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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | bitsp1 12701 | The 𝑀 + 1-th bit of 𝑁 is the 𝑀-th bit of ⌊(𝑁 / 2). (Contributed by Mario Carneiro, 5-Sep-2016.) |
| ⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → ((𝑀 + 1) ∈ (bits‘𝑁) ↔ 𝑀 ∈ (bits‘(⌊‘(𝑁 / 2))))) | ||
| Theorem | bitsp1e 12702 | The 𝑀 + 1-th bit of 2𝑁 is the 𝑀-th bit of 𝑁. (Contributed by Mario Carneiro, 5-Sep-2016.) |
| ⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → ((𝑀 + 1) ∈ (bits‘(2 · 𝑁)) ↔ 𝑀 ∈ (bits‘𝑁))) | ||
| Theorem | bitsp1o 12703 | The 𝑀 + 1-th bit of 2𝑁 + 1 is the 𝑀-th bit of 𝑁. (Contributed by Mario Carneiro, 5-Sep-2016.) |
| ⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → ((𝑀 + 1) ∈ (bits‘((2 · 𝑁) + 1)) ↔ 𝑀 ∈ (bits‘𝑁))) | ||
| Theorem | bitsfzolem 12704* | Lemma for bitsfzo 12705. (Contributed by Mario Carneiro, 5-Sep-2016.) (Revised by AV, 1-Oct-2020.) |
| ⊢ (𝜑 → 𝑁 ∈ ℕ0) & ⊢ (𝜑 → 𝑀 ∈ ℕ0) & ⊢ (𝜑 → (bits‘𝑁) ⊆ (0..^𝑀)) & ⊢ 𝑆 = inf({𝑛 ∈ ℕ0 ∣ 𝑁 < (2↑𝑛)}, ℝ, < ) ⇒ ⊢ (𝜑 → 𝑁 ∈ (0..^(2↑𝑀))) | ||
| Theorem | bitsfzo 12705 | The bits of a number are all at positions less than 𝑀 iff the number is nonnegative and less than 2↑𝑀. (Contributed by Mario Carneiro, 5-Sep-2016.) (Proof shortened by AV, 1-Oct-2020.) |
| ⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → (𝑁 ∈ (0..^(2↑𝑀)) ↔ (bits‘𝑁) ⊆ (0..^𝑀))) | ||
| Theorem | bitsmod 12706 | Truncating the bit sequence after some 𝑀 is equivalent to reducing the argument mod 2↑𝑀. (Contributed by Mario Carneiro, 6-Sep-2016.) |
| ⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → (bits‘(𝑁 mod (2↑𝑀))) = ((bits‘𝑁) ∩ (0..^𝑀))) | ||
| Theorem | bitsfi 12707 | Every number is associated with a finite set of bits. (Contributed by Mario Carneiro, 5-Sep-2016.) |
| ⊢ (𝑁 ∈ ℕ0 → (bits‘𝑁) ∈ Fin) | ||
| Theorem | bitscmp 12708 | The bit complement of 𝑁 is -𝑁 − 1. (Thus, by bitsfi 12707, all negative numbers have cofinite bits representations.) (Contributed by Mario Carneiro, 5-Sep-2016.) |
| ⊢ (𝑁 ∈ ℤ → (ℕ0 ∖ (bits‘𝑁)) = (bits‘(-𝑁 − 1))) | ||
| Theorem | 0bits 12709 | The bits of zero. (Contributed by Mario Carneiro, 6-Sep-2016.) |
| ⊢ (bits‘0) = ∅ | ||
| Theorem | m1bits 12710 | The bits of negative one. (Contributed by Mario Carneiro, 5-Sep-2016.) |
| ⊢ (bits‘-1) = ℕ0 | ||
| Theorem | bitsinv1lem 12711 | Lemma for bitsinv1 12712. (Contributed by Mario Carneiro, 22-Sep-2016.) |
| ⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → (𝑁 mod (2↑(𝑀 + 1))) = ((𝑁 mod (2↑𝑀)) + if(𝑀 ∈ (bits‘𝑁), (2↑𝑀), 0))) | ||
| Theorem | bitsinv1 12712* | There is an explicit inverse to the bits function for nonnegative integers (which can be extended to negative integers using bitscmp 12708), part 1. (Contributed by Mario Carneiro, 7-Sep-2016.) |
| ⊢ (𝑁 ∈ ℕ0 → Σ𝑛 ∈ (bits‘𝑁)(2↑𝑛) = 𝑁) | ||
| Syntax | cgcd 12713 | Extend the definition of a class to include the greatest common divisor operator. |
