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Theorem List for Intuitionistic Logic Explorer - 13201-13253   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremnninff 13201 An element of is a sequence of zeroes and ones. (Contributed by Jim Kingdon, 4-Aug-2022.)
(𝐴 ∈ ℕ𝐴:ω⟶2o)
 
Theoremnnsf 13202* Domain and range of 𝑆. Part of Definition 3.3 of [PradicBrown2022], p. 5. (Contributed by Jim Kingdon, 30-Jul-2022.)
𝑆 = (𝑝 ∈ ℕ ↦ (𝑖 ∈ ω ↦ if(𝑖 = ∅, 1o, (𝑝 𝑖))))       𝑆:ℕ⟶ℕ
 
Theorempeano4nninf 13203* The successor function on is one to one. Half of Lemma 3.4 of [PradicBrown2022], p. 5. (Contributed by Jim Kingdon, 31-Jul-2022.)
𝑆 = (𝑝 ∈ ℕ ↦ (𝑖 ∈ ω ↦ if(𝑖 = ∅, 1o, (𝑝 𝑖))))       𝑆:ℕ1-1→ℕ
 
Theorempeano3nninf 13204* The successor function on is never zero. Half of Lemma 3.4 of [PradicBrown2022], p. 5. (Contributed by Jim Kingdon, 1-Aug-2022.)
𝑆 = (𝑝 ∈ ℕ ↦ (𝑖 ∈ ω ↦ if(𝑖 = ∅, 1o, (𝑝 𝑖))))       (𝐴 ∈ ℕ → (𝑆𝐴) ≠ (𝑥 ∈ ω ↦ ∅))
 
Theoremnninfalllemn 13205* Lemma for nninfall 13207. Mapping of a natural number to an element of . (Contributed by Jim Kingdon, 4-Aug-2022.)
(𝜑𝑃 ∈ ℕ)    &   (𝜑𝑁 ∈ ω)    &   (𝜑 → ∀𝑥𝑁 (𝑃𝑥) = 1o)    &   (𝜑 → (𝑃𝑁) = ∅)       (𝜑𝑃 = (𝑖 ∈ ω ↦ if(𝑖𝑁, 1o, ∅)))
 
Theoremnninfalllem1 13206* Lemma for nninfall 13207. (Contributed by Jim Kingdon, 1-Aug-2022.)
(𝜑𝑄 ∈ (2o𝑚))    &   (𝜑 → (𝑄‘(𝑥 ∈ ω ↦ 1o)) = 1o)    &   (𝜑 → ∀𝑛 ∈ ω (𝑄‘(𝑖 ∈ ω ↦ if(𝑖𝑛, 1o, ∅))) = 1o)    &   (𝜑𝑃 ∈ ℕ)    &   (𝜑 → (𝑄𝑃) = ∅)       (𝜑 → ∀𝑛 ∈ ω (𝑃𝑛) = 1o)
 
Theoremnninfall 13207* Given a decidable predicate on , showing it holds for natural numbers and the point at infinity suffices to show it holds everywhere. The sense in which 𝑄 is a decidable predicate is that it assigns a value of either or 1o (which can be thought of as false and true) to every element of . Lemma 3.5 of [PradicBrown2022], p. 5. (Contributed by Jim Kingdon, 1-Aug-2022.)
(𝜑𝑄 ∈ (2o𝑚))    &   (𝜑 → (𝑄‘(𝑥 ∈ ω ↦ 1o)) = 1o)    &   (𝜑 → ∀𝑛 ∈ ω (𝑄‘(𝑖 ∈ ω ↦ if(𝑖𝑛, 1o, ∅))) = 1o)       (𝜑 → ∀𝑝 ∈ ℕ (𝑄𝑝) = 1o)
 
Theoremnninfex 13208 is a set. (Contributed by Jim Kingdon, 10-Aug-2022.)
∈ V
 
Theoremnninfsellemdc 13209* Lemma for nninfself 13212. Showing that the selection function is well defined. (Contributed by Jim Kingdon, 8-Aug-2022.)
((𝑄 ∈ (2o𝑚) ∧ 𝑁 ∈ ω) → DECID𝑘 ∈ suc 𝑁(𝑄‘(𝑖 ∈ ω ↦ if(𝑖𝑘, 1o, ∅))) = 1o)
 
