P. 164 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 16301-16400   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	qnnen 16301	The rational numbers are countable. This proof does not use the Axiom of Choice, even though it uses an onto function, because the base set (ℤ × ℕ) is numerable. Exercise 2 of [Enderton] p. 133. For purposes of the Metamath 100 list, we are considering Mario Carneiro's revision as the date this proof was completed. This is Metamath 100 proof #3. (Contributed by NM, 31-Jul-2004.) (Revised by Mario Carneiro, 3-Mar-2013.)
⊢ ℚ ≈ ℕ
5.12.2  The reals are uncountable
Theorem	rpnnen2lem1 16302*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    ⇒   ⊢ ((𝐴 ⊆ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐹‘𝐴)‘𝑁) = if(𝑁 ∈ 𝐴, ((1 / 3)↑𝑁), 0))
Theorem	rpnnen2lem2 16303*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 23-Aug-2014.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    ⇒   ⊢ (𝐴 ⊆ ℕ → (𝐹‘𝐴):ℕ⟶ℝ)
Theorem	rpnnen2lem3 16304*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    ⇒   ⊢ seq1( + , (𝐹‘ℕ)) ⇝ (1 / 2)
Theorem	rpnnen2lem4 16305*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 31-Aug-2014.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    ⇒   ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ ℕ ∧ 𝑘 ∈ ℕ) → (0 ≤ ((𝐹‘𝐴)‘𝑘) ∧ ((𝐹‘𝐴)‘𝑘) ≤ ((𝐹‘𝐵)‘𝑘)))
Theorem	rpnnen2lem5 16306*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    ⇒   ⊢ ((𝐴 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → seq𝑀( + , (𝐹‘𝐴)) ∈ dom ⇝ )
Theorem	rpnnen2lem6 16307*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    ⇒   ⊢ ((𝐴 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → Σ𝑘 ∈ (ℤ≥‘𝑀)((𝐹‘𝐴)‘𝑘) ∈ ℝ)
Theorem	rpnnen2lem7 16308*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    ⇒   ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → Σ𝑘 ∈ (ℤ≥‘𝑀)((𝐹‘𝐴)‘𝑘) ≤ Σ𝑘 ∈ (ℤ≥‘𝑀)((𝐹‘𝐵)‘𝑘))
Theorem	rpnnen2lem8 16309*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    ⇒   ⊢ ((𝐴 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → Σ𝑘 ∈ ℕ ((𝐹‘𝐴)‘𝑘) = (Σ𝑘 ∈ (1...(𝑀 − 1))((𝐹‘𝐴)‘𝑘) + Σ𝑘 ∈ (ℤ≥‘𝑀)((𝐹‘𝐴)‘𝑘)))
Theorem	rpnnen2lem9 16310*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    ⇒   ⊢ (𝑀 ∈ ℕ → Σ𝑘 ∈ (ℤ≥‘𝑀)((𝐹‘(ℕ ∖ {𝑀}))‘𝑘) = (0 + (((1 / 3)↑(𝑀 + 1)) / (1 − (1 / 3)))))
Theorem	rpnnen2lem10 16311*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    &   ⊢ (𝜑 → 𝐴 ⊆ ℕ)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ)    &   ⊢ (𝜑 → 𝑚 ∈ (𝐴 ∖ 𝐵))    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ (𝑛 < 𝑚 → (𝑛 ∈ 𝐴 ↔ 𝑛 ∈ 𝐵)))    &   ⊢ (𝜓 ↔ Σ𝑘 ∈ ℕ ((𝐹‘𝐴)‘𝑘) = Σ𝑘 ∈ ℕ ((𝐹‘𝐵)‘𝑘))    ⇒   ⊢ ((𝜑 ∧ 𝜓) → Σ𝑘 ∈ (ℤ≥‘𝑚)((𝐹‘𝐴)‘𝑘) = Σ𝑘 ∈ (ℤ≥‘𝑚)((𝐹‘𝐵)‘𝑘))
Theorem	rpnnen2lem11 16312*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    &   ⊢ (𝜑 → 𝐴 ⊆ ℕ)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ)    &   ⊢ (𝜑 → 𝑚 ∈ (𝐴 ∖ 𝐵))    &   ⊢ (𝜑 → ∀𝑛 ∈ ℕ (𝑛 < 𝑚 → (𝑛 ∈ 𝐴 ↔ 𝑛 ∈ 𝐵)))    &   ⊢ (𝜓 ↔ Σ𝑘 ∈ ℕ ((𝐹‘𝐴)‘𝑘) = Σ𝑘 ∈ ℕ ((𝐹‘𝐵)‘𝑘))    ⇒   ⊢ (𝜑 → ¬ 𝜓)
Theorem	rpnnen2lem12 16313*	Lemma for rpnnen2 16314. (Contributed by Mario Carneiro, 13-May-2013.)