| class gcd | ||
| Definition | df-gcd 12714* | Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 16728). (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ gcd = (𝑥 ∈ ℤ, 𝑦 ∈ ℤ ↦ if((𝑥 = 0 ∧ 𝑦 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦)}, ℝ, < ))) | ||
| Theorem | gcdmndc 12715 | Decidablity lemma used in various proofs related to gcd. (Contributed by Jim Kingdon, 12-Dec-2021.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → DECID (𝑀 = 0 ∧ 𝑁 = 0)) | ||
| Theorem | dvdsbnd 12716* | There is an upper bound to the divisors of a nonzero integer. (Contributed by Jim Kingdon, 11-Dec-2021.) |
| ⊢ ((𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) → ∃𝑛 ∈ ℕ ∀𝑚 ∈ (ℤ≥‘𝑛) ¬ 𝑚 ∥ 𝐴) | ||
| Theorem | gcdsupex 12717* | Existence of the supremum used in defining gcd. (Contributed by Jim Kingdon, 12-Dec-2021.) |
| ⊢ (((𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ) ∧ ¬ (𝑋 = 0 ∧ 𝑌 = 0)) → ∃𝑥 ∈ ℤ (∀𝑦 ∈ {𝑛 ∈ ℤ ∣ (𝑛 ∥ 𝑋 ∧ 𝑛 ∥ 𝑌)} ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧 ∈ {𝑛 ∈ ℤ ∣ (𝑛 ∥ 𝑋 ∧ 𝑛 ∥ 𝑌)}𝑦 < 𝑧))) | ||
| Theorem | gcdsupcl 12718* | Closure of the supremum used in defining gcd. A lemma for gcdval 12719 and gcdn0cl 12722. (Contributed by Jim Kingdon, 11-Dec-2021.) |
| ⊢ (((𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ) ∧ ¬ (𝑋 = 0 ∧ 𝑌 = 0)) → sup({𝑛 ∈ ℤ ∣ (𝑛 ∥ 𝑋 ∧ 𝑛 ∥ 𝑌)}, ℝ, < ) ∈ ℕ) | ||
| Theorem | gcdval 12719* | 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 ∣ (𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁)}, ℝ, < ))) | ||
| Theorem | gcd0val 12720 | The value, by convention, of the gcd operator when both operands are 0. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ (0 gcd 0) = 0 | ||
| Theorem | gcdn0val 12721* | 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 ∣ (𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁)}, ℝ, < )) | ||
| Theorem | gcdn0cl 12722 | Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → (𝑀 gcd 𝑁) ∈ ℕ) | ||
| Theorem | gcddvds 12723 | 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)) | ||
| Theorem | dvdslegcd 12724 | 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)) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ≤ (𝑀 gcd 𝑁))) | ||
| Theorem | nndvdslegcd 12725 | 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) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ≤ (𝑀 gcd 𝑁))) | ||
| Theorem | gcdcl 12726 | Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0) | ||
| Theorem | gcdnncl 12727 | Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.) |
| ⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ) | ||
| Theorem | gcdcld 12728 | Closure of the gcd operator. (Contributed by Mario Carneiro, 29-May-2016.) |
| ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) ⇒ ⊢ (𝜑 → (𝑀 gcd 𝑁) ∈ ℕ0) | ||
| Theorem | gcd2n0cl 12729 | 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) | ||
| Theorem | zeqzmulgcd 12730* | 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))) | ||
| Theorem | divgcdz 12731 | 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) | ||
| Theorem | gcdf 12732 | 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 | ||
| Theorem | gcdcom 12733 | 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)) | ||
| Theorem | gcdcomd 12734 | The gcd operator is commutative, deduction version. (Contributed by SN, 24-Aug-2024.) |
| ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) ⇒ ⊢ (𝜑 → (𝑀 gcd 𝑁) = (𝑁 gcd 𝑀)) | ||
| Theorem | divgcdnn 12735 | 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) | ||
| Theorem | divgcdnnr 12736 | 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) | ||
| Theorem | gcdeq0 12737 | 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))) | ||
| Theorem | gcdn0gt0 12738 | 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))) | ||
| Theorem | gcd0id 12739 | 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)) | ||
| Theorem | gcdid0 12740 | 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)) | ||
| Theorem | nn0gcdid0 12741 | The gcd of a nonnegative integer with 0 is itself. (Contributed by Paul Chapman, 31-Mar-2011.) |
| ⊢ (𝑁 ∈ ℕ0 → (𝑁 gcd 0) = 𝑁) | ||
| Theorem | gcdneg 12742 | 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)) | ||
| Theorem | neggcd 12743 | 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)) | ||
| Theorem | gcdaddm 12744 | 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)))) | ||
| Theorem | gcdadd 12745 | 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))) | ||
| Theorem | gcdid 12746 | 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)) | ||
| Theorem | gcd1 12747 | 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) | ||
| Theorem | gcdabs 12748 | 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)) | ||
| Theorem | gcdabs1 12749 | 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)) | ||
| Theorem | gcdabs2 12750 | 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)) | ||
| Theorem | modgcd 12751 | 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)) | ||
| Theorem | 1gcd 12752 | 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) | ||
| Theorem | gcdmultipled 12753 | The greatest common divisor of a nonnegative integer 𝑀 and a multiple of it is 𝑀 itself. (Contributed by Rohan Ridenour, 3-Aug-2023.) |
| ⊢ (𝜑 → 𝑀 ∈ ℕ0) & ⊢ (𝜑 → 𝑁 ∈ ℤ) ⇒ ⊢ (𝜑 → (𝑀 gcd (𝑁 · 𝑀)) = 𝑀) | ||
| Theorem | dvdsgcdidd 12754 | The greatest common divisor of a positive integer and another integer it divides is itself. (Contributed by Rohan Ridenour, 3-Aug-2023.) |
| ⊢ (𝜑 → 𝑀 ∈ ℕ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) & ⊢ (𝜑 → 𝑀 ∥ 𝑁) ⇒ ⊢ (𝜑 → (𝑀 gcd 𝑁) = 𝑀) | ||
| Theorem | 6gcd4e2 12755 | 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 | ||