Theoremnninfsellemcl 13210* Lemma for nninfself 13212. (Contributed by Jim Kingdon, 8-Aug-2022.)
((𝑄 ∈ (2o𝑚) ∧ 𝑁 ∈ ω) → if(∀𝑘 ∈ suc 𝑁(𝑄‘(𝑖 ∈ ω ↦ if(𝑖𝑘, 1o, ∅))) = 1o, 1o, ∅) ∈ 2o)
 
Theoremnninfsellemsuc 13211* Lemma for nninfself 13212. (Contributed by Jim Kingdon, 6-Aug-2022.)
((𝑄 ∈ (2o𝑚) ∧ 𝐽 ∈ ω) → if(∀𝑘 ∈ suc suc 𝐽(𝑄‘(𝑖 ∈ ω ↦ if(𝑖𝑘, 1o, ∅))) = 1o, 1o, ∅) ⊆ if(∀𝑘 ∈ suc 𝐽(𝑄‘(𝑖 ∈ ω ↦ if(𝑖𝑘, 1o, ∅))) = 1o, 1o, ∅))
 
Theoremnninfself 13212* Domain and range of the selection function for . (Contributed by Jim Kingdon, 6-Aug-2022.)
𝐸 = (𝑞 ∈ (2o𝑚) ↦ (𝑛 ∈ ω ↦ if(∀𝑘 ∈ suc 𝑛(𝑞‘(𝑖 ∈ ω ↦ if(𝑖𝑘, 1o, ∅))) = 1o, 1o, ∅)))       𝐸:(2o𝑚)⟶ℕ
 
Theoremnninfsellemeq 13213* Lemma for nninfsel 13216. (Contributed by Jim Kingdon, 9-Aug-2022.)
𝐸 = (𝑞 ∈ (2o𝑚) ↦ (𝑛 ∈ ω ↦ if(∀𝑘 ∈ suc 𝑛(𝑞‘(𝑖 ∈ ω ↦ if(𝑖𝑘, 1o, ∅))) = 1o, 1o, ∅)))    &   (𝜑𝑄 ∈ (2o𝑚))    &   (𝜑 → (𝑄‘(𝐸𝑄)) = 1o)    &   (𝜑𝑁 ∈ ω)    &   (𝜑 → ∀𝑘𝑁 (𝑄‘(𝑖 ∈ ω ↦ if(𝑖𝑘, 1o, ∅))) = 1o)    &   (𝜑 → (𝑄‘(𝑖 ∈ ω ↦ if(𝑖𝑁, 1o, ∅))) = ∅)       (𝜑 → (𝐸𝑄) = (𝑖 ∈ ω ↦ if(𝑖𝑁, 1o, ∅)))
 
Theoremnninfsellemqall 13214* Lemma for nninfsel 13216. (Contributed by Jim Kingdon, 9-Aug-2022.)
𝐸 = (𝑞 ∈ (2o𝑚) ↦ (𝑛 ∈ ω ↦ if(∀𝑘 ∈ suc 𝑛(𝑞‘(𝑖 ∈ ω ↦ if(𝑖𝑘, 1o, ∅))) = 1o, 1o, ∅)))    &   (𝜑𝑄 ∈ (2o𝑚))    &   (𝜑 → (𝑄‘(𝐸𝑄)) = 1o)    &   (𝜑𝑁 ∈ ω)       (𝜑 → (𝑄‘(𝑖 ∈ ω ↦ if(𝑖𝑁, 1o, ∅))) = 1o)
 
Theoremnninfsellemeqinf 13215* Lemma for nninfsel 13216. (Contributed by Jim Kingdon, 9-Aug-2022.)
𝐸 = (𝑞 ∈ (2o𝑚) ↦ (𝑛 ∈ ω ↦ if(∀𝑘 ∈ suc 𝑛(𝑞‘(𝑖 ∈ ω ↦ if(𝑖𝑘, 1o, ∅))) = 1o, 1o, ∅)))    &   (𝜑𝑄 ∈ (2o𝑚))    &   (𝜑 → (𝑄‘(𝐸𝑄)) = 1o)       (𝜑 → (𝐸𝑄) = (𝑖 ∈ ω ↦ 1o))
 
Theoremnninfsel 13216* 𝐸 is a selection function for . Theorem 3.6 of [PradicBrown2022], p. 5. (Contributed by Jim Kingdon, 9-Aug-2022.)
𝐸 = (𝑞 ∈ (2o𝑚) ↦ (𝑛 ∈ ω ↦ if(∀𝑘 ∈ suc 𝑛(𝑞‘(𝑖 ∈ ω ↦ if(𝑖𝑘, 1o, ∅))) = 1o, 1o, ∅)))    &   (𝜑𝑄 ∈ (2o𝑚))    &   (𝜑 → (𝑄‘(𝐸𝑄)) = 1o)       (𝜑 → ∀𝑝 ∈ ℕ (𝑄𝑝) = 1o)
 
Theoremnninfomnilem 13217* Lemma for nninfomni 13218. (Contributed by Jim Kingdon, 10-Aug-2022.)
𝐸 = (𝑞 ∈ (2o𝑚) ↦ (𝑛 ∈ ω ↦ if(∀𝑘 ∈ suc 𝑛(𝑞‘(𝑖 ∈ ω ↦ if(𝑖𝑘, 1o, ∅))) = 1o, 1o, ∅)))        ∈ Omni
 
Theoremnninfomni 13218 is omniscient. Corollary 3.7 of [PradicBrown2022], p. 5. (Contributed by Jim Kingdon, 10-Aug-2022.)
∈ Omni
 
Theoremnninffeq 13219* Equality of two functions on which agree at every integer and at the point at infinity. From an online post by Martin Escardo. (Contributed by Jim Kingdon, 4-Aug-2023.)
(𝜑𝐹:ℕ⟶ℕ0)    &   (𝜑𝐺:ℕ⟶ℕ0)    &   (𝜑 → (𝐹‘(𝑥 ∈ ω ↦ 1o)) = (𝐺‘(𝑥 ∈ ω ↦ 1o)))    &   (𝜑 → ∀𝑛 ∈ ω (𝐹‘(𝑖 ∈ ω ↦ if(𝑖𝑛, 1o, ∅))) = (𝐺‘(𝑖 ∈ ω ↦ if(𝑖𝑛, 1o, ∅))))       (𝜑𝐹 = 𝐺)
 
11.3.4  Schroeder-Bernstein Theorem
 
Theoremexmidsbthrlem 13220* Lemma for exmidsbthr 13221. (Contributed by Jim Kingdon, 11-Aug-2022.)
𝑆 = (𝑝 ∈ ℕ ↦ (𝑖 ∈ ω ↦ if(𝑖 = ∅, 1o, (𝑝 𝑖))))       (∀𝑥𝑦((𝑥𝑦𝑦𝑥) → 𝑥𝑦) → EXMID)
 
Theoremexmidsbthr 13221* The Schroeder-Bernstein Theorem implies excluded middle. Theorem 1 of [PradicBrown2022], p. 1. (Contributed by Jim Kingdon, 11-Aug-2022.)
(∀𝑥𝑦((𝑥𝑦𝑦𝑥) → 𝑥𝑦) → EXMID)
 
Theoremexmidsbth 13222* The Schroeder-Bernstein Theorem is equivalent to excluded middle. This is Metamath 100 proof #25. The forward direction (isbth 6855) is the proof of the Schroeder-Bernstein Theorem from the Metamath Proof Explorer database (in which excluded middle holds), but adapted to use EXMID as an antecedent rather than being unconditionally true, as in the non-intuitionist proof at https://us.metamath.org/mpeuni/sbth.html 6855.

The reverse direction (exmidsbthr 13221) is the one which establishes that Schroeder-Bernstein implies excluded middle. This resolves the question of whether we will be able to prove Schroeder-Bernstein from our axioms in the negative. (Contributed by Jim Kingdon, 13-Aug-2022.)

(EXMID ↔ ∀𝑥𝑦((𝑥𝑦𝑦𝑥) → 𝑥𝑦))
 
Theoremsbthomlem 13223 Lemma for sbthom 13224. (Contributed by Mario Carneiro and Jim Kingdon, 13-Jul-2023.)
(𝜑 → ω ∈ Omni)    &   (𝜑𝑌 ⊆ {∅})    &   (𝜑𝐹:ω–1-1-onto→(𝑌 ⊔ ω))       (𝜑 → (𝑌 = ∅ ∨ 𝑌 = {∅}))
 
Theoremsbthom 13224 Schroeder-Bernstein is not possible even for ω. We know by exmidsbth 13222 that full Schroeder-Bernstein will not be provable but what about the case where one of the sets is ω? That case plus the Limited Principle of Omniscience (LPO) implies excluded middle, so we will not be able to prove it. (Contributed by Mario Carneiro and Jim Kingdon, 10-Jul-2023.)
((∀𝑥((𝑥 ≼ ω ∧ ω ≼ 𝑥) → 𝑥 ≈ ω) ∧ ω ∈ Omni) → EXMID)
 
11.3.5  Real and complex numbers
 
Theoremqdencn 13225* 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 10977 (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)
 
Theoremrefeq 13226* Equality of two real functions which agree at negative numbers, positive numbers, and zero. This holds even without real trichotomy. From an online post by Martin Escardo. (Contributed by Jim Kingdon, 9-Jul-2023.)
(𝜑𝐹:ℝ⟶ℝ)    &   (𝜑𝐺:ℝ⟶ℝ)    &   (𝜑 → ∀𝑥 ∈ ℝ (𝑥 < 0 → (𝐹𝑥) = (𝐺𝑥)))    &   (𝜑 → ∀𝑥 ∈ ℝ (0 < 𝑥 → (𝐹𝑥) = (𝐺𝑥)))    &   (𝜑 → (𝐹‘0) = (𝐺‘0))       (𝜑𝐹 = 𝐺)
 
Theoremtriap 13227 Two ways of stating real number trichotomy. (Contributed by Jim Kingdon, 23-Aug-2023.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((𝐴 < 𝐵𝐴 = 𝐵𝐵 < 𝐴) ↔ DECID 𝐴 # 𝐵))
 
Theoremisomninnlem 13228* Lemma for isomninn 13229. The result, with a hypothesis to provide a convenient notation. (Contributed by Jim Kingdon, 30-Aug-2023.)
𝐺 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)       (𝐴𝑉 → (𝐴 ∈ Omni ↔ ∀𝑓 ∈ ({0, 1} ↑𝑚 𝐴)(∃𝑥𝐴 (𝑓𝑥) = 0 ∨ ∀𝑥𝐴 (𝑓𝑥) = 1)))
 
Theoremisomninn 13229* Omniscience stated in terms of natural numbers. Similar to isomnimap 7009 but it will sometimes be more convenient to use 0 and 1 rather than and 1o. (Contributed by Jim Kingdon, 30-Aug-2023.)
(𝐴𝑉 → (𝐴 ∈ Omni ↔ ∀𝑓 ∈ ({0, 1} ↑𝑚 𝐴)(∃𝑥𝐴 (𝑓𝑥) = 0 ∨ ∀𝑥𝐴 (𝑓𝑥) = 1)))
 
Theoremcvgcmp2nlemabs 13230* Lemma for cvgcmp2n 13231. The partial sums get closer to each other as we go further out. The proof proceeds by rewriting (seq1( + , 𝐺)‘𝑁) as the sum of (seq1( + , 𝐺)‘𝑀) and a term which gets smaller as 𝑀 gets large. (Contributed by Jim Kingdon, 25-Aug-2023.)
((𝜑𝑘 ∈ ℕ) → (𝐺𝑘) ∈ ℝ)    &   ((𝜑𝑘 ∈ ℕ) → 0 ≤ (𝐺𝑘))    &   ((𝜑𝑘 ∈ ℕ) → (𝐺𝑘) ≤ (1 / (2↑𝑘)))    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁 ∈ (ℤ𝑀))       (𝜑 → (abs‘((seq1( + , 𝐺)‘𝑁) − (seq1( + , 𝐺)‘𝑀))) < (2 / 𝑀))
 
Theoremcvgcmp2n 13231* A comparison test for convergence of a real infinite series. (Contributed by Jim Kingdon, 25-Aug-2023.)
((𝜑𝑘 ∈ ℕ) → (𝐺𝑘) ∈ ℝ)    &   ((𝜑𝑘 ∈ ℕ) → 0 ≤ (𝐺𝑘))    &   ((𝜑𝑘 ∈ ℕ) → (𝐺𝑘) ≤ (1 / (2↑𝑘)))       (𝜑 → seq1( + , 𝐺) ∈ dom ⇝ )
 
Theoremtrilpolemclim 13232* Lemma for trilpo 13239. Convergence of the series. (Contributed by Jim Kingdon, 24-Aug-2023.)
(𝜑𝐹:ℕ⟶{0, 1})    &   𝐺 = (𝑛 ∈ ℕ ↦ ((1 / (2↑𝑛)) · (𝐹𝑛)))       (𝜑 → seq1( + , 𝐺) ∈ dom ⇝ )
 
Theoremtrilpolemcl 13233* Lemma for trilpo 13239. The sum exists. (Contributed by Jim Kingdon, 23-Aug-2023.)
(𝜑𝐹:ℕ⟶{0, 1})    &   𝐴 = Σ𝑖 ∈ ℕ ((1 / (2↑𝑖)) · (𝐹𝑖))       (𝜑𝐴 ∈ ℝ)
 
Theoremtrilpolemisumle 13234* Lemma for trilpo 13239. An upper bound for the sum of the digits beyond a certain point. (Contributed by Jim Kingdon, 28-Aug-2023.)
(𝜑𝐹:ℕ⟶{0, 1})    &   𝐴 = Σ𝑖 ∈ ℕ ((1 / (2↑𝑖)) · (𝐹𝑖))    &   𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℕ)       (𝜑 → Σ𝑖𝑍 ((1 / (2↑𝑖)) · (𝐹𝑖)) ≤ Σ𝑖𝑍 (1 / (2↑𝑖)))
 
Theoremtrilpolemgt1 13235* Lemma for trilpo 13239. The 1 < 𝐴 case. (Contributed by Jim Kingdon, 23-Aug-2023.)
(𝜑𝐹:ℕ⟶{0, 1})    &   𝐴 = Σ𝑖 ∈ ℕ ((1 / (2↑𝑖)) · (𝐹𝑖))       (𝜑 → ¬ 1 < 𝐴)
 
Theoremtrilpolemeq1 13236* Lemma for trilpo 13239. The 𝐴 = 1 case. This is proved by noting that if any (𝐹𝑥) is zero, then the infinite sum 𝐴 is less than one based on the term which is zero. We are using the fact that the 𝐹 sequence is decidable (in the sense that each element is either zero or one). (Contributed by Jim Kingdon, 23-Aug-2023.)
(𝜑𝐹:ℕ⟶{0, 1})    &   𝐴 = Σ𝑖 ∈ ℕ ((1 / (2↑𝑖)) · (𝐹𝑖))    &   (𝜑𝐴 = 1)       (𝜑 → ∀𝑥 ∈ ℕ (𝐹𝑥) = 1)
 
Theoremtrilpolemlt1 13237* Lemma for trilpo 13239. The 𝐴 < 1 case. We can use the distance between 𝐴 and one (that is, 1 − 𝐴) to find a position in the sequence 𝑛 where terms after that point will not add up to as much as 1 − 𝐴. By finomni 7012 we know the terms up to 𝑛 either contain a zero or are all one. But if they are all one that contradicts the way we constructed 𝑛, so we know that the sequence contains a zero. (Contributed by Jim Kingdon, 23-Aug-2023.)
(𝜑𝐹:ℕ⟶{0, 1})    &   𝐴 = Σ𝑖 ∈ ℕ ((1 / (2↑𝑖)) · (𝐹𝑖))    &   (𝜑𝐴 < 1)       (𝜑 → ∃𝑥 ∈ ℕ (𝐹𝑥) = 0)
 
Theoremtrilpolemres 13238* Lemma for trilpo 13239. The result. (Contributed by Jim Kingdon, 23-Aug-2023.)
(𝜑𝐹:ℕ⟶{0, 1})    &   𝐴 = Σ𝑖 ∈ ℕ ((1 / (2↑𝑖)) · (𝐹𝑖))    &   (𝜑 → (𝐴 < 1 ∨ 𝐴 = 1 ∨ 1 < 𝐴))       (𝜑 → (∃𝑥 ∈ ℕ (𝐹𝑥) = 0 ∨ ∀𝑥 ∈ ℕ (𝐹𝑥) = 1))
 
Theoremtrilpo 13239* Real number trichotomy implies the Limited Principle of Omniscience (LPO). We expect that we'd need some form of countable choice to prove the converse.

Here's the outline of the proof. Given an infinite sequence F of zeroes and ones, we need to show the sequence contains a zero or it is all ones.

Construct a real number A whose representation in base two consists of a zero, a decimal point, and then the numbers of the sequence. Compare it with one using trichotomy. The three cases from trichotomy are trilpolemlt1 13237 (which means the sequence contains a zero), trilpolemeq1 13236 (which means the sequence is all ones), and trilpolemgt1 13235 (which is not possible). (Contributed by Jim Kingdon, 23-Aug-2023.)

(∀𝑥 ∈ ℝ ∀𝑦 ∈ ℝ (𝑥 < 𝑦𝑥 = 𝑦𝑦 < 𝑥) → ω ∈ Omni)
 
11.3.6  Supremum and infimum
 
Theoremsupfz 13240 The supremum of a finite sequence of integers. (Contributed by Scott Fenton, 8-Aug-2013.) (Revised by Jim Kingdon, 15-Oct-2022.)
(𝑁 ∈ (ℤ𝑀) → sup((𝑀...𝑁), ℤ, < ) = 𝑁)
 
Theoreminffz 13241 The infimum of a finite sequence of integers. (Contributed by Scott Fenton, 8-Aug-2013.) (Revised by Jim Kingdon, 15-Oct-2022.)
(𝑁 ∈ (ℤ𝑀) → inf((𝑀...𝑁), ℤ, < ) = 𝑀)
 
11.3.6.1  Circle constant
 
Theoremtaupi 13242 Relationship between τ and π. This can be seen as connecting the ratio of a circle's circumference to its radius and the ratio of a circle's circumference to its diameter. (Contributed by Jim Kingdon, 19-Feb-2019.) (Revised by AV, 1-Oct-2020.)
τ = (2 · π)
 
11.4  Mathbox for Mykola Mostovenko
 
Theoremax1hfs 13243 Heyting's formal system Axiom #1 from [Heyting] p. 127. (Contributed by MM, 11-Aug-2018.)
(𝜑 → (𝜑𝜑))
 
11.5  Mathbox for David A. Wheeler
 
11.5.1  Testable propositions
 
Theoremdftest 13244 A proposition is testable iff its negative or double-negative is true. See Chapter 2 [Moschovakis] p. 2.

We do not formally define testability with a new token, but instead use DECID ¬ before the formula in question. For example, DECID ¬ 𝑥 = 𝑦 corresponds to "𝑥 = 𝑦 is testable". (Contributed by David A. Wheeler, 13-Aug-2018.) For statements about testable propositions, search for the keyword "testable" in the comments of statements, for instance using the Metamath command "MM> SEARCH * "testable" / COMMENTS". (New usage is discouraged.)

(DECID ¬ 𝜑 ↔ (¬ 𝜑 ∨ ¬ ¬ 𝜑))
 
11.5.2  Allsome quantifier

These are definitions and proofs involving an experimental "allsome" quantifier (aka "all some").

In informal language, statements like "All Martians are green" imply that there is at least one Martian. But it's easy to mistranslate informal language into formal notations because similar statements like 𝑥𝜑𝜓 do not imply that 𝜑 is ever true, leading to vacuous truths. Some systems include a mechanism to counter this, e.g., PVS allows types to be appended with "+" to declare that they are nonempty. This section presents a different solution to the same problem.

The "allsome" quantifier expressly includes the notion of both "all" and "there exists at least one" (aka some), and is defined to make it easier to more directly express both notions. The hope is that if a quantifier more directly expresses this concept, it will be used instead and reduce the risk of creating formal expressions that look okay but in fact are mistranslations. The term "allsome" was chosen because it's short, easy to say, and clearly hints at the two concepts it combines.

I do not expect this to be used much in metamath, because in metamath there's a general policy of avoiding the use of new definitions unless there are very strong reasons to do so. Instead, my goal is to rigorously define this quantifier and demonstrate a few basic properties of it.

The syntax allows two forms that look like they would be problematic, but they are fine. When applied to a top-level implication we allow ∀!𝑥(𝜑𝜓), and when restricted (applied to a class) we allow ∀!𝑥𝐴𝜑. The first symbol after the setvar variable must always be if it is the form applied to a class, and since cannot begin a wff, it is unambiguous. The looks like it would be a problem because 𝜑 or 𝜓 might include implications, but any implication arrow within any wff must be surrounded by parentheses, so only the implication arrow of ∀! can follow the wff. The implication syntax would work fine without the parentheses, but I added the parentheses because it makes things clearer inside larger complex expressions, and it's also more consistent with the rest of the syntax.

For more, see "The Allsome Quantifier" by David A. Wheeler at https://dwheeler.com/essays/allsome.html I hope that others will eventually agree that allsome is awesome.

 
Syntaxwalsi 13245 Extend wff definition to include "all some" applied to a top-level implication, which means 𝜓 is true whenever 𝜑 is true, and there is at least least one 𝑥 where 𝜑 is true. (Contributed by David A. Wheeler, 20-Oct-2018.)
wff ∀!𝑥(𝜑𝜓)
 
Syntaxwalsc 13246 Extend wff definition to include "all some" applied to a class, which means 𝜑 is true for all 𝑥 in 𝐴, and there is at least one 𝑥 in 𝐴. (Contributed by David A. Wheeler, 20-Oct-2018.)
wff ∀!𝑥𝐴𝜑
 
Definitiondf-alsi 13247 Define "all some" applied to a top-level implication, which means 𝜓 is true whenever 𝜑 is true and there is at least one 𝑥 where 𝜑 is true. (Contributed by David A. Wheeler, 20-Oct-2018.)
(∀!𝑥(𝜑𝜓) ↔ (∀𝑥(𝜑𝜓) ∧ ∃𝑥𝜑))
 
Definitiondf-alsc 13248 Define "all some" applied to a class, which means 𝜑 is true for all 𝑥 in 𝐴 and there is at least one 𝑥 in 𝐴. (Contributed by David A. Wheeler, 20-Oct-2018.)
(∀!𝑥𝐴𝜑 ↔ (∀𝑥𝐴 𝜑 ∧ ∃𝑥 𝑥𝐴))
 
Theoremalsconv 13249 There is an equivalence between the two "all some" forms. (Contributed by David A. Wheeler, 22-Oct-2018.)
(∀!𝑥(𝑥𝐴𝜑) ↔ ∀!𝑥𝐴𝜑)
 
Theoremalsi1d 13250 Deduction rule: Given "all some" applied to a top-level inference, you can extract the "for all" part. (Contributed by David A. Wheeler, 20-Oct-2018.)
(𝜑 → ∀!𝑥(𝜓𝜒))       (𝜑 → ∀𝑥(𝜓𝜒))
 
Theoremalsi2d 13251 Deduction rule: Given "all some" applied to a top-level inference, you can extract the "exists" part. (Contributed by David A. Wheeler, 20-Oct-2018.)
(𝜑 → ∀!𝑥(𝜓𝜒))       (𝜑 → ∃𝑥𝜓)
 
Theoremalsc1d 13252 Deduction rule: Given "all some" applied to a class, you can extract the "for all" part. (Contributed by David A. Wheeler, 20-Oct-2018.)
(𝜑 → ∀!𝑥𝐴𝜓)       (𝜑 → ∀𝑥𝐴 𝜓)
 
Theoremalsc2d 13253 Deduction rule: Given "all some" applied to a class, you can extract the "there exists" part. (Contributed by David A. Wheeler, 20-Oct-2018.)
(𝜑 → ∀!𝑥𝐴𝜓)       (𝜑 → ∃𝑥 𝑥𝐴)
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