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Most recent proofs    These are the 100 (Unicode, GIF) or 1000 (Unicode, GIF) most recent proofs in the iset.mm database for the Intuitionistic Logic Explorer. The iset.mm database is maintained on GitHub with master (stable) and develop (development) versions. This page was created from the commit given on the MPE Most Recent Proofs page. The database from that commit is also available here: iset.mm.

See the MPE Most Recent Proofs page for news and some useful links.

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Last updated on 18-Jan-2022 at 1:34 AM ET.
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DateLabelDescription
Theorem

5-Dec-2021divalglemqt 10231 Lemma for divalg 10236. The 𝑄 = 𝑇 case involved in showing uniqueness. (Contributed by Jim Kingdon, 5-Dec-2021.)
(𝜑𝐷 ∈ ℤ)    &   (𝜑𝑅 ∈ ℤ)    &   (𝜑𝑆 ∈ ℤ)    &   (𝜑𝑄 ∈ ℤ)    &   (𝜑𝑇 ∈ ℤ)    &   (𝜑𝑄 = 𝑇)    &   (𝜑 → ((𝑄 · 𝐷) + 𝑅) = ((𝑇 · 𝐷) + 𝑆))       (𝜑𝑅 = 𝑆)

4-Dec-2021divalglemeuneg 10235 Lemma for divalg 10236. Uniqueness for a negative denominator. (Contributed by Jim Kingdon, 4-Dec-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 < 0) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))

4-Dec-2021divalglemeunn 10233 Lemma for divalg 10236. Uniqueness for a positive denominator. (Contributed by Jim Kingdon, 4-Dec-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))

4-Dec-2021divalglemnqt 10232 Lemma for divalg 10236. The 𝑄 < 𝑇 case involved in showing uniqueness. (Contributed by Jim Kingdon, 4-Dec-2021.)
(𝜑𝐷 ∈ ℕ)    &   (𝜑𝑅 ∈ ℤ)    &   (𝜑𝑆 ∈ ℤ)    &   (𝜑𝑄 ∈ ℤ)    &   (𝜑𝑇 ∈ ℤ)    &   (𝜑 → 0 ≤ 𝑆)    &   (𝜑𝑅 < 𝐷)    &   (𝜑 → ((𝑄 · 𝐷) + 𝑅) = ((𝑇 · 𝐷) + 𝑆))       (𝜑 → ¬ 𝑄 < 𝑇)

2-Dec-2021rspc2gv 2684 Restricted specialization with two quantifiers, using implicit substitution. (Contributed by BJ, 2-Dec-2021.)
((𝑥 = 𝐴𝑦 = 𝐵) → (𝜑𝜓))       ((𝐴𝑉𝐵𝑊) → (∀𝑥𝑉𝑦𝑊 𝜑𝜓))

30-Nov-2021divalglemex 10234 Lemma for divalg 10236. The quotient and remainder exist. (Contributed by Jim Kingdon, 30-Nov-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → ∃𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))

30-Nov-2021divalglemnn 10230 Lemma for divalg 10236. Existence for a positive denominator. (Contributed by Jim Kingdon, 30-Nov-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))

27-Nov-2021ioom 9217 An open interval of extended reals is inhabited iff the lower argument is less than the upper argument. (Contributed by Jim Kingdon, 27-Nov-2021.)
((𝐴 ∈ ℝ*𝐵 ∈ ℝ*) → (∃𝑥 𝑥 ∈ (𝐴(,)𝐵) ↔ 𝐴 < 𝐵))

26-Nov-2021supisoti 6414 Image of a supremum under an isomorphism. (Contributed by Jim Kingdon, 26-Nov-2021.)
(𝜑𝐹 Isom 𝑅, 𝑆 (𝐴, 𝐵))    &   (𝜑𝐶𝐴)    &   (𝜑 → ∃𝑥𝐴 (∀𝑦𝐶 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐶 𝑦𝑅𝑧)))    &   ((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))       (𝜑 → sup((𝐹𝐶), 𝐵, 𝑆) = (𝐹‘sup(𝐶, 𝐴, 𝑅)))

26-Nov-2021isoti 6411 An isomorphism preserves tightness. (Contributed by Jim Kingdon, 26-Nov-2021.)
(𝐹 Isom 𝑅, 𝑆 (𝐴, 𝐵) → (∀𝑢𝐴𝑣𝐴 (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)) ↔ ∀𝑢𝐵𝑣𝐵 (𝑢 = 𝑣 ↔ (¬ 𝑢𝑆𝑣 ∧ ¬ 𝑣𝑆𝑢))))

26-Nov-2021isotilem 6410 Lemma for isoti 6411. (Contributed by Jim Kingdon, 26-Nov-2021.)
(𝐹 Isom 𝑅, 𝑆 (𝐴, 𝐵) → (∀𝑥𝐵𝑦𝐵 (𝑥 = 𝑦 ↔ (¬ 𝑥𝑆𝑦 ∧ ¬ 𝑦𝑆𝑥)) → ∀𝑢𝐴𝑣𝐴 (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢))))