| Theorem | bezoutlemnewy 12756* | Lemma for Bézout's identity. The is-bezout predicate holds for (𝑦 mod 𝑊). (Contributed by Jim Kingdon, 6-Jan-2022.) |
| ⊢ (𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡))) & ⊢ (𝜃 → 𝐴 ∈ ℕ0) & ⊢ (𝜃 → 𝐵 ∈ ℕ0) & ⊢ (𝜃 → 𝑊 ∈ ℕ) & ⊢ (𝜃 → [𝑦 / 𝑟]𝜑) & ⊢ (𝜃 → 𝑦 ∈ ℕ0) & ⊢ (𝜃 → [𝑊 / 𝑟]𝜑) ⇒ ⊢ (𝜃 → [(𝑦 mod 𝑊) / 𝑟]𝜑) | ||
| Theorem | bezoutlemstep 12757* | Lemma for Bézout's identity. This is the induction step for the proof by induction. (Contributed by Jim Kingdon, 3-Jan-2022.) |
| ⊢ (𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡))) & ⊢ (𝜃 → 𝐴 ∈ ℕ0) & ⊢ (𝜃 → 𝐵 ∈ ℕ0) & ⊢ (𝜃 → 𝑊 ∈ ℕ) & ⊢ (𝜃 → [𝑦 / 𝑟]𝜑) & ⊢ (𝜃 → 𝑦 ∈ ℕ0) & ⊢ (𝜃 → [𝑊 / 𝑟]𝜑) & ⊢ (𝜓 ↔ ∀𝑧 ∈ ℕ0 (𝑧 ∥ 𝑟 → (𝑧 ∥ 𝑥 ∧ 𝑧 ∥ 𝑦))) & ⊢ ((𝜃 ∧ [(𝑦 mod 𝑊) / 𝑟]𝜑) → ∃𝑟 ∈ ℕ0 ([(𝑦 mod 𝑊) / 𝑥][𝑊 / 𝑦]𝜓 ∧ 𝜑)) & ⊢ Ⅎ𝑥𝜃 & ⊢ Ⅎ𝑟𝜃 ⇒ ⊢ (𝜃 → ∃𝑟 ∈ ℕ0 ([𝑊 / 𝑥]𝜓 ∧ 𝜑)) | ||
| Theorem | bezoutlemmain 12758* | Lemma for Bézout's identity. This is the main result which we prove by induction and which represents the application of the Extended Euclidean algorithm. (Contributed by Jim Kingdon, 30-Dec-2021.) |
| ⊢ (𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡))) & ⊢ (𝜓 ↔ ∀𝑧 ∈ ℕ0 (𝑧 ∥ 𝑟 → (𝑧 ∥ 𝑥 ∧ 𝑧 ∥ 𝑦))) & ⊢ (𝜃 → 𝐴 ∈ ℕ0) & ⊢ (𝜃 → 𝐵 ∈ ℕ0) ⇒ ⊢ (𝜃 → ∀𝑥 ∈ ℕ0 ([𝑥 / 𝑟]𝜑 → ∀𝑦 ∈ ℕ0 ([𝑦 / 𝑟]𝜑 → ∃𝑟 ∈ ℕ0 (𝜓 ∧ 𝜑)))) | ||
| Theorem | bezoutlema 12759* | Lemma for Bézout's identity. The is-bezout condition is satisfied by 𝐴. (Contributed by Jim Kingdon, 30-Dec-2021.) |
| ⊢ (𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡))) & ⊢ (𝜃 → 𝐴 ∈ ℕ0) & ⊢ (𝜃 → 𝐵 ∈ ℕ0) ⇒ ⊢ (𝜃 → [𝐴 / 𝑟]𝜑) | ||
| Theorem | bezoutlemb 12760* | Lemma for Bézout's identity. The is-bezout condition is satisfied by 𝐵. (Contributed by Jim Kingdon, 30-Dec-2021.) |
| ⊢ (𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡))) & ⊢ (𝜃 → 𝐴 ∈ ℕ0) & ⊢ (𝜃 → 𝐵 ∈ ℕ0) ⇒ ⊢ (𝜃 → [𝐵 / 𝑟]𝜑) | ||
| Theorem | bezoutlemex 12761* | Lemma for Bézout's identity. Existence of a number which we will later show to be the greater common divisor and its decomposition into cofactors. (Contributed by Mario Carneiro and Jim Kingdon, 3-Jan-2022.) |
| ⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℕ0 (𝑧 ∥ 𝑑 → (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦)))) | ||
| Theorem | bezoutlemzz 12762* | Lemma for Bézout's identity. Like bezoutlemex 12761 but where ' z ' is any integer, not just a nonnegative one. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.) |
| ⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 → (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦)))) | ||
| Theorem | bezoutlemaz 12763* | Lemma for Bézout's identity. Like bezoutlemzz 12762 but where ' A ' can be any integer, not just a nonnegative one. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.) |
| ⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ0) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 → (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦)))) | ||
