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Theorem List for Intuitionistic Logic Explorer - 6301-6400   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremxpcomen 6301 Commutative law for equinumerosity of Cartesian product. Proposition 4.22(d) of [Mendelson] p. 254. (Contributed by NM, 5-Jan-2004.) (Revised by Mario Carneiro, 15-Nov-2014.)
𝐴 ∈ V    &   𝐵 ∈ V       (𝐴 × 𝐵) ≈ (𝐵 × 𝐴)

Theoremxpcomeng 6302 Commutative law for equinumerosity of Cartesian product. Proposition 4.22(d) of [Mendelson] p. 254. (Contributed by NM, 27-Mar-2006.)
((𝐴𝑉𝐵𝑊) → (𝐴 × 𝐵) ≈ (𝐵 × 𝐴))

Theoremxpsnen2g 6303 A set is equinumerous to its Cartesian product with a singleton on the left. (Contributed by Stefan O'Rear, 21-Nov-2014.)
((𝐴𝑉𝐵𝑊) → ({𝐴} × 𝐵) ≈ 𝐵)

Theoremxpassen 6304 Associative law for equinumerosity of Cartesian product. Proposition 4.22(e) of [Mendelson] p. 254. (Contributed by NM, 22-Jan-2004.) (Revised by Mario Carneiro, 15-Nov-2014.)
𝐴 ∈ V    &   𝐵 ∈ V    &   𝐶 ∈ V       ((𝐴 × 𝐵) × 𝐶) ≈ (𝐴 × (𝐵 × 𝐶))

Theoremxpdom2 6305 Dominance law for Cartesian product. Proposition 10.33(2) of [TakeutiZaring] p. 92. (Contributed by NM, 24-Jul-2004.) (Revised by Mario Carneiro, 15-Nov-2014.)
𝐶 ∈ V       (𝐴𝐵 → (𝐶 × 𝐴) ≼ (𝐶 × 𝐵))

Theoremxpdom2g 6306 Dominance law for Cartesian product. Theorem 6L(c) of [Enderton] p. 149. (Contributed by Mario Carneiro, 26-Apr-2015.)
((𝐶𝑉𝐴𝐵) → (𝐶 × 𝐴) ≼ (𝐶 × 𝐵))

Theoremxpdom1g 6307 Dominance law for Cartesian product. Theorem 6L(c) of [Enderton] p. 149. (Contributed by NM, 25-Mar-2006.) (Revised by Mario Carneiro, 26-Apr-2015.)
((𝐶𝑉𝐴𝐵) → (𝐴 × 𝐶) ≼ (𝐵 × 𝐶))

Theoremxpdom3m 6308* A set is dominated by its Cartesian product with an inhabited set. Exercise 6 of [Suppes] p. 98. (Contributed by Jim Kingdon, 15-Apr-2020.)
((𝐴𝑉𝐵𝑊 ∧ ∃𝑥 𝑥𝐵) → 𝐴 ≼ (𝐴 × 𝐵))

Theoremxpdom1 6309 Dominance law for Cartesian product. Theorem 6L(c) of [Enderton] p. 149. (Contributed by NM, 28-Sep-2004.) (Revised by NM, 29-Mar-2006.) (Revised by Mario Carneiro, 7-May-2015.)
𝐶 ∈ V       (𝐴𝐵 → (𝐴 × 𝐶) ≼ (𝐵 × 𝐶))

Theoremfopwdom 6310 Covering implies injection on power sets. (Contributed by Stefan O'Rear, 6-Nov-2014.) (Revised by Mario Carneiro, 24-Jun-2015.)
((𝐹 ∈ V ∧ 𝐹:𝐴onto𝐵) → 𝒫 𝐵 ≼ 𝒫 𝐴)

Theoremenen1 6311 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))

Theoremenen2 6312 Equality-like theorem for equinumerosity. (Contributed by NM, 18-Dec-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))

Theoremdomen1 6313 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐴𝐶𝐵𝐶))

Theoremdomen2 6314 Equality-like theorem for equinumerosity and dominance. (Contributed by NM, 8-Nov-2003.)
(𝐴𝐵 → (𝐶𝐴𝐶𝐵))

2.6.26  Pigeonhole Principle

Theoremphplem1 6315 Lemma for Pigeonhole Principle. If we join a natural number to itself minus an element, we end up with its successor minus the same element. (Contributed by NM, 25-May-1998.)
((𝐴 ∈ ω ∧ 𝐵𝐴) → ({𝐴} ∪ (𝐴 ∖ {𝐵})) = (suc 𝐴 ∖ {𝐵}))

Theoremphplem2 6316 Lemma for Pigeonhole Principle. A natural number is equinumerous to its successor minus one of its elements. (Contributed by NM, 11-Jun-1998.) (Revised by Mario Carneiro, 16-Nov-2014.)
𝐴 ∈ V    &   𝐵 ∈ V       ((𝐴 ∈ ω ∧ 𝐵𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))

Theoremphplem3 6317 Lemma for Pigeonhole Principle. A natural number is equinumerous to its successor minus any element of the successor. For a version without the redundant hypotheses, see phplem3g 6319. (Contributed by NM, 26-May-1998.)
𝐴 ∈ V    &   𝐵 ∈ V       ((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))

Theoremphplem4 6318 Lemma for Pigeonhole Principle. Equinumerosity of successors implies equinumerosity of the original natural numbers. (Contributed by NM, 28-May-1998.) (Revised by Mario Carneiro, 24-Jun-2015.)
𝐴 ∈ V    &   𝐵 ∈ V       ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 ≈ suc 𝐵𝐴𝐵))

Theoremphplem3g 6319 A natural number is equinumerous to its successor minus any element of the successor. Version of phplem3 6317 with unnecessary hypotheses removed. (Contributed by Jim Kingdon, 1-Sep-2021.)
((𝐴 ∈ ω ∧ 𝐵 ∈ suc 𝐴) → 𝐴 ≈ (suc 𝐴 ∖ {𝐵}))

Theoremnneneq 6320 Two equinumerous natural numbers are equal. Proposition 10.20 of [TakeutiZaring] p. 90 and its converse. Also compare Corollary 6E of [Enderton] p. 136. (Contributed by NM, 28-May-1998.)
((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴𝐵𝐴 = 𝐵))

Theoremphp5 6321 A natural number is not equinumerous to its successor. Corollary 10.21(1) of [TakeutiZaring] p. 90. (Contributed by NM, 26-Jul-2004.)
(𝐴 ∈ ω → ¬ 𝐴 ≈ suc 𝐴)

Theoremsnnen2og 6322 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a proper class, see snnen2oprc 6323. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴𝑉 → ¬ {𝐴} ≈ 2𝑜)

Theoremsnnen2oprc 6323 A singleton {𝐴} is never equinumerous with the ordinal number 2. If 𝐴 is a set, see snnen2og 6322. (Contributed by Jim Kingdon, 1-Sep-2021.)
𝐴 ∈ V → ¬ {𝐴} ≈ 2𝑜)

Theoremphplem4dom 6324 Dominance of successors implies dominance of the original natural numbers. (Contributed by Jim Kingdon, 1-Sep-2021.)
((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 ≼ suc 𝐵𝐴𝐵))

Theoremphp5dom 6325 A natural number does not dominate its successor. (Contributed by Jim Kingdon, 1-Sep-2021.)
(𝐴 ∈ ω → ¬ suc 𝐴𝐴)

Theoremnndomo 6326 Cardinal ordering agrees with natural number ordering. Example 3 of [Enderton] p. 146. (Contributed by NM, 17-Jun-1998.)
((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐴𝐵𝐴𝐵))

Theoremphpm 6327* Pigeonhole Principle. A natural number is not equinumerous to a proper subset of itself. By "proper subset" here we mean that there is an element which is in the natural number and not in the subset, or in symbols 𝑥𝑥 ∈ (𝐴𝐵) (which is stronger than not being equal in the absence of excluded middle). Theorem (Pigeonhole Principle) of [Enderton] p. 134. The theorem is so-called because you can't put n + 1 pigeons into n holes (if each hole holds only one pigeon). The proof consists of lemmas phplem1 6315 through phplem4 6318, nneneq 6320, and this final piece of the proof. (Contributed by NM, 29-May-1998.)
((𝐴 ∈ ω ∧ 𝐵𝐴 ∧ ∃𝑥 𝑥 ∈ (𝐴𝐵)) → ¬ 𝐴𝐵)

Theoremphpelm 6328 Pigeonhole Principle. A natural number is not equinumerous to an element of itself. (Contributed by Jim Kingdon, 6-Sep-2021.)
((𝐴 ∈ ω ∧ 𝐵𝐴) → ¬ 𝐴𝐵)

Theoremphplem4on 6329 Equinumerosity of successors of an ordinal and a natural number implies equinumerosity of the originals. (Contributed by Jim Kingdon, 5-Sep-2021.)
((𝐴 ∈ On ∧ 𝐵 ∈ ω) → (suc 𝐴 ≈ suc 𝐵𝐴𝐵))

2.6.27  Finite sets

Theoremfidceq 6330 Equality of members of a finite set is decidable. This may be counterintuitive: cannot any two sets be elements of a finite set? Well, to show, for example, that {𝐵, 𝐶} is finite would require showing it is equinumerous to 1𝑜 or to 2𝑜 but to show that you'd need to know 𝐵 = 𝐶 or ¬ 𝐵 = 𝐶, respectively. (Contributed by Jim Kingdon, 5-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵𝐴𝐶𝐴) → DECID 𝐵 = 𝐶)

Theoremfidifsnen 6331 All decrements of a finite set are equinumerous. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝑋 ∈ Fin ∧ 𝐴𝑋𝐵𝑋) → (𝑋 ∖ {𝐴}) ≈ (𝑋 ∖ {𝐵}))

Theoremfidifsnid 6332 If we remove a single element from a finite set then put it back in, we end up with the original finite set. This strengthens difsnss 3510 from subset to equality when the set is finite. (Contributed by Jim Kingdon, 9-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵𝐴) → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = 𝐴)

Theoremnnfi 6333 Natural numbers are finite sets. (Contributed by Stefan O'Rear, 21-Mar-2015.)
(𝐴 ∈ ω → 𝐴 ∈ Fin)

Theoremenfi 6334 Equinumerous sets have the same finiteness. (Contributed by NM, 22-Aug-2008.)
(𝐴𝐵 → (𝐴 ∈ Fin ↔ 𝐵 ∈ Fin))

Theoremenfii 6335 A set equinumerous to a finite set is finite. (Contributed by Mario Carneiro, 12-Mar-2015.)
((𝐵 ∈ Fin ∧ 𝐴𝐵) → 𝐴 ∈ Fin)

Theoremssfiexmid 6336* If any subset of a finite set is finite, excluded middle follows. One direction of Theorem 2.1 of [Bauer], p. 485. (Contributed by Jim Kingdon, 19-May-2020.)
𝑥𝑦((𝑥 ∈ Fin ∧ 𝑦𝑥) → 𝑦 ∈ Fin)       (𝜑 ∨ ¬ 𝜑)

Theoremdif1en 6337 If a set 𝐴 is equinumerous to the successor of a natural number 𝑀, then 𝐴 with an element removed is equinumerous to 𝑀. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Stefan O'Rear, 16-Aug-2015.)
((𝑀 ∈ ω ∧ 𝐴 ≈ suc 𝑀𝑋𝐴) → (𝐴 ∖ {𝑋}) ≈ 𝑀)

Theoremfiunsnnn 6338 Adding one element to a finite set which is equinumerous to a natural number. (Contributed by Jim Kingdon, 13-Sep-2021.)
(((𝐴 ∈ Fin ∧ 𝐵 ∈ (V ∖ 𝐴)) ∧ (𝑁 ∈ ω ∧ 𝐴𝑁)) → (𝐴 ∪ {𝐵}) ≈ suc 𝑁)

Theoremphp5fin 6339 A finite set is not equinumerous to a set which adds one element. (Contributed by Jim Kingdon, 13-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ (V ∖ 𝐴)) → ¬ 𝐴 ≈ (𝐴 ∪ {𝐵}))

Theoremfisbth 6340 Schroeder-Bernstein Theorem for finite sets. (Contributed by Jim Kingdon, 12-Sep-2021.)
(((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) ∧ (𝐴𝐵𝐵𝐴)) → 𝐴𝐵)

Theorem0fin 6341 The empty set is finite. (Contributed by FL, 14-Jul-2008.)
∅ ∈ Fin

Theoremfin0 6342* A nonempty finite set has at least one element. (Contributed by Jim Kingdon, 10-Sep-2021.)
(𝐴 ∈ Fin → (𝐴 ≠ ∅ ↔ ∃𝑥 𝑥𝐴))