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0)))    ⇒   ⊢ 𝒫 ℕ ≼ (0[,]1)
Theorem	rpnnen2 16314	The other half of rpnnen 16315, where we show an injection from sets of positive integers to real numbers. The obvious choice for this is binary expansion, but it has the unfortunate property that it does not produce an injection on numbers which end with all 0's or all 1's (the more well-known decimal version of this is 0.999... 15970). Instead, we opt for a ternary expansion, which produces (a scaled version of) the Cantor set. Since the Cantor set is riddled with gaps, we can show that any two sequences that are not equal must differ somewhere, and when they do, they are placed a finite distance apart, thus ensuring that the map is injective. Our map assigns to each subset 𝐴 of the positive integers the number Σ𝑘 ∈ 𝐴(3↑-𝑘) = Σ𝑘 ∈ ℕ((𝐹‘𝐴)‘𝑘), where ((𝐹‘𝐴)‘𝑘) = if(𝑘 ∈ 𝐴, (3↑-𝑘), 0)) (rpnnen2lem1 16302). This is an infinite sum of real numbers (rpnnen2lem2 16303), and since 𝐴 ⊆ 𝐵 implies (𝐹‘𝐴) ≤ (𝐹‘𝐵) (rpnnen2lem4 16305) and (𝐹‘ℕ) converges to 1 / 2 (rpnnen2lem3 16304) by geoisum1 15968, the sum is convergent to some real (rpnnen2lem5 16306 and rpnnen2lem6 16307) by the comparison test for convergence cvgcmp 15903. The comparison test also tells us that 𝐴 ⊆ 𝐵 implies Σ(𝐹‘𝐴) ≤ Σ(𝐹‘𝐵) (rpnnen2lem7 16308). Putting it all together, if we have two sets 𝑥 ≠ 𝑦, there must differ somewhere, and so there must be an 𝑚 such that ∀𝑛 < 𝑚(𝑛 ∈ 𝑥 ↔ 𝑛 ∈ 𝑦) but 𝑚 ∈ (𝑥 ∖ 𝑦) or vice versa. In this case, we split off the first 𝑚 − 1 terms (rpnnen2lem8 16309) and cancel them (rpnnen2lem10 16311), since these are the same for both sets. For the remaining terms, we use the subset property to establish that Σ(𝐹‘𝑦) ≤ Σ(𝐹‘(ℕ ∖ {𝑚})) and Σ(𝐹‘{𝑚}) ≤ Σ(𝐹‘𝑥) (where these sums are only over (ℤ≥‘𝑚)), and since Σ(𝐹‘(ℕ ∖ {𝑚})) = (3↑-𝑚) / 2 (rpnnen2lem9 16310) and Σ(𝐹‘{𝑚}) = (3↑-𝑚), we establish that Σ(𝐹‘𝑦) < Σ(𝐹‘𝑥) (rpnnen2lem11 16312) so that they must be different. By contraposition (rpnnen2lem12 16313), we find that this map is an injection. (Contributed by Mario Carneiro, 13-May-2013.) (Proof shortened by Mario Carneiro, 30-Apr-2014.) (Revised by NM, 17-Aug-2021.)
⊢ 𝒫 ℕ ≼ (0[,]1)
Theorem	rpnnen 16315	The cardinality of the continuum is the same as the powerset of ω. This is a stronger statement than ruc 16331, which only asserts that ℝ is uncountable, i.e. has a cardinality larger than ω. The main proof is in two parts, rpnnen1 13033 and rpnnen2 16314, each showing an injection in one direction, and this last part uses sbth 9095 to prove that the sets are equinumerous. By constructing explicit injections, we avoid the use of AC. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 23-Aug-2014.)
⊢ ℝ ≈ 𝒫 ℕ
Theorem	rexpen 16316	The real numbers are equinumerous to their own Cartesian product, even though it is not necessarily true that ℝ is well-orderable (so we cannot use infxpidm2 10020 directly). (Contributed by NM, 30-Jul-2004.) (Revised by Mario Carneiro, 16-Jun-2013.)
⊢ (ℝ × ℝ) ≈ ℝ
Theorem	cpnnen 16317	The complex numbers are equinumerous to the powerset of the positive integers. (Contributed by Mario Carneiro, 16-Jun-2013.)
⊢ ℂ ≈ 𝒫 ℕ
Theorem	rucALT 16318	Alternate proof of ruc 16331. This proof is a simple corollary of rpnnen 16315, which determines the exact cardinality of the reals. For an alternate proof discussed at mmcomplex.html#uncountable 16315, see ruc 16331. (Contributed by NM, 13-Oct-2004.) (Revised by Mario Carneiro, 13-May-2013.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ℕ ≺ ℝ
Theorem	ruclem1 16319*	Lemma for ruc 16331 (the reals are uncountable). Substitutions for the function 𝐷. (Contributed by Mario Carneiro, 28-May-2014.) (Revised by Fan Zheng, 6-Jun-2016.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ)    &   ⊢ (𝜑 → 𝑀 ∈ ℝ)    &   ⊢ 𝑋 = (1st ‘(⟨𝐴, 𝐵⟩𝐷𝑀))    &   ⊢ 𝑌 = (2nd ‘(⟨𝐴, 𝐵⟩𝐷𝑀))    ⇒   ⊢ (𝜑 → ((⟨𝐴, 𝐵⟩𝐷𝑀) ∈ (ℝ × ℝ) ∧ 𝑋 = if(((𝐴 + 𝐵) / 2) < 𝑀, 𝐴, ((((𝐴 + 𝐵) / 2) + 𝐵) / 2)) ∧ 𝑌 = if(((𝐴 + 𝐵) / 2) < 𝑀, ((𝐴 + 𝐵) / 2), 𝐵)))
Theorem	ruclem2 16320*	Lemma for ruc 16331. Ordering property for the input to 𝐷. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ)    &   ⊢ (𝜑 → 𝑀 ∈ ℝ)    &   ⊢ 𝑋 = (1st ‘(⟨𝐴, 𝐵⟩𝐷𝑀))    &   ⊢ 𝑌 = (2nd ‘(⟨𝐴, 𝐵⟩𝐷𝑀))    &   ⊢ (𝜑 → 𝐴 < 𝐵)    ⇒   ⊢ (𝜑 → (𝐴 ≤ 𝑋 ∧ 𝑋 < 𝑌 ∧ 𝑌 ≤ 𝐵))
Theorem	ruclem3 16321*	Lemma for ruc 16331. The constructed interval [𝑋, 𝑌] always excludes 𝑀. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ)    &   ⊢ (𝜑 → 𝑀 ∈ ℝ)    &   ⊢ 𝑋 = (1st ‘(⟨𝐴, 𝐵⟩𝐷𝑀))    &   ⊢ 𝑌 = (2nd ‘(⟨𝐴, 𝐵⟩𝐷𝑀))    &   ⊢ (𝜑 → 𝐴 < 𝐵)    ⇒   ⊢ (𝜑 → (𝑀 < 𝑋 ∨ 𝑌 < 𝑀))
Theorem	ruclem4 16322*	Lemma for ruc 16331. Initial value of the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    ⇒   ⊢ (𝜑 → (𝐺‘0) = ⟨0, 1⟩)
Theorem	ruclem6 16323*	Lemma for ruc 16331. Domain and codomain of the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    ⇒   ⊢ (𝜑 → 𝐺:ℕ0⟶(ℝ × ℝ))
Theorem	ruclem7 16324*	Lemma for ruc 16331. Successor value for the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ0) → (𝐺‘(𝑁 + 1)) = ((𝐺‘𝑁)𝐷(𝐹‘(𝑁 + 1))))
Theorem	ruclem8 16325*	Lemma for ruc 16331. The intervals of the 𝐺 sequence are all nonempty. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ0) → (1st ‘(𝐺‘𝑁)) < (2nd ‘(𝐺‘𝑁)))
Theorem	ruclem9 16326*	Lemma for ruc 16331. The first components of the 𝐺 sequence are increasing, and the second components are decreasing. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    ⇒   ⊢ (𝜑 → ((1st ‘(𝐺‘𝑀)) ≤ (1st ‘(𝐺‘𝑁)) ∧ (2nd ‘(𝐺‘𝑁)) ≤ (2nd ‘(𝐺‘𝑀))))
Theorem	ruclem10 16327*	Lemma for ruc 16331. Every first component of the 𝐺 sequence is less than every second component. That is, the sequences form a chain a1 < a2 <... < b2 < b1, where ai are the first components and bi are the second components. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (1st ‘(𝐺‘𝑀)) < (2nd ‘(𝐺‘𝑁)))
Theorem	ruclem11 16328*	Lemma for ruc 16331. Closure lemmas for supremum. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    ⇒   ⊢ (𝜑 → (ran (1st ∘ 𝐺) ⊆ ℝ ∧ ran (1st ∘ 𝐺) ≠ ∅ ∧ ∀𝑧 ∈ ran (1st ∘ 𝐺)𝑧 ≤ 1))
Theorem	ruclem12 16329*	Lemma for ruc 16331. The supremum of the increasing sequence 1st ∘ 𝐺 is a real number that is not in the range of 𝐹. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    &   ⊢ 𝑆 = sup(ran (1st ∘ 𝐺), ℝ, < )    ⇒   ⊢ (𝜑 → 𝑆 ∈ (ℝ ∖ ran 𝐹))
Theorem	ruclem13 16330	Lemma for ruc 16331. There is no function that maps ℕ onto ℝ. (Use nex 1833 if you want this in the form ¬ ∃𝑓𝑓:ℕ–onto→ℝ.) (Contributed by NM, 14-Oct-2004.) (Proof shortened by Fan Zheng, 6-Jun-2016.)
⊢ ¬ 𝐹:ℕ–onto→ℝ
Theorem	ruc 16331	The set of positive integers is strictly dominated by the set of real numbers, i.e. the real numbers are uncountable. The proof consists of lemmas ruclem1 16319 through ruclem13 16330 and this final piece. Our proof is based on the proof of Theorem 5.18 of [Truss] p. 114. See ruclem13 16330 for the function existence version of this theorem. For an informal discussion of this proof, see mmcomplex.html#uncountable 16330. For an alternate proof see rucALT 16318. This is Metamath 100 proof #22. (Contributed by NM, 13-Oct-2004.)
⊢ ℕ ≺ ℝ
Theorem	resdomq 16332	The set of rationals is strictly less equinumerous than the set of reals (ℝ strictly dominates ℚ). (Contributed by NM, 18-Dec-2004.)
⊢ ℚ ≺ ℝ
Theorem	aleph1re 16333	There are at least aleph-one real numbers. (Contributed by NM, 2-Feb-2005.)
⊢ (ℵ‘1o) ≼ ℝ
Theorem	aleph1irr 16334	There are at least aleph-one irrationals. (Contributed by NM, 2-Feb-2005.)
⊢ (ℵ‘1o) ≼ (ℝ ∖ ℚ)
Theorem	cnso 16335	The complex numbers can be linearly ordered. (Contributed by Stefan O'Rear, 16-Nov-2014.)
⊢ ∃𝑥 𝑥 Or ℂ
PART 6  ELEMENTARY NUMBER THEORY Here we introduce elementary number theory, in particular the elementary properties of divisibility and elementary prime number theory.
6.1  Elementary properties of divisibility
6.1.1  Irrationality of square root of 2
Theorem	sqrt2irrlem 16336	Lemma for sqrt2irr 16337. This is the core of the proof: if 𝐴 / 𝐵 = √(2), then 𝐴 and 𝐵 are even, so 𝐴 / 2 and 𝐵 / 2 are smaller representatives, which is absurd by the method of infinite descent (here implemented by strong induction). This is Metamath 100 proof #1. (Contributed by NM, 20-Aug-2001.) (Revised by Mario Carneiro, 12-Sep-2015.) (Proof shortened by JV, 4-Jan-2022.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → (√‘2) = (𝐴 / 𝐵))    ⇒   ⊢ (𝜑 → ((𝐴 / 2) ∈ ℤ ∧ (𝐵 / 2) ∈ ℕ))
Theorem	sqrt2irr 16337	The square root of 2 is irrational. See zsqrtelqelz 16849 for a generalization to all non-square integers. The proof's core is proven in sqrt2irrlem 16336, which shows that if 𝐴 / 𝐵 = √(2), then 𝐴 and 𝐵 are even, so 𝐴 / 2 and 𝐵 / 2 are smaller representatives, which is absurd. An older version of this proof was included in The Seventeen Provers of the World compiled by Freek Wiedijk. It is also the first of the "top 100" mathematical theorems whose formalization is tracked by Freek Wiedijk on his Formalizing 100 Theorems page at http://www.cs.ru.nl/~freek/100/ 16336. (Contributed by NM, 8-Jan-2002.) (Proof shortened by Mario Carneiro, 12-Sep-2015.)
⊢ (√‘2) ∉ ℚ
Theorem	sqrt2re 16338	The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.)
⊢ (√‘2) ∈ ℝ
Theorem	sqrt2irr0 16339	The square root of 2 is an irrational number. (Contributed by AV, 23-Dec-2022.)
⊢ (√‘2) ∈ (ℝ ∖ ℚ)
6.1.2  Some Number sets are chains of proper subsets
Theorem	nthruc 16340	The sequence ℕ, ℤ, ℚ, ℝ, and ℂ forms a chain of proper subsets. In each case the proper subset relationship is shown by demonstrating a number that belongs to one set but not the other. We show that zero belongs to ℤ but not ℕ, one-half belongs to ℚ but not ℤ, the square root of 2 belongs to ℝ but not ℚ, and finally that the imaginary number i belongs to ℂ but not ℝ. See nthruz 16341 for a further refinement. (Contributed by NM, 12-Jan-2002.)
⊢ ((ℕ ⊊ ℤ ∧ ℤ ⊊ ℚ) ∧ (ℚ ⊊ ℝ ∧ ℝ ⊊ ℂ))
Theorem	nthruz 16341	The sequence ℕ, ℕ0, and ℤ forms a chain of proper subsets. In each case the proper subset relationship is shown by demonstrating a number that belongs to one set but not the other. We show that zero belongs to ℕ0 but not ℕ and minus one belongs to ℤ but not ℕ0. This theorem refines the chain of proper subsets nthruc 16340. (Contributed by NM, 9-May-2004.)
⊢ (ℕ ⊊ ℕ0 ∧ ℕ0 ⊊ ℤ)
6.1.3  The divides relation
Syntax	cdvds 16342	Extend the definition of a class to include the divides relation. See df-dvds 16343.
class ∥
Definition	df-dvds 16343*	Define the divides relation, see definition in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ∥ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ) ∧ ∃𝑛 ∈ ℤ (𝑛 · 𝑥) = 𝑦)}
Theorem	divides 16344*	Define the divides relation. 𝑀 ∥ 𝑁 means 𝑀 divides into 𝑁 with no remainder. For example, 3 ∥ 6 (ex-dvds 30936). As proven in dvdsval3 16346, 𝑀 ∥ 𝑁 ↔ (𝑁 mod 𝑀) = 0. See divides 16344 and dvdsval2 16345 for other equivalent expressions. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = 𝑁))
Theorem	dvdsval2 16345	One nonzero integer divides another integer if and only if their quotient is an integer. (Contributed by Jeff Hankins, 29-Sep-2013.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑀 ≠ 0 ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ (𝑁 / 𝑀) ∈ ℤ))
Theorem	dvdsval3 16346	One nonzero integer divides another integer if and only if the remainder upon division is zero, see remark in [ApostolNT] p. 106. (Contributed by Mario Carneiro, 22-Feb-2014.) (Revised by Mario Carneiro, 15-Jul-2014.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ (𝑁 mod 𝑀) = 0))
Theorem	dvdszrcl 16347	Reverse closure for the divisibility relation. (Contributed by Stefan O'Rear, 5-Sep-2015.)
⊢ (𝑋 ∥ 𝑌 → (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ))
Theorem	dvdsmod0 16348	If a positive integer divides another integer, then the remainder upon division is zero. (Contributed by AV, 3-Mar-2022.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑀 ∥ 𝑁) → (𝑁 mod 𝑀) = 0)
Theorem	p1modz1 16349	If a number greater than 1 divides another number, the second number increased by 1 is 1 modulo the first number. (Contributed by AV, 19-Mar-2022.)
⊢ ((𝑀 ∥ 𝐴 ∧ 1 < 𝑀) → ((𝐴 + 1) mod 𝑀) = 1)
Theorem	dvdsmodexp 16350	If a positive integer divides another integer, this other integer is equal to its positive powers modulo the positive integer. (Formerly part of the proof for fermltl 16875). (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by AV, 19-Mar-2022.)
⊢ ((𝑁 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∥ 𝐴) → ((𝐴↑𝐵) mod 𝑁) = (𝐴 mod 𝑁))
Theorem	nndivdvds 16351	Strong form of dvdsval2 16345 for positive integers. (Contributed by Stefan O'Rear, 13-Sep-2014.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵 ∥ 𝐴 ↔ (𝐴 / 𝐵) ∈ ℕ))
Theorem	nndivides 16352*	Definition of the divides relation for positive integers. (Contributed by AV, 26-Jul-2021.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 ∥ 𝑁 ↔ ∃𝑛 ∈ ℕ (𝑛 · 𝑀) = 𝑁))
Theorem	moddvds 16353	Two ways to say 𝐴≡𝐵 (mod 𝑁), see also definition in [ApostolNT] p. 106. (Contributed by Mario Carneiro, 18-Feb-2014.)
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((𝐴 mod 𝑁) = (𝐵 mod 𝑁) ↔ 𝑁 ∥ (𝐴 − 𝐵)))
Theorem	modm1div 16354	An integer greater than one divides another integer minus one iff the second integer modulo the first integer is one. (Contributed by AV, 30-May-2023.)
⊢ ((𝑁 ∈ (ℤ≥‘2) ∧ 𝐴 ∈ ℤ) → ((𝐴 mod 𝑁) = 1 ↔ 𝑁 ∥ (𝐴 − 1)))
Theorem	addmulmodb 16355	An integer plus a product is itself modulo a positive integer iff the product is divisible by the positive integer. (Contributed by AV, 8-Sep-2025.)
⊢ ((𝑁 ∈ ℕ ∧ (𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ)) → (𝑁 ∥ (𝐵 · 𝐶) ↔ ((𝐴 + (𝐵 · 𝐶)) mod 𝑁) = (𝐴 mod 𝑁)))
Theorem	dvds0lem 16356	A lemma to assist theorems of ∥ with no antecedents. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (𝐾 · 𝑀) = 𝑁) → 𝑀 ∥ 𝑁)
Theorem	dvds1lem 16357*	A lemma to assist theorems of ∥ with one antecedent. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝜑 → (𝐽 ∈ ℤ ∧ 𝐾 ∈ ℤ))    &   ⊢ (𝜑 → (𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ))    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ℤ) → 𝑍 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ℤ) → ((𝑥 · 𝐽) = 𝐾 → (𝑍 · 𝑀) = 𝑁))    ⇒   ⊢ (𝜑 → (𝐽 ∥ 𝐾 → 𝑀 ∥ 𝑁))
Theorem	dvds2lem 16358*	A lemma to assist theorems of ∥ with two antecedents. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝜑 → (𝐼 ∈ ℤ ∧ 𝐽 ∈ ℤ))    &   ⊢ (𝜑 → (𝐾 ∈ ℤ ∧ 𝐿 ∈ ℤ))    &   ⊢ (𝜑 → (𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ))    &   ⊢ ((𝜑 ∧ (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ)) → 𝑍 ∈ ℤ)    &   ⊢ ((𝜑 ∧ (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ)) → (((𝑥 · 𝐼) = 𝐽 ∧ (𝑦 · 𝐾) = 𝐿) → (𝑍 · 𝑀) = 𝑁))    ⇒   ⊢ (𝜑 → ((𝐼 ∥ 𝐽 ∧ 𝐾 ∥ 𝐿) → 𝑀 ∥ 𝑁))
Theorem	iddvds 16359	An integer divides itself. Theorem 1.1(a) in [ApostolNT] p. 14 (reflexive property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → 𝑁 ∥ 𝑁)
Theorem	1dvds 16360	1 divides any integer. Theorem 1.1(f) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → 1 ∥ 𝑁)
Theorem	dvds0 16361	Any integer divides 0. Theorem 1.1(g) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → 𝑁 ∥ 0)
Theorem	negdvdsb 16362	An integer divides another iff its negation does. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ -𝑀 ∥ 𝑁))
Theorem	dvdsnegb 16363	An integer divides another iff it divides its negation. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ -𝑁))
Theorem	absdvdsb 16364	An integer divides another iff its absolute value does. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ (abs‘𝑀) ∥ 𝑁))
Theorem	dvdsabsb 16365	An integer divides another iff it divides its absolute value. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (abs‘𝑁)))
Theorem	0dvds 16366	Only 0 is divisible by 0. Theorem 1.1(h) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → (0 ∥ 𝑁 ↔ 𝑁 = 0))
Theorem	dvdsmul1 16367	An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑀 ∥ (𝑀 · 𝑁))
Theorem	dvdsmul2 16368	An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑁 ∥ (𝑀 · 𝑁))
Theorem	iddvdsexp 16369	An integer divides a positive integer power of itself. (Contributed by Paul Chapman, 26-Oct-2012.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → 𝑀 ∥ (𝑀↑𝑁))
Theorem	muldvds1 16370	If a product divides an integer, so does one of its factors. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) ∥ 𝑁 → 𝐾 ∥ 𝑁))
Theorem	muldvds2 16371	If a product divides an integer, so does one of its factors. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) ∥ 𝑁 → 𝑀 ∥ 𝑁))
Theorem	dvdscmul 16372	Multiplication by a constant maintains the divides relation. Theorem 1.1(d) in [ApostolNT] p. 14 (multiplication property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝐾 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝐾 · 𝑀) ∥ (𝐾 · 𝑁)))
Theorem	dvdsmulc 16373	Multiplication by a constant maintains the divides relation. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝐾 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝑀 · 𝐾) ∥ (𝑁 · 𝐾)))
Theorem	dvdscmulr 16374	Cancellation law for the divides relation. Theorem 1.1(e) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ (𝐾 ∈ ℤ ∧ 𝐾 ≠ 0)) → ((𝐾 · 𝑀) ∥ (𝐾 · 𝑁) ↔ 𝑀 ∥ 𝑁))
Theorem	dvdsmulcr 16375	Cancellation law for the divides relation. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ (𝐾 ∈ ℤ ∧ 𝐾 ≠ 0)) → ((𝑀 · 𝐾) ∥ (𝑁 · 𝐾) ↔ 𝑀 ∥ 𝑁))
Theorem	summodnegmod 16376	The sum of two integers modulo a positive integer equals zero iff the first of the two integers equals the negative of the other integer modulo the positive integer. (Contributed by AV, 25-Jul-2021.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (((𝐴 + 𝐵) mod 𝑁) = 0 ↔ (𝐴 mod 𝑁) = (-𝐵 mod 𝑁)))
Theorem	difmod0 16377	The difference of two integers modulo a positive integer equals zero iff the two integers are equal modulo the positive integer. (Contributed by AV, 15-Nov-2025.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (((𝐴 − 𝐵) mod 𝑁) = 0 ↔ (𝐴 mod 𝑁) = (𝐵 mod 𝑁)))
Theorem	modmulconst 16378	Constant multiplication in a modulo operation, see theorem 5.3 in [ApostolNT] p. 108. (Contributed by AV, 21-Jul-2021.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ) ∧ 𝑀 ∈ ℕ) → ((𝐴 mod 𝑀) = (𝐵 mod 𝑀) ↔ ((𝐶 · 𝐴) mod (𝐶 · 𝑀)) = ((𝐶 · 𝐵) mod (𝐶 · 𝑀))))
Theorem	dvds2ln 16379	If an integer divides each of two other integers, it divides any linear combination of them. Theorem 1.1(c) in [ApostolNT] p. 14 (linearity property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (((𝐼 ∈ ℤ ∧ 𝐽 ∈ ℤ) ∧ (𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ)) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ ((𝐼 · 𝑀) + (𝐽 · 𝑁))))
Theorem	dvds2add 16380	If an integer divides each of two other integers, it divides their sum. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 + 𝑁)))
Theorem	dvds2sub 16381	If an integer divides each of two other integers, it divides their difference. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 − 𝑁)))
Theorem	dvds2addd 16382	Deduction form of dvds2add 16380. (Contributed by SN, 21-Aug-2024.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑀)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑁)    ⇒   ⊢ (𝜑 → 𝐾 ∥ (𝑀 + 𝑁))
Theorem	dvds2subd 16383	Deduction form of dvds2sub 16381. (Contributed by Stanislas Polu, 9-Mar-2020.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑀)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑁)    ⇒   ⊢ (𝜑 → 𝐾 ∥ (𝑀 − 𝑁))
