fix missing lemma docs from statement names

2023-04-03 11:53:48 +02:00
parent c2b298e650
commit f636a8d833
27 changed files with 56 additions and 36 deletions
@@ -464,3 +464,21 @@ LemmaDoc Iff.symm as "Iff.symm" in "Logic"

 * Mathlib Doc: [#Iff.symm](https://leanprover-community.github.io/mathlib4_docs/Init/Core.html#Iff.symm)
 "
+
+LemmaDoc not_imp_not as "not_imp_not" in "Logic"
+"
+`theorem not_imp_not {a : Prop} {b : Prop} : ¬a → ¬b ↔ b → a`
+
+## Eigenschaften
+
+* Mathlib Doc: [#not_imp_not](https://leanprover-community.github.io/mathlib4_docs/Mathlib/Logic/Basic.html#not_imp_not)
+"
+
+LemmaDoc Set.nonempty_iff_ne_empty as "nonempty_iff_ne_empty" in "Logic"
+"
+`theorem Set.nonempty_iff_ne_empty {α : Type u} {s : Set α} : Set.Nonempty s ↔ s ≠ ∅`
+
+## Eigenschaften
+
+* Mathlib Doc: [#nonempty_iff_ne_empty](https://leanprover-community.github.io/mathlib4_docs/Mathlib/Data/Set/Basic.html#Set.nonempty_iff_ne_empty)
+"
@@ -20,7 +20,8 @@ aber er möchte, dass du ihm das hier und jetzt nochmals von Grund auf zeigst.
 open Function

 --TODO: This is a really hard proof
-Statement bijective_iff_has_inverse "" {A B : Type} (f : A → B) :
+-- bijective_iff_has_inverse
+Statement "" {A B : Type} (f : A → B) :
    Bijective f ↔ ∃ g, LeftInverse g f ∧ RightInverse g f := by
  constructor
  intro h
@@ -19,7 +19,9 @@ aber er möchte, dass du ihm das hier und jetzt nochmals von Grund auf zeigst.

 open Function
 set_option pp.rawOnError true
-Statement bijective_iff_has_inverse "" {A B : Type} (f : A → B) :
+
+-- bijective_iff_has_inverse
+Statement {A B : Type} (f : A → B) :
    Bijective f ↔ ∃ g, LeftInverse g f ∧ RightInverse g f := by
  Branch
    exfalso
@@ -27,7 +27,8 @@ local notation "ℝ²" => Fin 2 → ℝ

 open Submodule Set Finsupp

-Statement mem_span_of_mem
+-- mem_span_of_mem
+Statement
  "Zeige, dass $x \\in M$ auch Element von der $K$-linearen Hülle
  \\langle M \\rangle ist."
    {V K : Type _} [Field K] [AddCommMonoid V]
@@ -19,7 +19,7 @@ Introduction
 Sofort taucht das nächste Blatt auf.  Anscheinend hatten sie sich auf einen Kompromiss geeinigt.
 "

-Statement odd_square (n : ℕ) (h : Odd n) : Odd (n ^ 2) := by
+Statement (n : ℕ) (h : Odd n) : Odd (n ^ 2) := by
  Hint (hidden := true) "**Robo**: mit `rcases h with ⟨r, hr⟩` kannst du wieder
  das `r` nehmen, das laut Annahme existieren muss.

@@ -32,7 +32,7 @@ erklärt mir doch folgendes:

 open Set

-Statement mem_univ "" {A : Type} (x : A) : x ∈ (univ : Set A) := by
+Statement Set.mem_univ "" {A : Type} (x : A) : x ∈ (univ : Set A) := by
  Hint "**Du**: Also `A` ist ein `Type`, `x` ist ein Element in `A`…

  **Robo** … und `univ` ist die Menge aller Elemente in `A`.
@@ -18,7 +18,7 @@ Ihr zieht also durch die Gegend und redet mit den Leuten. Ein Junge rennt zu euc

 open Set

-Statement not_mem_empty "" {A : Type} (x : A) :
+Statement Set.not_mem_empty "" {A : Type} (x : A) :
    x ∉ (∅ : Set A) := by
  Hint "**Du**: Kein Element ist in der leeren Menge enthalten? Das ist ja alles
  tautologisches Zeugs...
@@ -31,7 +31,7 @@ aufteilen.

 open Set Subset

-Statement subset_empty_iff {A : Type _} (s : Set A) :
+Statement Set.subset_empty_iff {A : Type _} (s : Set A) :
    s ⊆ ∅ ↔ s = ∅ := by
  Hint "**Du**: Ja, die einzige Teilmenge der leeren Menge ist die leere Menge.
  Das ist doch eine Tautologie?
@@ -22,7 +22,7 @@ Zeige folgendes Lemma, welches wir gleich brauchen werden:
 open Set


-Statement eq_empty_iff_forall_not_mem
+Statement Set.eq_empty_iff_forall_not_mem
 ""
    {A : Type _} (s : Set A) :
    s = ∅ ↔ ∀ x, x ∉ s := by
@@ -21,8 +21,7 @@ Zeige dass die beiden Ausdrücke äquivalent sind:

 open Set

-Statement nonempty_iff_ne_empty
-""
+Statement Set.nonempty_iff_ne_empty
    {A : Type _} (s : Set A) :
    s.Nonempty ↔ s ≠ ∅ := by
  Hint "Am besten fängst du mit `unfold Set.Nonempty` an."
@@ -42,7 +42,6 @@ open Set
 #check mem_compl_iff