26-Nov-2021supsnti 6409 The supremum of a singleton. (Contributed by Jim Kingdon, 26-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))    &   (𝜑𝐵𝐴)       (𝜑 → sup({𝐵}, 𝐴, 𝑅) = 𝐵)

24-Nov-2021supmaxti 6408 The greatest element of a set is its supremum. Note that the converse is not true; the supremum might not be an element of the set considered. (Contributed by Jim Kingdon, 24-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))    &   (𝜑𝐶𝐴)    &   (𝜑𝐶𝐵)    &   ((𝜑𝑦𝐵) → ¬ 𝐶𝑅𝑦)       (𝜑 → sup(𝐵, 𝐴, 𝑅) = 𝐶)

24-Nov-2021suplubti 6406 A supremum is the least upper bound. See also supclti 6404 and supubti 6405. (Contributed by Jim Kingdon, 24-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))    &   (𝜑 → ∃𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))       (𝜑 → ((𝐶𝐴𝐶𝑅sup(𝐵, 𝐴, 𝑅)) → ∃𝑧𝐵 𝐶𝑅𝑧))

24-Nov-2021supubti 6405 A supremum is an upper bound. See also supclti 6404 and suplubti 6406.

This proof demonstrates how to expand an iota-based definition (df-iota 4895) using riotacl2 5509.

(Contributed by Jim Kingdon, 24-Nov-2021.)

((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))    &   (𝜑 → ∃𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))       (𝜑 → (𝐶𝐵 → ¬ sup(𝐵, 𝐴, 𝑅)𝑅𝐶))

24-Nov-2021supclti 6404 A supremum belongs to its base class (closure law). See also supubti 6405 and suplubti 6406. (Contributed by Jim Kingdon, 24-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))    &   (𝜑 → ∃𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))       (𝜑 → sup(𝐵, 𝐴, 𝑅) ∈ 𝐴)

24-Nov-2021eqsuptid 6403 Sufficient condition for an element to be equal to the supremum. (Contributed by Jim Kingdon, 24-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))    &   (𝜑𝐶𝐴)    &   ((𝜑𝑦𝐵) → ¬ 𝐶𝑅𝑦)    &   ((𝜑 ∧ (𝑦𝐴𝑦𝑅𝐶)) → ∃𝑧𝐵 𝑦𝑅𝑧)       (𝜑 → sup(𝐵, 𝐴, 𝑅) = 𝐶)

23-Nov-2021eqsupti 6402 Sufficient condition for an element to be equal to the supremum. (Contributed by Jim Kingdon, 23-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))       (𝜑 → ((𝐶𝐴 ∧ ∀𝑦𝐵 ¬ 𝐶𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝐶 → ∃𝑧𝐵 𝑦𝑅𝑧)) → sup(𝐵, 𝐴, 𝑅) = 𝐶))

23-Nov-2021supval2ti 6401 Alternate expression for the supremum. (Contributed by Jim Kingdon, 23-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))    &   (𝜑 → ∃𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))       (𝜑 → sup(𝐵, 𝐴, 𝑅) = (𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))))

23-Nov-2021supeuti 6400 A supremum is unique. Similar to Theorem I.26 of [Apostol] p. 24 (but for suprema in general). (Contributed by Jim Kingdon, 23-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))    &   (𝜑 → ∃𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))       (𝜑 → ∃!𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))

23-Nov-2021supmoti 6399 Any class 𝐵 has at most one supremum in 𝐴 (where 𝑅 is interpreted as 'less than'). The hypothesis is satisfied by real numbers (see lttri3 7157) or other orders which correspond to tight apartnesses. (Contributed by Jim Kingdon, 23-Nov-2021.)
((𝜑 ∧ (𝑢𝐴𝑣𝐴)) → (𝑢 = 𝑣 ↔ (¬ 𝑢𝑅𝑣 ∧ ¬ 𝑣𝑅𝑢)))       (𝜑 → ∃*𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))

17-Nov-2021oddpwdclemndvds 10259 Lemma for oddpwdc 10262. A natural number is not divisible by one more than the highest power of two which divides it. (Contributed by Jim Kingdon, 17-Nov-2021.)
(𝐴 ∈ ℕ → ¬ (2↑((𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴)) + 1)) ∥ 𝐴)

17-Nov-2021oddpwdclemdvds 10258 Lemma for oddpwdc 10262. A natural number is divisible by the highest power of two which divides it. (Contributed by Jim Kingdon, 17-Nov-2021.)
(𝐴 ∈ ℕ → (2↑(𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴))) ∥ 𝐴)

17-Nov-2021pw2dvdseulemle 10255 Lemma for pw2dvdseu 10256. Powers of two which do and do not divide a natural number. (Contributed by Jim Kingdon, 17-Nov-2021.)
(𝜑𝑁 ∈ ℕ)    &   (𝜑𝐴 ∈ ℕ0)    &   (𝜑𝐵 ∈ ℕ0)    &   (𝜑 → (2↑𝐴) ∥ 𝑁)    &   (𝜑 → ¬ (2↑(𝐵 + 1)) ∥ 𝑁)       (𝜑𝐴𝐵)

16-Nov-2021oddpwdclemdc 10261 Lemma for oddpwdc 10262. Decomposing a number into odd and even parts. (Contributed by Jim Kingdon, 16-Nov-2021.)
((((𝑋 ∈ ℕ ∧ ¬ 2 ∥ 𝑋) ∧ 𝑌 ∈ ℕ0) ∧ 𝐴 = ((2↑𝑌) · 𝑋)) ↔ (𝐴 ∈ ℕ ∧ (𝑋 = (𝐴 / (2↑(𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴)))) ∧ 𝑌 = (𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴)))))

16-Nov-2021oddpwdclemodd 10260 Lemma for oddpwdc 10262. Removing the powers of two from a natural number produces an odd number. (Contributed by Jim Kingdon, 16-Nov-2021.)
(𝐴 ∈ ℕ → ¬ 2 ∥ (𝐴 / (2↑(𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴)))))

16-Nov-2021oddpwdclemxy 10257 Lemma for oddpwdc 10262. Another way of stating that decomposing a natural number into a power of two and an odd number is unique. (Contributed by Jim Kingdon, 16-Nov-2021.)
((((𝑋 ∈ ℕ ∧ ¬ 2 ∥ 𝑋) ∧ 𝑌 ∈ ℕ0) ∧ 𝐴 = ((2↑𝑌) · 𝑋)) → (𝑋 = (𝐴 / (2↑(𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴)))) ∧ 𝑌 = (𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴))))

16-Nov-2021pw2dvdseu 10256 A natural number has a unique highest power of two which divides it. (Contributed by Jim Kingdon, 16-Nov-2021.)
(𝑁 ∈ ℕ → ∃!𝑚 ∈ ℕ0 ((2↑𝑚) ∥ 𝑁 ∧ ¬ (2↑(𝑚 + 1)) ∥ 𝑁))

14-Nov-2021pw2dvds 10254 A natural number has a highest power of two which divides it. (Contributed by Jim Kingdon, 14-Nov-2021.)
(𝑁 ∈ ℕ → ∃𝑚 ∈ ℕ0 ((2↑𝑚) ∥ 𝑁 ∧ ¬ (2↑(𝑚 + 1)) ∥ 𝑁))

14-Nov-2021pw2dvdslemn 10253 Lemma for pw2dvds 10254. If a natural number has some power of two which does not divide it, there is a highest power of two which does divide it. (Contributed by Jim Kingdon, 14-Nov-2021.)
((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ ¬ (2↑𝐴) ∥ 𝑁) → ∃𝑚 ∈ ℕ0 ((2↑𝑚) ∥ 𝑁 ∧ ¬ (2↑(𝑚 + 1)) ∥ 𝑁))

14-Nov-2021dvdsdc 10116 Divisibility is decidable. (Contributed by Jim Kingdon, 14-Nov-2021.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ) → DECID 𝑀𝑁)

11-Nov-2021fz01or 10190 An integer is in the integer range from zero to one iff it is either zero or one. (Contributed by Jim Kingdon, 11-Nov-2021.)
(𝐴 ∈ (0...1) ↔ (𝐴 = 0 ∨ 𝐴 = 1))

11-Nov-2021muldivdirap 7758 Distribution of division over addition with a multiplication. (Contributed by Jim Kingdon, 11-Nov-2021.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → (((𝐶 · 𝐴) + 𝐵) / 𝐶) = (𝐴 + (𝐵 / 𝐶)))

10-Nov-2021zeoxor 10180 An integer is even or odd but not both. (Contributed by Jim Kingdon, 10-Nov-2021.)
(𝑁 ∈ ℤ → (2 ∥ 𝑁 ⊻ ¬ 2 ∥ 𝑁))

9-Nov-2021qlttri2 8673 Apartness is equivalent to not equal for rationals. (Contributed by Jim Kingdon, 9-Nov-2021.)
((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ) → (𝐴𝐵 ↔ (𝐴 < 𝐵𝐵 < 𝐴)))

8-Nov-2021dvdslelemd 10155 Lemma for dvdsle 10156. (Contributed by Jim Kingdon, 8-Nov-2021.)
(𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝐾 ∈ ℤ)    &   (𝜑𝑁 < 𝑀)       (𝜑 → (𝐾 · 𝑀) ≠ 𝑁)

8-Nov-2021qabsord 9903 The absolute value of a rational number is either that number or its negative. (Contributed by Jim Kingdon, 8-Nov-2021.)
(𝜑𝐴 ∈ ℚ)       (𝜑 → ((abs‘𝐴) = 𝐴 ∨ (abs‘𝐴) = -𝐴))

8-Nov-2021qabsor 9902 The absolute value of a rational number is either that number or its negative. (Contributed by Jim Kingdon, 8-Nov-2021.)
(𝐴 ∈ ℚ → ((abs‘𝐴) = 𝐴 ∨ (abs‘𝐴) = -𝐴))

6-Nov-2021ibcval5 9631 Write out the top and bottom parts of the binomial coefficient (𝑁C𝐾) = (𝑁 · (𝑁 − 1) · ... · ((𝑁𝐾) + 1)) / 𝐾! explicitly. In this form, it is valid even for 𝑁 < 𝐾, although it is no longer valid for nonpositive 𝐾. (Contributed by Jim Kingdon, 6-Nov-2021.)
((𝑁 ∈ ℕ0𝐾 ∈ ℕ) → (𝑁C𝐾) = ((seq((𝑁𝐾) + 1)( · , I , ℂ)‘𝑁) / (!‘𝐾)))

3-Nov-2021qavgle 9215 The average of two rational numbers is less than or equal to at least one of them. (Contributed by Jim Kingdon, 3-Nov-2021.)
((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ) → (((𝐴 + 𝐵) / 2) ≤ 𝐴 ∨ ((𝐴 + 𝐵) / 2) ≤ 𝐵))

31-Oct-2021nn0opth2d 9591 An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. See comments for nn0opthd 9590. (Contributed by Jim Kingdon, 31-Oct-2021.)
(𝜑𝐴 ∈ ℕ0)    &   (𝜑𝐵 ∈ ℕ0)    &   (𝜑𝐶 ∈ ℕ0)    &   (𝜑𝐷 ∈ ℕ0)       (𝜑 → ((((𝐴 + 𝐵)↑2) + 𝐵) = (((𝐶 + 𝐷)↑2) + 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))

31-Oct-2021nn0opthd 9590 An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. We can represent an ordered pair of nonnegative integers 𝐴 and 𝐵 by (((𝐴 + 𝐵) · (𝐴 + 𝐵)) + 𝐵). If two such ordered pairs are equal, their first elements are equal and their second elements are equal. Contrast this ordered pair representation with the standard one df-op 3412 that works for any set. (Contributed by Jim Kingdon, 31-Oct-2021.)
(𝜑𝐴 ∈ ℕ0)    &   (𝜑𝐵 ∈ ℕ0)    &   (𝜑𝐶 ∈ ℕ0)    &   (𝜑𝐷 ∈ ℕ0)       (𝜑 → ((((𝐴 + 𝐵) · (𝐴 + 𝐵)) + 𝐵) = (((𝐶 + 𝐷) · (𝐶 + 𝐷)) + 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))

31-Oct-2021nn0opthlem2d 9589 Lemma for nn0opth2 9592. (Contributed by Jim Kingdon, 31-Oct-2021.)
(𝜑𝐴 ∈ ℕ0)    &   (𝜑𝐵 ∈ ℕ0)    &   (𝜑𝐶 ∈ ℕ0)    &   (𝜑𝐷 ∈ ℕ0)       (𝜑 → ((𝐴 + 𝐵) < 𝐶 → ((𝐶 · 𝐶) + 𝐷) ≠ (((𝐴 + 𝐵) · (𝐴 + 𝐵)) + 𝐵)))

31-Oct-2021nn0opthlem1d 9588 A rather pretty lemma for nn0opth2 9592. (Contributed by Jim Kingdon, 31-Oct-2021.)
(𝜑𝐴 ∈ ℕ0)    &   (𝜑𝐶 ∈ ℕ0)       (𝜑 → (𝐴 < 𝐶 ↔ ((𝐴 · 𝐴) + (2 · 𝐴)) < (𝐶 · 𝐶)))

31-Oct-2021nn0le2msqd 9587 The square function on nonnegative integers is monotonic. (Contributed by Jim Kingdon, 31-Oct-2021.)
(𝜑𝐴 ∈ ℕ0)    &   (𝜑𝐵 ∈ ℕ0)       (𝜑 → (𝐴𝐵 ↔ (𝐴 · 𝐴) ≤ (𝐵 · 𝐵)))

28-Oct-2021mulle0r 7985 Multiplying a nonnegative number by a nonpositive number yields a nonpositive number. (Contributed by Jim Kingdon, 28-Oct-2021.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (𝐴 ≤ 0 ∧ 0 ≤ 𝐵)) → (𝐴 · 𝐵) ≤ 0)

26-Oct-2021modqeqmodmin 9344 A rational number equals the difference of the rational number and a modulus modulo the modulus. (Contributed by Jim Kingdon, 26-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝑀 ∈ ℚ ∧ 0 < 𝑀) → (𝐴 mod 𝑀) = ((𝐴𝑀) mod 𝑀))

26-Oct-2021modqsubdir 9343 Distribute the modulo operation over a subtraction. (Contributed by Jim Kingdon, 26-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ) ∧ (𝐶 ∈ ℚ ∧ 0 < 𝐶)) → ((𝐵 mod 𝐶) ≤ (𝐴 mod 𝐶) ↔ ((𝐴𝐵) mod 𝐶) = ((𝐴 mod 𝐶) − (𝐵 mod 𝐶))))

26-Oct-2021modqdi 9342 Distribute multiplication over a modulo operation. (Contributed by Jim Kingdon, 26-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 0 < 𝐴) ∧ 𝐵 ∈ ℚ ∧ (𝐶 ∈ ℚ ∧ 0 < 𝐶)) → (𝐴 · (𝐵 mod 𝐶)) = ((𝐴 · 𝐵) mod (𝐴 · 𝐶)))

26-Oct-2021modqaddmulmod 9341 The sum of a rational number and the product of a second rational number modulo a modulus and an integer equals the sum of the rational number and the product of the other rational number and the integer modulo the modulus. (Contributed by Jim Kingdon, 26-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 𝐶 ∈ ℤ) ∧ (𝑀 ∈ ℚ ∧ 0 < 𝑀)) → ((𝐴 + ((𝐵 mod 𝑀) · 𝐶)) mod 𝑀) = ((𝐴 + (𝐵 · 𝐶)) mod 𝑀))

26-Oct-2021modqmulmodr 9340 The product of an integer and a rational number modulo a modulus equals the product of the integer and the rational number modulo the modulus. (Contributed by Jim Kingdon, 26-Oct-2021.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℚ) ∧ (𝑀 ∈ ℚ ∧ 0 < 𝑀)) → ((𝐴 · (𝐵 mod 𝑀)) mod 𝑀) = ((𝐴 · 𝐵) mod 𝑀))

25-Oct-2021modqmulmod 9339 The product of a rational number modulo a modulus and an integer equals the product of the rational number and the integer modulo the modulus. (Contributed by Jim Kingdon, 25-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℤ) ∧ (𝑀 ∈ ℚ ∧ 0 < 𝑀)) → (((𝐴 mod 𝑀) · 𝐵) mod 𝑀) = ((𝐴 · 𝐵) mod 𝑀))

25-Oct-2021q2submod 9335 If a number is between a modulus and twice the modulus, the first number modulo the modulus equals the first number minus the modulus. (Contributed by Jim Kingdon, 25-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 0 < 𝐵) ∧ (𝐵𝐴𝐴 < (2 · 𝐵))) → (𝐴 mod 𝐵) = (𝐴𝐵))

25-Oct-2021q2txmodxeq0 9334 Two times a positive number modulo the number is zero. (Contributed by Jim Kingdon, 25-Oct-2021.)
((𝑋 ∈ ℚ ∧ 0 < 𝑋) → ((2 · 𝑋) mod 𝑋) = 0)

25-Oct-2021modqsubmodmod 9333 The difference of a number modulo a modulus and another number modulo the same modulus equals the difference of the two numbers modulo the modulus. (Contributed by Jim Kingdon, 25-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ) ∧ (𝑀 ∈ ℚ ∧ 0 < 𝑀)) → (((𝐴 mod 𝑀) − (𝐵 mod 𝑀)) mod 𝑀) = ((𝐴𝐵) mod 𝑀))

25-Oct-2021modqsubmod 9332 The difference of a number modulo a modulus and another number equals the difference of the two numbers modulo the modulus. (Contributed by Jim Kingdon, 25-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ) ∧ (𝑀 ∈ ℚ ∧ 0 < 𝑀)) → (((𝐴 mod 𝑀) − 𝐵) mod 𝑀) = ((𝐴𝐵) mod 𝑀))

25-Oct-2021modqsub12d 9331 Subtraction property of the modulo operation. (Contributed by Jim Kingdon, 25-Oct-2021.)
(𝜑𝐴 ∈ ℚ)    &   (𝜑𝐵 ∈ ℚ)    &   (𝜑𝐶 ∈ ℚ)    &   (𝜑𝐷 ∈ ℚ)    &   (𝜑𝐸 ∈ ℚ)    &   (𝜑 → 0 < 𝐸)    &   (𝜑 → (𝐴 mod 𝐸) = (𝐵 mod 𝐸))    &   (𝜑 → (𝐶 mod 𝐸) = (𝐷 mod 𝐸))       (𝜑 → ((𝐴𝐶) mod 𝐸) = ((𝐵𝐷) mod 𝐸))

25-Oct-2021modqadd12d 9330 Additive property of the modulo operation. (Contributed by Jim Kingdon, 25-Oct-2021.)
(𝜑𝐴 ∈ ℚ)    &   (𝜑𝐵 ∈ ℚ)    &   (𝜑𝐶 ∈ ℚ)    &   (𝜑𝐷 ∈ ℚ)    &   (𝜑𝐸 ∈ ℚ)    &   (𝜑 → 0 < 𝐸)    &   (𝜑 → (𝐴 mod 𝐸) = (𝐵 mod 𝐸))    &   (𝜑 → (𝐶 mod 𝐸) = (𝐷 mod 𝐸))       (𝜑 → ((𝐴 + 𝐶) mod 𝐸) = ((𝐵 + 𝐷) mod 𝐸))

25-Oct-2021syl11 31 A syllogism inference. Commuted form of an instance of syl 14. (Contributed by BJ, 25-Oct-2021.)
(𝜑 → (𝜓𝜒))    &   (𝜃𝜑)       (𝜓 → (𝜃𝜒))

24-Oct-2021ax-ddkcomp 10501 Axiom of Dedekind completeness for Dedekind real numbers: every inhabited upper-bounded located set of reals has a real upper bound. Ideally, this axiom should be "proved" as "axddkcomp" for the real numbers constructed from IZF, and then the axiom ax-ddkcomp 10501 should be used in place of construction specific results. In particular, axcaucvg 7032 should be proved from it. (Contributed by BJ, 24-Oct-2021.)
(((𝐴 ⊆ ℝ ∧ ∃𝑥 𝑥𝐴) ∧ ∃𝑥 ∈ ℝ ∀𝑦𝐴 𝑦 < 𝑥 ∧ ∀𝑥 ∈ ℝ ∀𝑦 ∈ ℝ (𝑥 < 𝑦 → (∃𝑧𝐴 𝑥 < 𝑧 ∨ ∀𝑧𝐴 𝑧 < 𝑦))) → ∃𝑥 ∈ ℝ (∀𝑦𝐴 𝑦𝑥 ∧ ((𝐵𝑅 ∧ ∀𝑦𝐴 𝑦𝐵) → 𝑥𝐵)))

24-Oct-2021modqnegd 9329 Negation property of the modulo operation. (Contributed by Jim Kingdon, 24-Oct-2021.)
(𝜑𝐴 ∈ ℚ)    &   (𝜑𝐵 ∈ ℚ)    &   (𝜑𝐶 ∈ ℚ)    &   (𝜑 → 0 < 𝐶)    &   (𝜑 → (𝐴 mod 𝐶) = (𝐵 mod 𝐶))       (𝜑 → (-𝐴 mod 𝐶) = (-𝐵 mod 𝐶))

24-Oct-2021modqmul12d 9328 Multiplication property of the modulo operation, see theorem 5.2(b) in [ApostolNT] p. 107. (Contributed by Jim Kingdon, 24-Oct-2021.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ ℤ)    &   (𝜑𝐸 ∈ ℚ)    &   (𝜑 → 0 < 𝐸)    &   (𝜑 → (𝐴 mod 𝐸) = (𝐵 mod 𝐸))    &   (𝜑 → (𝐶 mod 𝐸) = (𝐷 mod 𝐸))       (𝜑 → ((𝐴 · 𝐶) mod 𝐸) = ((𝐵 · 𝐷) mod 𝐸))

24-Oct-2021modqmul1 9327 Multiplication property of the modulo operation. Note that the multiplier 𝐶 must be an integer. (Contributed by Jim Kingdon, 24-Oct-2021.)
(𝜑𝐴 ∈ ℚ)    &   (𝜑𝐵 ∈ ℚ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ ℚ)    &   (𝜑 → 0 < 𝐷)    &   (𝜑 → (𝐴 mod 𝐷) = (𝐵 mod 𝐷))       (𝜑 → ((𝐴 · 𝐶) mod 𝐷) = ((𝐵 · 𝐶) mod 𝐷))

24-Oct-2021modqltm1p1mod 9326 If a number modulo a modulus is less than the modulus decreased by 1, the first number increased by 1 modulo the modulus equals the first number modulo the modulus, increased by 1. (Contributed by Jim Kingdon, 24-Oct-2021.)
(((𝐴 ∈ ℚ ∧ (𝐴 mod 𝑀) < (𝑀 − 1)) ∧ (𝑀 ∈ ℚ ∧ 0 < 𝑀)) → ((𝐴 + 1) mod 𝑀) = ((𝐴 mod 𝑀) + 1))

24-Oct-2021modqm1p1mod0 9325 If a number modulo a modulus equals the modulus decreased by 1, the first number increased by 1 modulo the modulus equals 0. (Contributed by Jim Kingdon, 24-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝑀 ∈ ℚ ∧ 0 < 𝑀) → ((𝐴 mod 𝑀) = (𝑀 − 1) → ((𝐴 + 1) mod 𝑀) = 0))

24-Oct-2021modqadd2mod 9324 The sum of a number modulo a modulus and another number equals the sum of the two numbers modulo the modulus. (Contributed by Jim Kingdon, 24-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ) ∧ (𝑀 ∈ ℚ ∧ 0 < 𝑀)) → ((𝐵 + (𝐴 mod 𝑀)) mod 𝑀) = ((𝐵 + 𝐴) mod 𝑀))

24-Oct-2021qnegmod 9319 The negation of a number modulo a positive number is equal to the difference of the modulus and the number modulo the modulus. (Contributed by Jim Kingdon, 24-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝑁 ∈ ℚ ∧ 0 < 𝑁) → (-𝐴 mod 𝑁) = ((𝑁𝐴) mod 𝑁))

23-Oct-2021modqmuladdnn0 9318 Implication of a decomposition of a nonnegative integer into a multiple of a modulus and a remainder. (Contributed by Jim Kingdon, 23-Oct-2021.)
((𝐴 ∈ ℕ0𝑀 ∈ ℚ ∧ 0 < 𝑀) → ((𝐴 mod 𝑀) = 𝐵 → ∃𝑘 ∈ ℕ0 𝐴 = ((𝑘 · 𝑀) + 𝐵)))

23-Oct-2021modqmuladdim 9317 Implication of a decomposition of an integer into a multiple of a modulus and a remainder. (Contributed by Jim Kingdon, 23-Oct-2021.)
((𝐴 ∈ ℤ ∧ 𝑀 ∈ ℚ ∧ 0 < 𝑀) → ((𝐴 mod 𝑀) = 𝐵 → ∃𝑘 ∈ ℤ 𝐴 = ((𝑘 · 𝑀) + 𝐵)))

23-Oct-2021modqmuladd 9316 Decomposition of an integer into a multiple of a modulus and a remainder. (Contributed by Jim Kingdon, 23-Oct-2021.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℚ)    &   (𝜑𝐵 ∈ (0[,)𝑀))    &   (𝜑𝑀 ∈ ℚ)    &   (𝜑 → 0 < 𝑀)       (𝜑 → ((𝐴 mod 𝑀) = 𝐵 ↔ ∃𝑘 ∈ ℤ 𝐴 = ((𝑘 · 𝑀) + 𝐵)))

23-Oct-2021mulqaddmodid 9314 The sum of a positive rational number less than an upper bound and the product of an integer and the upper bound is the positive rational number modulo the upper bound. (Contributed by Jim Kingdon, 23-Oct-2021.)
(((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℚ) ∧ (𝐴 ∈ ℚ ∧ 𝐴 ∈ (0[,)𝑀))) → (((𝑁 · 𝑀) + 𝐴) mod 𝑀) = 𝐴)

23-Oct-2021modqaddmod 9313 The sum of a number modulo a modulus and another number equals the sum of the two numbers modulo the same modulus. (Contributed by Jim Kingdon, 23-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ) ∧ (𝑀 ∈ ℚ ∧ 0 < 𝑀)) → (((𝐴 mod 𝑀) + 𝐵) mod 𝑀) = ((𝐴 + 𝐵) mod 𝑀))

22-Oct-2021modqaddabs 9312 Absorption law for modulo. (Contributed by Jim Kingdon, 22-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ) ∧ (𝐶 ∈ ℚ ∧ 0 < 𝐶)) → (((𝐴 mod 𝐶) + (𝐵 mod 𝐶)) mod 𝐶) = ((𝐴 + 𝐵) mod 𝐶))

22-Oct-2021modqadd1 9311 Addition property of the modulo operation. (Contributed by Jim Kingdon, 22-Oct-2021.)
(𝜑𝐴 ∈ ℚ)    &   (𝜑𝐵 ∈ ℚ)    &   (𝜑𝐶 ∈ ℚ)    &   (𝜑𝐷 ∈ ℚ)    &   (𝜑 → 0 < 𝐷)    &   (𝜑 → (𝐴 mod 𝐷) = (𝐵 mod 𝐷))       (𝜑 → ((𝐴 + 𝐶) mod 𝐷) = ((𝐵 + 𝐶) mod 𝐷))

21-Oct-2021modqcyc2 9310 The modulo operation is periodic. (Contributed by Jim Kingdon, 21-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝑁 ∈ ℤ) ∧ (𝐵 ∈ ℚ ∧ 0 < 𝐵)) → ((𝐴 − (𝐵 · 𝑁)) mod 𝐵) = (𝐴 mod 𝐵))

21-Oct-2021modqcyc 9309 The modulo operation is periodic. (Contributed by Jim Kingdon, 21-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝑁 ∈ ℤ) ∧ (𝐵 ∈ ℚ ∧ 0 < 𝐵)) → ((𝐴 + (𝑁 · 𝐵)) mod 𝐵) = (𝐴 mod 𝐵))

21-Oct-2021modqabs2 9308 Absorption law for modulo. (Contributed by Jim Kingdon, 21-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 0 < 𝐵) → ((𝐴 mod 𝐵) mod 𝐵) = (𝐴 mod 𝐵))

21-Oct-2021modqabs 9307 Absorption law for modulo. (Contributed by Jim Kingdon, 21-Oct-2021.)
(𝜑𝐴 ∈ ℚ)    &   (𝜑𝐵 ∈ ℚ)    &   (𝜑 → 0 < 𝐵)    &   (𝜑𝐶 ∈ ℚ)    &   (𝜑𝐵𝐶)       (𝜑 → ((𝐴 mod 𝐵) mod 𝐶) = (𝐴 mod 𝐵))

21-Oct-2021q1mod 9306 Special case: 1 modulo a real number greater than 1 is 1. (Contributed by Jim Kingdon, 21-Oct-2021.)
((𝑁 ∈ ℚ ∧ 1 < 𝑁) → (1 mod 𝑁) = 1)

21-Oct-2021q0mod 9305 Special case: 0 modulo a positive real number is 0. (Contributed by Jim Kingdon, 21-Oct-2021.)
((𝑁 ∈ ℚ ∧ 0 < 𝑁) → (0 mod 𝑁) = 0)

21-Oct-2021modqid2 9301 Identity law for modulo. (Contributed by Jim Kingdon, 21-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 0 < 𝐵) → ((𝐴 mod 𝐵) = 𝐴 ↔ (0 ≤ 𝐴𝐴 < 𝐵)))

21-Oct-2021modqid0 9300 A positive real number modulo itself is 0. (Contributed by Jim Kingdon, 21-Oct-2021.)
((𝑁 ∈ ℚ ∧ 0 < 𝑁) → (𝑁 mod 𝑁) = 0)

21-Oct-2021modqid 9299 Identity law for modulo. (Contributed by Jim Kingdon, 21-Oct-2021.)
(((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ) ∧ (0 ≤ 𝐴𝐴 < 𝐵)) → (𝐴 mod 𝐵) = 𝐴)

20-Oct-2021modqvalp1 9293 The value of the modulo operation (expressed with sum of denominator and nominator). (Contributed by Jim Kingdon, 20-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 0 < 𝐵) → ((𝐴 + 𝐵) − (((⌊‘(𝐴 / 𝐵)) + 1) · 𝐵)) = (𝐴 mod 𝐵))

20-Oct-2021onunsnss 6386 Adding a singleton to create an ordinal. (Contributed by Jim Kingdon, 20-Oct-2021.)
((𝐵𝑉 ∧ (𝐴 ∪ {𝐵}) ∈ On) → 𝐵𝐴)

19-Oct-2021snon0 6387 An ordinal which is a singleton is {∅}. (Contributed by Jim Kingdon, 19-Oct-2021.)
((𝐴𝑉 ∧ {𝐴} ∈ On) → 𝐴 = ∅)

18-Oct-2021qdencn 10511 The set of complex numbers whose real and imaginary parts are rational is dense in the complex plane. This is a two dimensional analogue to qdenre 10029 (and also would hold for ℝ × ℝ with the usual metric; this is not about complex numbers in particular). (Contributed by Jim Kingdon, 18-Oct-2021.)
𝑄 = {𝑧 ∈ ℂ ∣ ((ℜ‘𝑧) ∈ ℚ ∧ (ℑ‘𝑧) ∈ ℚ)}       ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℝ+) → ∃𝑥𝑄 (abs‘(𝑥𝐴)) < 𝐵)

18-Oct-2021modqmulnn 9292 Move a positive integer in and out of a floor in the first argument of a modulo operation. (Contributed by Jim Kingdon, 18-Oct-2021.)
((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℚ ∧ 𝑀 ∈ ℕ) → ((𝑁 · (⌊‘𝐴)) mod (𝑁 · 𝑀)) ≤ ((⌊‘(𝑁 · 𝐴)) mod (𝑁 · 𝑀)))

18-Oct-2021intqfrac 9289 Break a number into its integer part and its fractional part. (Contributed by Jim Kingdon, 18-Oct-2021.)
(𝐴 ∈ ℚ → 𝐴 = ((⌊‘𝐴) + (𝐴 mod 1)))

18-Oct-2021flqmod 9288 The floor function expressed in terms of the modulo operation. (Contributed by Jim Kingdon, 18-Oct-2021.)
(𝐴 ∈ ℚ → (⌊‘𝐴) = (𝐴 − (𝐴 mod 1)))

18-Oct-2021modqfrac 9287 The fractional part of a number is the number modulo 1. (Contributed by Jim Kingdon, 18-Oct-2021.)
(𝐴 ∈ ℚ → (𝐴 mod 1) = (𝐴 − (⌊‘𝐴)))

18-Oct-2021modqdifz 9286 The modulo operation differs from 𝐴 by an integer multiple of 𝐵. (Contributed by Jim Kingdon, 18-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 0 < 𝐵) → ((𝐴 − (𝐴 mod 𝐵)) / 𝐵) ∈ ℤ)

18-Oct-2021modqdiffl 9285 The modulo operation differs from 𝐴 by an integer multiple of 𝐵. (Contributed by Jim Kingdon, 18-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 0 < 𝐵) → ((𝐴 − (𝐴 mod 𝐵)) / 𝐵) = (⌊‘(𝐴 / 𝐵)))

18-Oct-2021modqelico 9284 Modular reduction produces a half-open interval. (Contributed by Jim Kingdon, 18-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 0 < 𝐵) → (𝐴 mod 𝐵) ∈ (0[,)𝐵))

18-Oct-2021modqlt 9283 The modulo operation is less than its second argument. (Contributed by Jim Kingdon, 18-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 0 < 𝐵) → (𝐴 mod 𝐵) < 𝐵)

18-Oct-2021modqge0 9282 The modulo operation is nonnegative. (Contributed by Jim Kingdon, 18-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 0 < 𝐵) → 0 ≤ (𝐴 mod 𝐵))

18-Oct-2021negqmod0 9281 𝐴 is divisible by 𝐵 iff its negative is. (Contributed by Jim Kingdon, 18-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 0 < 𝐵) → ((𝐴 mod 𝐵) = 0 ↔ (-𝐴 mod 𝐵) = 0))

18-Oct-2021mulqmod0 9280 The product of an integer and a positive rational number is 0 modulo the positive real number. (Contributed by Jim Kingdon, 18-Oct-2021.)
((𝐴 ∈ ℤ ∧ 𝑀 ∈ ℚ ∧ 0 < 𝑀) → ((𝐴 · 𝑀) mod 𝑀) = 0)

18-Oct-2021flqdiv 9271 Cancellation of the embedded floor of a real divided by an integer. (Contributed by Jim Kingdon, 18-Oct-2021.)
((𝐴 ∈ ℚ ∧ 𝑁 ∈ ℕ) → (⌊‘((⌊‘𝐴) / 𝑁)) = (⌊‘(𝐴 / 𝑁)))

18-Oct-2021intqfrac2 9269 Decompose a real into integer and fractional parts. (Contributed by Jim Kingdon, 18-Oct-2021.)
𝑍 = (⌊‘𝐴)    &   𝐹 = (𝐴𝑍)       (𝐴 ∈ ℚ → (0 ≤ 𝐹𝐹 < 1 ∧ 𝐴 = (𝑍 + 𝐹)))

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