| Theorem | bezoutlembz 12764* | Lemma for Bézout's identity. Like bezoutlemaz 12763 but where ' B ' can be any integer, not just a nonnegative one. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.) |
| ⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 → (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦)))) | ||
| Theorem | bezoutlembi 12765* | Lemma for Bézout's identity. Like bezoutlembz 12764 but the greatest common divisor condition is a biconditional, not just an implication. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.) |
| ⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦)))) | ||
| Theorem | bezoutlemmo 12766* | Lemma for Bézout's identity. There is at most one nonnegative integer meeting the greatest common divisor condition. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.) |
| ⊢ (𝜑 → 𝐴 ∈ ℤ) & ⊢ (𝜑 → 𝐵 ∈ ℤ) & ⊢ (𝜑 → 𝐷 ∈ ℕ0) & ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) & ⊢ (𝜑 → 𝐸 ∈ ℕ0) & ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐸 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) ⇒ ⊢ (𝜑 → 𝐷 = 𝐸) | ||
| Theorem | bezoutlemeu 12767* | Lemma for Bézout's identity. There is exactly one nonnegative integer meeting the greatest common divisor condition. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.) |
| ⊢ (𝜑 → 𝐴 ∈ ℤ) & ⊢ (𝜑 → 𝐵 ∈ ℤ) & ⊢ (𝜑 → 𝐷 ∈ ℕ0) & ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) ⇒ ⊢ (𝜑 → ∃!𝑑 ∈ ℕ0 ∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) | ||
| Theorem | bezoutlemle 12768* | Lemma for Bézout's identity. The number satisfying the greatest common divisor condition is the largest number which divides both 𝐴 and 𝐵. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.) |
| ⊢ (𝜑 → 𝐴 ∈ ℤ) & ⊢ (𝜑 → 𝐵 ∈ ℤ) & ⊢ (𝜑 → 𝐷 ∈ ℕ0) & ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) & ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0)) ⇒ ⊢ (𝜑 → ∀𝑧 ∈ ℤ ((𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵) → 𝑧 ≤ 𝐷)) | ||
| Theorem | bezoutlemsup 12769* | Lemma for Bézout's identity. The number satisfying the greatest common divisor condition is the supremum of divisors of both 𝐴 and 𝐵. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.) |
| ⊢ (𝜑 → 𝐴 ∈ ℤ) & ⊢ (𝜑 → 𝐵 ∈ ℤ) & ⊢ (𝜑 → 𝐷 ∈ ℕ0) & ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) & ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0)) ⇒ ⊢ (𝜑 → 𝐷 = sup({𝑧 ∈ ℤ ∣ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)}, ℝ, < )) | ||
| Theorem | dfgcd3 12770* | Alternate definition of the gcd operator. (Contributed by Jim Kingdon, 31-Dec-2021.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (℩𝑑 ∈ ℕ0 ∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 ↔ (𝑧 ∥ 𝑀 ∧ 𝑧 ∥ 𝑁)))) | ||
| Theorem | bezout 12771* |
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.