Theoremfin0or 6343* A finite set is either empty or inhabited. (Contributed by Jim Kingdon, 30-Sep-2021.)
(𝐴 ∈ Fin → (𝐴 = ∅ ∨ ∃𝑥 𝑥𝐴))

Theoremdiffitest 6344* If subtracting any set from a finite set gives a finite set, any proposition of the form ¬ 𝜑 is decidable. This is not a proof of full excluded middle, but it is close enough to show we won't be able to prove 𝐴 ∈ Fin → (𝐴𝐵) ∈ Fin. (Contributed by Jim Kingdon, 8-Sep-2021.)
𝑎 ∈ Fin ∀𝑏(𝑎𝑏) ∈ Fin       𝜑 ∨ ¬ ¬ 𝜑)

Theoremfindcard 6345* Schema for induction on the cardinality of a finite set. The inductive hypothesis is that the result is true on the given set with any one element removed. The result is then proven to be true for all finite sets. (Contributed by Jeff Madsen, 2-Sep-2009.)
(𝑥 = ∅ → (𝜑𝜓))    &   (𝑥 = (𝑦 ∖ {𝑧}) → (𝜑𝜒))    &   (𝑥 = 𝑦 → (𝜑𝜃))    &   (𝑥 = 𝐴 → (𝜑𝜏))    &   𝜓    &   (𝑦 ∈ Fin → (∀𝑧𝑦 𝜒𝜃))       (𝐴 ∈ Fin → 𝜏)

Theoremfindcard2 6346* Schema for induction on the cardinality of a finite set. The inductive step shows that the result is true if one more element is added to the set. The result is then proven to be true for all finite sets. (Contributed by Jeff Madsen, 8-Jul-2010.)
(𝑥 = ∅ → (𝜑𝜓))    &   (𝑥 = 𝑦 → (𝜑𝜒))    &   (𝑥 = (𝑦 ∪ {𝑧}) → (𝜑𝜃))    &   (𝑥 = 𝐴 → (𝜑𝜏))    &   𝜓    &   (𝑦 ∈ Fin → (𝜒𝜃))       (𝐴 ∈ Fin → 𝜏)

Theoremfindcard2s 6347* Variation of findcard2 6346 requiring that the element added in the induction step not be a member of the original set. (Contributed by Paul Chapman, 30-Nov-2012.)
(𝑥 = ∅ → (𝜑𝜓))    &   (𝑥 = 𝑦 → (𝜑𝜒))    &   (𝑥 = (𝑦 ∪ {𝑧}) → (𝜑𝜃))    &   (𝑥 = 𝐴 → (𝜑𝜏))    &   𝜓    &   ((𝑦 ∈ Fin ∧ ¬ 𝑧𝑦) → (𝜒𝜃))       (𝐴 ∈ Fin → 𝜏)

Theoremfindcard2d 6348* Deduction version of findcard2 6346. If you also need 𝑦 ∈ Fin (which doesn't come for free due to ssfiexmid 6336), use findcard2sd 6349 instead. (Contributed by SO, 16-Jul-2018.)
(𝑥 = ∅ → (𝜓𝜒))    &   (𝑥 = 𝑦 → (𝜓𝜃))    &   (𝑥 = (𝑦 ∪ {𝑧}) → (𝜓𝜏))    &   (𝑥 = 𝐴 → (𝜓𝜂))    &   (𝜑𝜒)    &   ((𝜑 ∧ (𝑦𝐴𝑧 ∈ (𝐴𝑦))) → (𝜃𝜏))    &   (𝜑𝐴 ∈ Fin)       (𝜑𝜂)

Theoremfindcard2sd 6349* Deduction form of finite set induction . (Contributed by Jim Kingdon, 14-Sep-2021.)
(𝑥 = ∅ → (𝜓𝜒))    &   (𝑥 = 𝑦 → (𝜓𝜃))    &   (𝑥 = (𝑦 ∪ {𝑧}) → (𝜓𝜏))    &   (𝑥 = 𝐴 → (𝜓𝜂))    &   (𝜑𝜒)    &   (((𝜑𝑦 ∈ Fin) ∧ (𝑦𝐴𝑧 ∈ (𝐴𝑦))) → (𝜃𝜏))    &   (𝜑𝐴 ∈ Fin)       (𝜑𝜂)

Theoremdiffisn 6350 Subtracting a singleton from a finite set produces a finite set. (Contributed by Jim Kingdon, 11-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵𝐴) → (𝐴 ∖ {𝐵}) ∈ Fin)

Theoremdiffifi 6351 Subtracting one finite set from another produces a finite set. (Contributed by Jim Kingdon, 8-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ 𝐵𝐴) → (𝐴𝐵) ∈ Fin)

Theoremac6sfi 6352* Existence of a choice function for finite sets. (Contributed by Jeff Hankins, 26-Jun-2009.) (Proof shortened by Mario Carneiro, 29-Jan-2014.)
(𝑦 = (𝑓𝑥) → (𝜑𝜓))       ((𝐴 ∈ Fin ∧ ∀𝑥𝐴𝑦𝐵 𝜑) → ∃𝑓(𝑓:𝐴𝐵 ∧ ∀𝑥𝐴 𝜓))

Theoremfientri3 6353 Trichotomy of dominance for finite sets. (Contributed by Jim Kingdon, 15-Sep-2021.)
((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (𝐴𝐵𝐵𝐴))

Theoremnnwetri 6354* A natural number is well-ordered by E. More specifically, this order both satisfies We and is trichotomous. (Contributed by Jim Kingdon, 25-Sep-2021.)
(𝐴 ∈ ω → ( E We 𝐴 ∧ ∀𝑥𝐴𝑦𝐴 (𝑥 E 𝑦𝑥 = 𝑦𝑦 E 𝑥)))

Theoremonunsnss 6355 Adding a singleton to create an ordinal. (Contributed by Jim Kingdon, 20-Oct-2021.)
((𝐵𝑉 ∧ (𝐴 ∪ {𝐵}) ∈ On) → 𝐵𝐴)

Theoremsnon0 6356 An ordinal which is a singleton is {∅}. (Contributed by Jim Kingdon, 19-Oct-2021.)
((𝐴𝑉 ∧ {𝐴} ∈ On) → 𝐴 = ∅)

2.6.28  Ordinal isomorphism

Theoremordiso2 6357 Generalize ordiso 6358 to proper classes. (Contributed by Mario Carneiro, 24-Jun-2015.)
((𝐹 Isom E , E (𝐴, 𝐵) ∧ Ord 𝐴 ∧ Ord 𝐵) → 𝐴 = 𝐵)

Theoremordiso 6358* Order-isomorphic ordinal numbers are equal. (Contributed by Jeff Hankins, 16-Oct-2009.) (Proof shortened by Mario Carneiro, 24-Jun-2015.)
((𝐴 ∈ On ∧ 𝐵 ∈ On) → (𝐴 = 𝐵 ↔ ∃𝑓 𝑓 Isom E , E (𝐴, 𝐵)))

2.6.29  Cardinal numbers

Syntaxccrd 6359 Extend class definition to include the cardinal size function.
class card

Definitiondf-card 6360* Define the cardinal number function. The cardinal number of a set is the least ordinal number equinumerous to it. In other words, it is the "size" of the set. Definition of [Enderton] p. 197. Our notation is from Enderton. Other textbooks often use a double bar over the set to express this function. (Contributed by NM, 21-Oct-2003.)
card = (𝑥 ∈ V ↦ {𝑦 ∈ On ∣ 𝑦𝑥})

Theoremcardcl 6361* The cardinality of a well-orderable set is an ordinal. (Contributed by Jim Kingdon, 30-Aug-2021.)
(∃𝑦 ∈ On 𝑦𝐴 → (card‘𝐴) ∈ On)

Theoremisnumi 6362 A set equinumerous to an ordinal is numerable. (Contributed by Mario Carneiro, 29-Apr-2015.)
((𝐴 ∈ On ∧ 𝐴𝐵) → 𝐵 ∈ dom card)

Theoremfinnum 6363 Every finite set is numerable. (Contributed by Mario Carneiro, 4-Feb-2013.) (Revised by Mario Carneiro, 29-Apr-2015.)
(𝐴 ∈ Fin → 𝐴 ∈ dom card)

Theoremonenon 6364 Every ordinal number is numerable. (Contributed by Mario Carneiro, 29-Apr-2015.)
(𝐴 ∈ On → 𝐴 ∈ dom card)