Theorem	dvdstr 16384	The divides relation is transitive. Theorem 1.1(b) in [ApostolNT] p. 14 (transitive property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝑀 ∥ 𝑁) → 𝐾 ∥ 𝑁))
Theorem	dvdstrd 16385	The divides relation is transitive, a deduction version of dvdstr 16384. (Contributed by metakunt, 12-May-2024.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑀)    &   ⊢ (𝜑 → 𝑀 ∥ 𝑁)    ⇒   ⊢ (𝜑 → 𝐾 ∥ 𝑁)
Theorem	dvdsmultr1 16386	If an integer divides another, it divides a multiple of it. (Contributed by Paul Chapman, 17-Nov-2012.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐾 ∥ 𝑀 → 𝐾 ∥ (𝑀 · 𝑁)))
Theorem	dvdsmultr1d 16387	Deduction form of dvdsmultr1 16386. (Contributed by Stanislas Polu, 9-Mar-2020.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑀)    ⇒   ⊢ (𝜑 → 𝐾 ∥ (𝑀 · 𝑁))
Theorem	dvdsmultr2 16388	If an integer divides another, it divides a multiple of it. (Contributed by Paul Chapman, 17-Nov-2012.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐾 ∥ 𝑁 → 𝐾 ∥ (𝑀 · 𝑁)))
Theorem	dvdsmultr2d 16389	Deduction form of dvdsmultr2 16388. (Contributed by SN, 23-Aug-2024.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑁)    ⇒   ⊢ (𝜑 → 𝐾 ∥ (𝑀 · 𝑁))
Theorem	ordvdsmul 16390	If an integer divides either of two others, it divides their product. (Contributed by Paul Chapman, 17-Nov-2012.) (Proof shortened by Mario Carneiro, 17-Jul-2014.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∨ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 · 𝑁)))
Theorem	dvdssub2 16391	If an integer divides a difference, then it divides one term iff it divides the other. (Contributed by Mario Carneiro, 13-Jul-2014.)
⊢ (((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ 𝐾 ∥ (𝑀 − 𝑁)) → (𝐾 ∥ 𝑀 ↔ 𝐾 ∥ 𝑁))
Theorem	dvdsadd 16392	An integer divides another iff it divides their sum. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 13-Jul-2014.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑀 + 𝑁)))
Theorem	dvdsaddr 16393	An integer divides another iff it divides their sum. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑁 + 𝑀)))
Theorem	dvdssub 16394	An integer divides another iff it divides their difference. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑀 − 𝑁)))
Theorem	dvdssubr 16395	An integer divides another iff it divides their difference. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑁 − 𝑀)))
Theorem	dvdsadd2b 16396	Adding a multiple of the base does not affect divisibility. (Contributed by Stefan O'Rear, 23-Sep-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ (𝐶 ∈ ℤ ∧ 𝐴 ∥ 𝐶)) → (𝐴 ∥ 𝐵 ↔ 𝐴 ∥ (𝐶 + 𝐵)))
Theorem	dvdsaddre2b 16397	Adding a multiple of the base does not affect divisibility. Variant of dvdsadd2b 16396 only requiring 𝐵 to be a real number (not necessarily an integer). (Contributed by AV, 19-Jul-2021.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℝ ∧ (𝐶 ∈ ℤ ∧ 𝐴 ∥ 𝐶)) → (𝐴 ∥ 𝐵 ↔ 𝐴 ∥ (𝐶 + 𝐵)))
Theorem	fsumdvds 16398*	If every term in a sum is divisible by 𝑁, then so is the sum. (Contributed by Mario Carneiro, 17-Jan-2015.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝑁 ∥ 𝐵)    ⇒   ⊢ (𝜑 → 𝑁 ∥ Σ𝑘 ∈ 𝐴 𝐵)
Theorem	dvdslelem 16399	Lemma for dvdsle 16400. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑀 ∈ ℤ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐾 ∈ ℤ    ⇒   ⊢ (𝑁 < 𝑀 → (𝐾 · 𝑀) ≠ 𝑁)
Theorem	dvdsle 16400	The divisors of a positive integer are bounded by it. The proof does not use /. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (𝑀 ∥ 𝑁 → 𝑀 ≤ 𝑁))