 Statement
-""
    (A B C : Set ℕ) : (A \ B)ᶜ ∩ (C \ B)ᶜ = ((univ \ A) \ C) ∪ (univ \ Bᶜ) := by
  rw [←compl_union]
  rw [←union_diff_distrib]
@@ -22,7 +22,7 @@ open Fin

 open BigOperators

-Statement odd_arithmetic_sum
+Statement
    "$\\sum_{i = 0}^n (2n + 1) = n ^ 2$."
    (n : ℕ) : (∑ i : Fin n, (2 * (i : ℕ) + 1)) = n ^ 2 := by
  Hint "**Robo**: Das funktioniert genau gleich wie zuvor, viel Glück."
@@ -1,31 +1,31 @@
 import GameServer.Commands

 LemmaDoc MyNat.add_zero as "add_zero" in "Nat"
-""
+"(missing)"

 LemmaDoc MyNat.add_succ as "add_succ" in "Nat"
-""
+"(missing)"

 LemmaDoc MyNat.zero_add as "zero_add" in "Nat"
-""
+"(missing)"

 LemmaDoc MyNat.add_assoc as "add_assoc" in "Nat"
-""
+"(missing)"

 LemmaDoc MyNat.succ_add as "succ_add" in "Nat"
-""
+"(missing)"

 LemmaDoc MyNat.add_comm as "add_comm" in "Nat"
-""
+"(missing)"

 LemmaDoc MyNat.one_eq_succ_zero as "one_eq_succ_zero" in "Nat"
-""
+"(missing)"

 LemmaDoc not_iff_imp_false as "not_iff_imp_false" in "Prop"
-""
+"(missing)"

 LemmaDoc MyNat.succ_inj as "succ_inj" in "Nat"
-""
+"(missing)"

 LemmaDoc MyNat.zero_ne_succ as "zero_ne_succ" in "Nat"
-""
+"(missing)"
@@ -36,7 +36,7 @@ note that `zero_add` is about zero add something, and `add_zero` is about someth
 The names of the proofs tell you what the theorems are. Anyway, let's prove `0 + n = n`.
 "

-Statement zero_add
+Statement MyNat.zero_add
 "For all natural numbers $n$, we have $0 + n = n$."
    (n : ℕ) : 0 + n = n := by
  Hint "You can start a proof by induction over `n` by typing:
@@ -30,7 +30,7 @@ that addition, as defined the way we've defined it, is associative.
 See if you can prove associativity of addition.
 "

-Statement add_assoc
+Statement MyNat.add_assoc
 "On the set of natural numbers, addition is associative.
 In other words, for all natural numbers $a, b$ and $c$, we have
 $ (a + b) + c = a + (b + c). $"
@@ -42,7 +42,7 @@ that `a + succ(b) = succ(a + b)`. The tactic `rw [add_succ]` just says to Lean \
 what the variables are\".
 "

-Statement succ_add
+Statement MyNat.succ_add
 "For all natural numbers $a, b$, we have
 $ \\operatorname{succ}(a) + b = \\operatorname{succ}(a + b)$."
    (a b : ℕ) : succ a + b = succ (a + b)  := by
@@ -29,7 +29,7 @@ Look in your inventory to see the proofs you have available.
 These should be enough.
 "

-Statement add_comm
+Statement MyNat.add_comm
 "On the set of natural numbers, addition is commutative.
 In other words, for all natural numbers $a$ and $b$, we have
 $a + b = b + a$."
@@ -40,7 +40,7 @@ some theorems about $0$ (`zero_add`, `add_zero`), but, other than `1 = succ 0`,
 no theorems at all which mention $1$. Let's prove one now.
 "

-Statement succ_eq_add_one
+Statement --MyNat.succ_eq_add_one
 "For any natural number $n$, we have
 $ \\operatorname{succ}(n) = n+1$ ."
    (n : ℕ) : succ n = n + 1 := by
@@ -35,7 +35,7 @@ additions of the form `a + ?`, and `rw add_comm a b,` will only
 swap additions of the form `a + b`.
 "

-Statement add_right_comm
+Statement --add_right_comm
 "For all natural numbers $a, b$ and $c$, we have
 $a + b + c = a + c + b$."
    (a b c : ℕ) : a + b + c = a + c + b := by
@@ -17,7 +17,7 @@ theorem MyNat.succ_inj {a b : ℕ} : succ a = succ b → a = b := by simp only [

 theorem MyNat.zero_ne_succ (a : ℕ) : zero ≠ succ a := by simp only [ne_eq, not_false_iff]

-Statement succ_inj'
+Statement -- MyNat.succ_inj'
 ""
    {a b : ℕ} (hs : succ a = succ b) :  a = b := by
    exact succ_inj hs
@@ -16,7 +16,7 @@ Introduction

 set_option tactic.hygienic false

-Statement and_symm
+Statement --and_symm
 ""
    (P Q : Prop) : P ∧ Q → Q ∧ P :=  by
  intro h
@@ -14,7 +14,7 @@ Introduction

 "

-Statement and_trans
+Statement --and_trans
 ""
    (P Q R : Prop) : P ∧ Q → Q ∧ R → P ∧ R := by
  intro hpq
@@ -14,7 +14,7 @@ Introduction

 "

-Statement iff_trans
+Statement --iff_trans
 ""
    (P Q R : Prop) : (P ↔ Q) → (Q ↔ R) → (P ↔ R) := by
  intro hpq
@@ -14,7 +14,7 @@ Introduction

 "

-Statement iff_trans
+Statement --iff_trans
 ""
    (P Q R : Prop) : (P ↔ Q) → (Q ↔ R) → (P ↔ R) := by
  intro hpq hqr
@@ -15,7 +15,7 @@ Introduction

 "

-Statement or_symm
+Statement --or_symm
 ""
    (P Q : Prop) : P ∨ Q → Q ∨ P := by
  intro h
@@ -15,7 +15,7 @@ Introduction

 "

-Statement and_or_distrib_left
+Statement --and_or_distrib_left
 ""
    (P Q R : