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Theorem List for Metamath Proof Explorer - 15801-15900   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theorembitsinv1 15801* There is an explicit inverse to the bits function for nonnegative integers (which can be extended to negative integers using bitscmp 15797), part 1. (Contributed by Mario Carneiro, 7-Sep-2016.)
(𝑁 ∈ ℕ0 → Σ𝑛 ∈ (bits‘𝑁)(2↑𝑛) = 𝑁)

Theorembitsinv2 15802* There is an explicit inverse to the bits function for nonnegative integers, part 2. (Contributed by Mario Carneiro, 8-Sep-2016.)
(𝐴 ∈ (𝒫 ℕ0 ∩ Fin) → (bits‘Σ𝑛𝐴 (2↑𝑛)) = 𝐴)

Theorembitsf1ocnv 15803* The bits function restricted to nonnegative integers is a bijection from the integers to the finite sets of integers. It is in fact the inverse of the Ackermann bijection ackbijnn 15195. (Contributed by Mario Carneiro, 8-Sep-2016.)
((bits ↾ ℕ0):ℕ01-1-onto→(𝒫 ℕ0 ∩ Fin) ∧ (bits ↾ ℕ0) = (𝑥 ∈ (𝒫 ℕ0 ∩ Fin) ↦ Σ𝑛𝑥 (2↑𝑛)))

Theorembitsf1o 15804 The bits function restricted to nonnegative integers is a bijection from the integers to the finite sets of integers. It is in fact the inverse of the Ackermann bijection ackbijnn 15195. (Contributed by Mario Carneiro, 8-Sep-2016.)
(bits ↾ ℕ0):ℕ01-1-onto→(𝒫 ℕ0 ∩ Fin)

Theorembitsf1 15805 The bits function is an injection from to 𝒫 ℕ0. It is obviously not a bijection (by Cantor's theorem canth2 8672), and in fact its range is the set of finite and cofinite subsets of 0. (Contributed by Mario Carneiro, 22-Sep-2016.)
bits:ℤ–1-1→𝒫 ℕ0

Theorem2ebits 15806 The bits of a power of two. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℕ0 → (bits‘(2↑𝑁)) = {𝑁})

Theorembitsinv 15807* The inverse of the bits function. (Contributed by Mario Carneiro, 8-Sep-2016.)
𝐾 = (bits ↾ ℕ0)       (𝐴 ∈ (𝒫 ℕ0 ∩ Fin) → (𝐾𝐴) = Σ𝑘𝐴 (2↑𝑘))

Theorembitsinvp1 15808 Recursive definition of the inverse of the bits function. (Contributed by Mario Carneiro, 8-Sep-2016.)
𝐾 = (bits ↾ ℕ0)       ((𝐴 ⊆ ℕ0𝑁 ∈ ℕ0) → (𝐾‘(𝐴 ∩ (0..^(𝑁 + 1)))) = ((𝐾‘(𝐴 ∩ (0..^𝑁))) + if(𝑁𝐴, (2↑𝑁), 0)))

Theoremsadadd2lem2 15809 The core of the proof of sadadd2 15819. The intuitive justification for this is that cadd is true if at least two arguments are true, and hadd is true if an odd number of arguments are true, so altogether the result is 𝑛 · 𝐴 where 𝑛 is the number of true arguments, which is equivalently obtained by adding together one 𝐴 for each true argument, on the right side. (Contributed by Mario Carneiro, 8-Sep-2016.)
(𝐴 ∈ ℂ → (if(hadd(𝜑, 𝜓, 𝜒), 𝐴, 0) + if(cadd(𝜑, 𝜓, 𝜒), (2 · 𝐴), 0)) = ((if(𝜑, 𝐴, 0) + if(𝜓, 𝐴, 0)) + if(𝜒, 𝐴, 0)))

sadd = (𝑥 ∈ 𝒫 ℕ0, 𝑦 ∈ 𝒫 ℕ0 ↦ {𝑘 ∈ ℕ0 ∣ hadd(𝑘𝑥, 𝑘𝑦, ∅ ∈ (seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝑥, 𝑚𝑦, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘))})

(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑 → (𝐴 sadd 𝐵) = {𝑘 ∈ ℕ0 ∣ hadd(𝑘𝐴, 𝑘𝐵, ∅ ∈ (𝐶𝑘))})

Theoremsadcf 15812* The carry sequence is a sequence of elements of 2o encoding a "sequence of wffs". (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑𝐶:ℕ0⟶2o)

Theoremsadc0 15813* The initial element of the carry sequence is . (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑 → ¬ ∅ ∈ (𝐶‘0))

Theoremsadcp1 15814* The carry sequence (which is a sequence of wffs, encoded as 1o and ) is defined recursively as the carry operation applied to the previous carry and the two current inputs. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (∅ ∈ (𝐶‘(𝑁 + 1)) ↔ cadd(𝑁𝐴, 𝑁𝐵, ∅ ∈ (𝐶𝑁))))

Theoremsadval 15815* The full adder sequence is the half adder function applied to the inputs and the carry sequence. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝑁 ∈ (𝐴 sadd 𝐵) ↔ hadd(𝑁𝐴, 𝑁𝐵, ∅ ∈ (𝐶𝑁))))

(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   𝐾 = (bits ↾ ℕ0)    &   (𝜑 → (∅ ∈ (𝐶𝑁) ↔ (2↑𝑁) ≤ ((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁))))))       (𝜑 → (∅ ∈ (𝐶‘(𝑁 + 1)) ↔ (2↑(𝑁 + 1)) ≤ ((𝐾‘(𝐴 ∩ (0..^(𝑁 + 1)))) + (𝐾‘(𝐵 ∩ (0..^(𝑁 + 1)))))))

Theoremsadcadd 15817* Non-recursive definition of the carry sequence. (Contributed by Mario Carneiro, 8-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   𝐾 = (bits ↾ ℕ0)       (𝜑 → (∅ ∈ (𝐶𝑁) ↔ (2↑𝑁) ≤ ((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁))))))

(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   𝐾 = (bits ↾ ℕ0)    &   (𝜑 → ((𝐾‘((𝐴 sadd 𝐵) ∩ (0..^𝑁))) + if(∅ ∈ (𝐶𝑁), (2↑𝑁), 0)) = ((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁)))))       (𝜑 → ((𝐾‘((𝐴 sadd 𝐵) ∩ (0..^(𝑁 + 1)))) + if(∅ ∈ (𝐶‘(𝑁 + 1)), (2↑(𝑁 + 1)), 0)) = ((𝐾‘(𝐴 ∩ (0..^(𝑁 + 1)))) + (𝐾‘(𝐵 ∩ (0..^(𝑁 + 1))))))

(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   𝐾 = (bits ↾ ℕ0)       (𝜑 → ((𝐾‘((𝐴 sadd 𝐵) ∩ (0..^𝑁))) + if(∅ ∈ (𝐶𝑁), (2↑𝑁), 0)) = ((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁)))))

(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   𝐾 = (bits ↾ ℕ0)       (𝜑 → ((𝐾‘((𝐴 sadd 𝐵) ∩ (0..^𝑁))) mod (2↑𝑁)) = (((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁)))) mod (2↑𝑁)))

Theoremsadcl 15821 The sum of two sequences is a sequence. (Contributed by Mario Carneiro, 5-Sep-2016.)
((𝐴 ⊆ ℕ0𝐵 ⊆ ℕ0) → (𝐴 sadd 𝐵) ⊆ ℕ0)

Theoremsadcom 15822 The adder sequence function is commutative. (Contributed by Mario Carneiro, 5-Sep-2016.)
((𝐴 ⊆ ℕ0𝐵 ⊆ ℕ0) → (𝐴 sadd 𝐵) = (𝐵 sadd 𝐴))

(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑 → (𝐴𝐵) = ∅)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝑁 ∈ (𝐴 sadd 𝐵) ↔ 𝑁 ∈ (𝐴𝐵)))

Theoremsaddisj 15824 The sum of disjoint sequences is the union of the sequences. (In this case, there are no carried bits.) (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑 → (𝐴𝐵) = ∅)       (𝜑 → (𝐴 sadd 𝐵) = (𝐴𝐵))

𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ (bits‘𝐴), 𝑚 ∈ (bits‘𝐵), ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   𝐾 = (bits ↾ ℕ0)    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (((bits‘𝐴) sadd (bits‘𝐵)) ∩ (0..^𝑁)) = (bits‘((𝐴 + 𝐵) mod (2↑𝑁))))

Theoremsadadd 15826 For sequences that correspond to valid integers, the adder sequence function produces the sequence for the sum. This is effectively a proof of the correctness of the ripple carry adder, implemented with logic gates corresponding to df-had 1595 and df-cad 1609.

It is interesting to consider in what sense the sadd function can be said to be "adding" things outside the range of the bits function, that is, when adding sequences that are not eventually constant and so do not denote any integer. The correct interpretation is that the sequences are representations of 2-adic integers, which have a natural ring structure. (Contributed by Mario Carneiro, 9-Sep-2016.)

((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((bits‘𝐴) sadd (bits‘𝐵)) = (bits‘(𝐴 + 𝐵)))

Theoremsadid1 15827 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 15828 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 15831 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 15832 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 15833 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 15834* 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 15835* 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 15836* 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 15837* 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 15838* 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 15839* 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 15840* Define the addition of two bit sequences, using df-had 1595 and df-cad 1609 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 15841* 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 15842* 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 15843 The product of two sequences is a sequence. (Contributed by Mario Carneiro, 19-Sep-2016.)
((𝐴 ⊆ ℕ0𝐵 ⊆ ℕ0) → (𝐴 smul 𝐵) ⊆ ℕ0)

Theoremsmu01lem 15844* Lemma for smu01 15845 and smu02 15846. (Contributed by Mario Carneiro, 19-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   ((𝜑 ∧ (𝑘 ∈ ℕ0𝑛 ∈ ℕ0)) → ¬ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵))       (𝜑 → (𝐴 smul 𝐵) = ∅)

Theoremsmu01 15845 Multiplication of a sequence by 0 on the right. (Contributed by Mario Carneiro, 19-Sep-2016.)
(𝐴 ⊆ ℕ0 → (𝐴 smul ∅) = ∅)

Theoremsmu02 15846 Multiplication of a sequence by 0 on the left. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝐴 ⊆ ℕ0 → (∅ smul 𝐴) = ∅)

Theoremsmupval 15847* 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 15848* Rewrite smupp1 15839 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 15849* 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 15850 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 15851 Lemma for smumul 15852. (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 15852 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 15810, whose correctness is verified in sadadd 15826.

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 15853 Extend the definition of a class to include the greatest common divisor operator.
class gcd

Definitiondf-gcd 15854* Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 28286). For an alternate definition, based on the definition in [ApostolNT] p. 15, see dfgcd2 15904. (Contributed by Paul Chapman, 21-Mar-2011.)
gcd = (𝑥 ∈ ℤ, 𝑦 ∈ ℤ ↦ if((𝑥 = 0 ∧ 𝑦 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛𝑥𝑛𝑦)}, ℝ, < )))

Theoremgcdval 15855* 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 15856 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 15857* 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 15858* Lemma for gcdn0cl 15861, gcddvds 15862 and dvdslegcd 15863. (Contributed by Paul Chapman, 21-Mar-2011.)
𝑆 = {𝑧 ∈ ℤ ∣ ∀𝑛𝐴 𝑧𝑛}       ((𝐴 ⊆ ℤ ∧ ∃𝑛𝐴 𝑛 ≠ 0) → (𝑆 ≠ ∅ ∧ ∃𝑥 ∈ ℤ ∀𝑦𝑆 𝑦𝑥))

Theoremgcdcllem2 15859* Lemma for gcdn0cl 15861, gcddvds 15862 and dvdslegcd 15863. (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 15860* Lemma for gcdn0cl 15861, gcddvds 15862 and dvdslegcd 15863. (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 15861 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 15862 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 15863 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 15864 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 15865 Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)

Theoremgcdnncl 15866 Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)

Theoremgcdcld 15867 Closure of the gcd operator. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝑀 gcd 𝑁) ∈ ℕ0)

Theoremgcd2n0cl 15868 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 15869* 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 15870 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 15871 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 15872 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 15873 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 15874 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 15875 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 15876 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 15877 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 15878 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 15879 The gcd of a nonnegative integer with 0 is itself. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑁 ∈ ℕ0 → (𝑁 gcd 0) = 𝑁)

Theoremgcdneg 15880 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 15881 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 15883 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 15884 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 15885 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 15886 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 15887 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 15888 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 15889 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 15890 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 15891 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)

Theoremgcdmultipled 15892 The greatest common divisor of a nonnegative integer 𝑀 and a multiple of it is 𝑀 itself. (Contributed by Rohan Ridenour, 3-Aug-2023.)
(𝜑𝑀 ∈ ℕ0)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝑀 gcd (𝑁 · 𝑀)) = 𝑀)

Theoremgcdmultiplez 15893 The GCD of a multiple of an integer is the integer itself. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) (Proof shortened by AV, 12-Jan-2023.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)

Theoremgcdmultiple 15894 The GCD of a multiple of a positive integer is the positive integer itself. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) (Proof shortened by AV, 12-Jan-2023.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)

Theoremdvdsgcdidd 15895 The greatest common divisor of a positive integer and another integer it divides is itself. (Contributed by Rohan Ridenour, 3-Aug-2023.)
(𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑𝑀𝑁)       (𝜑 → (𝑀 gcd 𝑁) = 𝑀)

Theorem6gcd4e2 15896 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 15897* Lemma for bezout 15901. (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 15898* Lemma for bezout 15901. (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 15899* Lemma for bezout 15901. (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 15900* Lemma for bezout 15901. (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)

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