by Paul Chapman, 31-Mar-2011.)
𝐾 ∈ ℤ    &   𝑀 ∈ ℤ    &   𝑁 ∈ ℤ       (𝑀 gcd 𝑁) = (𝑀 gcd ((𝐾 · 𝑀) + 𝑁))
 
Theoremgcdaddm 15652 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 15653 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 15654 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 15655 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 15656 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 15657 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 15658 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 15659 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 15660 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 15661 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 15662* Lemma for bezout 15666. (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))
 
Theorembezoutlem2 15663* Lemma for bezout 15666. (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 15664* Lemma for bezout 15666. (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 15665* Lemma for bezout 15666. (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 15666* 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 15667 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 15668 Biconditional form of dvdsgcd 15667. (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 15669* 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 15670 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 15671 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 15672 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 15673 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 15674 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 15675 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 15676 Extend gcdmultiple 15675 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 15677 A positive integer 𝐴 is equal to its gcd with an integer 𝐵 if and only if 𝐴 divides 𝐵. Generalization of gcdeq 15678. (Contributed by AV, 1-Jul-2020.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵) = 𝐴𝐴𝐵))
 
Theoremgcdeq 15678 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 15679 Unidirectional form of dvdssq 15686. (Contributed by Scott Fenton, 19-Apr-2014.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁 → (𝑀↑2) ∥ (𝑁↑2)))
 
Theoremdvdsmulgcd 15680 A divisibility equivalent for odmulg 18357. (Contributed by Stefan O'Rear, 6-Sep-2015.)
((𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) → (𝐴 ∥ (𝐵 · 𝐶) ↔ 𝐴 ∥ (𝐵 · (𝐶 gcd 𝐴))))
 
Theoremrpmulgcd 15681 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 15682 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 15683 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 15684 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 15685 Lemma for dvdssq 15686. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))
 
Theoremdvdssq 15686 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 15687 Partial converse to bezout 15666. 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 15688 Converse of bezout 15666 for when the greater common divisor is one (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 15689* 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 15690 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 15691 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 15692 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)
 
Theoremalgrp1 15693 The value of the algorithm iterator 𝑅 at (𝐾 + 1). (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 27-Dec-2014.)
𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)       ((𝜑𝐾𝑍) → (𝑅‘(𝐾 + 1)) = (𝐹‘(𝑅𝐾)))
 
Theoremalginv 15694* If 𝐼 is an invariant of 𝐹, then its value is unchanged after any number of iterations of 𝐹. (Contributed by Paul Chapman, 31-Mar-2011.)
𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐹:𝑆𝑆    &   (𝑥𝑆 → (𝐼‘(𝐹𝑥)) = (𝐼𝑥))       ((𝐴𝑆𝐾 ∈ ℕ0) → (𝐼‘(𝑅𝐾)) = (𝐼‘(𝑅‘0)))
 
Theoremalgcvg 15695* One way to prove that an algorithm halts is to construct a countdown function 𝐶:𝑆⟶ℕ0 whose value is guaranteed to decrease for each iteration of 𝐹 until it reaches 0. That is, if 𝑋𝑆 is not a fixed point of 𝐹, then (𝐶‘(𝐹𝑋)) < (𝐶𝑋).

If 𝐶 is a countdown function for algorithm 𝐹, the sequence (𝐶‘(𝑅𝑘)) reaches 0 after at most 𝑁 steps, where 𝑁 is the value of 𝐶 for the initial state 𝐴. (Contributed by Paul Chapman, 22-Jun-2011.)

𝐹:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)       (𝐴𝑆 → (𝐶‘(𝑅𝑁)) = 0)
 
Theoremalgcvgblem 15696 Lemma for algcvgb 15697. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → ((𝑁 ≠ 0 → 𝑁 < 𝑀) ↔ ((𝑀 ≠ 0 → 𝑁 < 𝑀) ∧ (𝑀 = 0 → 𝑁 = 0))))
 
Theoremalgcvgb 15697 Two ways of expressing that 𝐶 is a countdown function for algorithm 𝐹. The first is used in these theorems. The second states the condition more intuitively as a conjunction: if the countdown function's value is currently nonzero, it must decrease at the next step; if it has reached zero, it must remain zero at the next step. (Contributed by Paul Chapman, 31-Mar-2011.)
𝐹:𝑆𝑆    &   𝐶:𝑆⟶ℕ0       (𝑋𝑆 → (((𝐶‘(𝐹𝑋)) ≠ 0 → (𝐶‘(𝐹𝑋)) < (𝐶𝑋)) ↔ (((𝐶𝑋) ≠ 0 → (𝐶‘(𝐹𝑋)) < (𝐶𝑋)) ∧ ((𝐶𝑋) = 0 → (𝐶‘(𝐹𝑋)) = 0))))
 
Theoremalgcvga 15698* The countdown function 𝐶 remains 0 after 𝑁 steps. (Contributed by Paul Chapman, 22-Jun-2011.)
𝐹:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)       (𝐴𝑆 → (𝐾 ∈ (ℤ𝑁) → (𝐶‘(𝑅𝐾)) = 0))
 
Theoremalgfx 15699* If 𝐹 reaches a fixed point when the countdown function 𝐶 reaches 0, 𝐹 remains fixed after 𝑁 steps. (Contributed by Paul Chapman, 22-Jun-2011.)
𝐹:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)    &   (𝑧𝑆 → ((𝐶𝑧) = 0 → (𝐹𝑧) = 𝑧))       (𝐴𝑆 → (𝐾 ∈ (ℤ𝑁) → (𝑅𝐾) = (𝑅𝑁)))
 
6.1.10  Euclid's Algorithm
 
Theoremeucalgval2 15700* The value of the step function 𝐸 for Euclid's Algorithm on an ordered pair. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       ((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀𝐸𝑁) = if(𝑁 = 0, ⟨𝑀, 𝑁⟩, ⟨𝑁, (𝑀 mod 𝑁)⟩))
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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 16701-16800 169 16801-16900 170 16901-17000 171 17001-17100 172 17101-17200 173 17201-17300 174 17301-17400 175 17401-17500 176 17501-17600 177 17601-17700 178 17701-17800 179 17801-17900 180 17901-18000 181 18001-18100 182 18101-18200 183 18201-18300 184 18301-18400 185 18401-18500 186 18501-18600 187 18601-18700 188 18701-18800 189 18801-18900 190 18901-19000 191 19001-19100 192 19101-19200 193 19201-19300 194 19301-19400 195 19401-19500 196 19501-19600 197 19601-19700 198 19701-19800 199 19801-19900 200 19901-20000 201 20001-20100 202 20101-20200 203 20201-20300 204 20301-20400 205 20401-20500 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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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