The proof is constructive, in the sense that it applies the Extended Euclidian Algorithm to constuct a number which can be shown to be (𝐴 gcd 𝐵) and which satisfies the rest of the theorem. In the presence of excluded middle, it is common to prove Bézout's identity by taking the smallest number which satisfies the Bézout condition, and showing it is the greatest common divisor. But we do not have the ability to show that number exists other than by providing a way to determine it. (Contributed by Mario Carneiro, 22-Feb-2014.) |
| ⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ (𝐴 gcd 𝐵) = ((𝐴 · 𝑥) + (𝐵 · 𝑦))) | ||
| Theorem | dvdsgcd 12772 | 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) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 gcd 𝑁))) | ||
| Theorem | dvdsgcdb 12773 | Biconditional form of dvdsgcd 12772. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) ↔ 𝐾 ∥ (𝑀 gcd 𝑁))) | ||
| Theorem | dfgcd2 12774* | 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 ∧ (𝐷 ∥ 𝑀 ∧ 𝐷 ∥ 𝑁) ∧ ∀𝑒 ∈ ℤ ((𝑒 ∥ 𝑀 ∧ 𝑒 ∥ 𝑁) → 𝑒 ∥ 𝐷)))) | ||
| Theorem | gcdass 12775 | 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))) | ||
| Theorem | mulgcd 12776 | Distribute multiplication by a nonnegative integer over gcd. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Mario Carneiro, 30-May-2014.) |
| ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) gcd (𝐾 · 𝑁)) = (𝐾 · (𝑀 gcd 𝑁))) | ||
| Theorem | absmulgcd 12777 | 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)))) | ||
| Theorem | mulgcdr 12778 | 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)) | ||
| Theorem | gcddiv 12779 | Division law for GCD. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ) ∧ (𝐶 ∥ 𝐴 ∧ 𝐶 ∥ 𝐵)) → ((𝐴 gcd 𝐵) / 𝐶) = ((𝐴 / 𝐶) gcd (𝐵 / 𝐶))) | ||
| Theorem | gcdmultiple 12780 | 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) | ||
| Theorem | gcdmultiplez 12781 | Extend gcdmultiple 12780 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) | ||
| Theorem | gcdzeq 12782 | A positive integer 𝐴 is equal to its gcd with an integer 𝐵 if and only if 𝐴 divides 𝐵. Generalization of gcdeq 12783. (Contributed by AV, 1-Jul-2020.) |
| ⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵) = 𝐴 ↔ 𝐴 ∥ 𝐵)) | ||
| Theorem | gcdeq 12783 | 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) = 𝐴 ↔ 𝐴 ∥ 𝐵)) | ||
| Theorem | dvdssqim 12784 | Unidirectional form of dvdssq 12791. (Contributed by Scott Fenton, 19-Apr-2014.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝑀↑2) ∥ (𝑁↑2))) | ||
| Theorem | dvdsmulgcd 12785 | Relationship between the order of an element and that of a multiple. (a divisibility equivalent). (Contributed by Stefan O'Rear, 6-Sep-2015.) |
| ⊢ ((𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) → (𝐴 ∥ (𝐵 · 𝐶) ↔ 𝐴 ∥ (𝐵 · (𝐶 gcd 𝐴)))) | ||
| Theorem | rpmulgcd 12786 | 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)) | ||
| Theorem | rplpwr 12787 | If 𝐴 and 𝐵 are relatively prime, then so are 𝐴↑𝑁 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 → ((𝐴↑𝑁) gcd 𝐵) = 1)) | ||
| Theorem | rppwr 12788 | If 𝐴 and 𝐵 are relatively prime, then so are 𝐴↑𝑁 and 𝐵↑𝑁. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → ((𝐴↑𝑁) gcd (𝐵↑𝑁)) = 1)) | ||
| Theorem | sqgcd 12789 | 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))) | ||
| Theorem | dvdssqlem 12790 | Lemma for dvdssq 12791. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 ∥ 𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2))) | ||
| Theorem | dvdssq 12791 | 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) → (𝑀 ∥ 𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2))) | ||
| Theorem | bezoutr 12792 | Partial converse to bezout 12771. 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))) | ||
| Theorem | bezoutr1 12793 | Converse of bezout 12771 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)) | ||
| Theorem | nnmindc 12794* | An inhabited decidable subset of the natural numbers has a minimum. (Contributed by Jim Kingdon, 23-Sep-2024.) |
| ⊢ ((𝐴 ⊆ ℕ ∧ ∀𝑥 ∈ ℕ DECID 𝑥 ∈ 𝐴 ∧ ∃𝑦 𝑦 ∈ 𝐴) → inf(𝐴, ℝ, < ) ∈ 𝐴) | ||
| Theorem | nnminle 12795* | The infimum of a decidable subset of the natural numbers is less than an element of the set. The infimum is also a minimum as shown at nnmindc 12794. (Contributed by Jim Kingdon, 26-Sep-2024.) |
| ⊢ ((𝐴 ⊆ ℕ ∧ ∀𝑥 ∈ ℕ DECID 𝑥 ∈ 𝐴 ∧ 𝐵 ∈ 𝐴) → inf(𝐴, ℝ, < ) ≤ 𝐵) | ||
| Theorem | nnwodc 12796* | Well-ordering principle: any inhabited decidable set of positive integers has a least element. Theorem I.37 (well-ordering principle) of [Apostol] p. 34. (Contributed by NM, 17-Aug-2001.) (Revised by Jim Kingdon, 23-Oct-2024.) |
| ⊢ ((𝐴 ⊆ ℕ ∧ ∃𝑤 𝑤 ∈ 𝐴 ∧ ∀𝑗 ∈ ℕ DECID 𝑗 ∈ 𝐴) → ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 𝑥 ≤ 𝑦) | ||
| Theorem | uzwodc 12797* | Well-ordering principle: any inhabited decidable subset of an upper set of integers has a least element. (Contributed by NM, 8-Oct-2005.) (Revised by Jim Kingdon, 22-Oct-2024.) |
| ⊢ ((𝑆 ⊆ (ℤ≥‘𝑀) ∧ ∃𝑥 𝑥 ∈ 𝑆 ∧ ∀𝑥 ∈ (ℤ≥‘𝑀)DECID 𝑥 ∈ 𝑆) → ∃𝑗 ∈ 𝑆 ∀𝑘 ∈ 𝑆 𝑗 ≤ 𝑘) | ||
| Theorem | nnwofdc 12798* | Well-ordering principle: any inhabited decidable set of positive integers has a least element. This version allows 𝑥 and 𝑦 to be present in 𝐴 as long as they are effectively not free. (Contributed by NM, 17-Aug-2001.) (Revised by Mario Carneiro, 15-Oct-2016.) |
| ⊢ Ⅎ𝑥𝐴 & ⊢ Ⅎ𝑦𝐴 ⇒ ⊢ ((𝐴 ⊆ ℕ ∧ ∃𝑧 𝑧 ∈ 𝐴 ∧ ∀𝑗 ∈ ℕ DECID 𝑗 ∈ 𝐴) → ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 𝑥 ≤ 𝑦) | ||
| Theorem | nnwosdc 12799* | Well-ordering principle: any inhabited decidable set of positive integers has a least element (schema form). (Contributed by NM, 17-Aug-2001.) (Revised by Jim Kingdon, 25-Oct-2024.) |
| ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓)) ⇒ ⊢ ((∃𝑥 ∈ ℕ 𝜑 ∧ ∀𝑥 ∈ ℕ DECID 𝜑) → ∃𝑥 ∈ ℕ (𝜑 ∧ ∀𝑦 ∈ ℕ (𝜓 → 𝑥 ≤ 𝑦))) | ||
| Theorem | nninfctlemfo 12800* | Lemma for nninfct 12801. (Contributed by Jim Kingdon, 10-Jul-2025.) |
| ⊢ 𝐺 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0) & ⊢ 𝐹 = (𝑛 ∈ ω ↦ (𝑖 ∈ ω ↦ if(𝑖 ∈ 𝑛, 1o, ∅))) & ⊢ 𝐼 = ((𝐹 ∘ ◡𝐺) ∪ {〈+∞, (ω × {1o})〉}) ⇒ ⊢ (ω ∈ Omni → 𝐼:ℕ0*–onto→ℕ∞) | ||
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