Theoremcardval3ex 6365* The value of (card‘𝐴). (Contributed by Jim Kingdon, 30-Aug-2021.)
(∃𝑥 ∈ On 𝑥𝐴 → (card‘𝐴) = {𝑦 ∈ On ∣ 𝑦𝐴})

Theoremoncardval 6366* The value of the cardinal number function with an ordinal number as its argument. (Contributed by NM, 24-Nov-2003.) (Revised by Mario Carneiro, 13-Sep-2013.)
(𝐴 ∈ On → (card‘𝐴) = {𝑥 ∈ On ∣ 𝑥𝐴})

Theoremcardonle 6367 The cardinal of an ordinal number is less than or equal to the ordinal number. Proposition 10.6(3) of [TakeutiZaring] p. 85. (Contributed by NM, 22-Oct-2003.)
(𝐴 ∈ On → (card‘𝐴) ⊆ 𝐴)

Theoremcard0 6368 The cardinality of the empty set is the empty set. (Contributed by NM, 25-Oct-2003.)
(card‘∅) = ∅

Theoremcarden2bex 6369* If two numerable sets are equinumerous, then they have equal cardinalities. (Contributed by Jim Kingdon, 30-Aug-2021.)
((𝐴𝐵 ∧ ∃𝑥 ∈ On 𝑥𝐴) → (card‘𝐴) = (card‘𝐵))

PART 3  REAL AND COMPLEX NUMBERS

This section derives the basics of real and complex numbers.

To construct the real numbers constructively, we follow two main sources. The first is Metamath Proof Explorer, which has the advantage of being already formalized in metamath. Its disadvantage, for our purposes, is that it assumes the law of the excluded middle throughout. Since we have already developed natural numbers ( for example, nna0 6053 and similar theorems ), going from there to positive integers (df-ni 6402) and then positive rational numbers (df-nqqs 6446) does not involve a major change in approach compared with the Metamath Proof Explorer.

It is when we proceed to Dedekind cuts that we bring in more material from Section 11.2 of [HoTT], which focuses on the aspects of Dedekind cuts which are different without excluded middle. With excluded middle, it is natural to define the cut as the lower set only (as Metamath Proof Explorer does), but we define the cut as a pair of both the lower and upper sets, as [HoTT] does. There are also differences in how we handle order and replacing "not equal to zero" with "apart from zero".

3.1  Construction and axiomatization of real and complex numbers

3.1.1  Dedekind-cut construction of real and complex numbers

Syntaxcnpi 6370 The set of positive integers, which is the set of natural numbers ω with 0 removed.

Note: This is the start of the Dedekind-cut construction of real and _complex numbers.

class N

class +N

Syntaxcmi 6372 Positive integer multiplication.
class ·N

Syntaxclti 6373 Positive integer ordering relation.
class <N

class +pQ

Syntaxcmpq 6375 Positive pre-fraction multiplication.
class ·pQ

Syntaxcltpq 6376 Positive pre-fraction ordering relation.
class <pQ

Syntaxceq 6377 Equivalence class used to construct positive fractions.
class ~Q

Syntaxcnq 6378 Set of positive fractions.
class Q

Syntaxc1q 6379 The positive fraction constant 1.
class 1Q

class +Q

Syntaxcmq 6381 Positive fraction multiplication.
class ·Q

Syntaxcrq 6382 Positive fraction reciprocal operation.
class *Q

Syntaxcltq 6383 Positive fraction ordering relation.
class <Q

Syntaxceq0 6384 Equivalence class used to construct non-negative fractions.
class ~Q0

Syntaxcnq0 6385 Set of non-negative fractions.
class Q0

Syntaxc0q0 6386 The non-negative fraction constant 0.
class 0Q0

class +Q0

Syntaxcmq0 6388 Non-negative fraction multiplication.
class ·Q0

Syntaxcnp 6389 Set of positive reals.
class P

Syntaxc1p 6390 Positive real constant 1.
class 1P

class +P

Syntaxcmp 6392 Positive real multiplication.
class ·P

Syntaxcltp 6393 Positive real ordering relation.
class <P

Syntaxcer 6394 Equivalence class used to construct signed reals.
class ~R

Syntaxcnr 6395 Set of signed reals.
class R

Syntaxc0r 6396 The signed real constant 0.
class 0R

Syntaxc1r 6397 The signed real constant 1.
class 1R

Syntaxcm1r 6398 The signed real constant -1.
class -1R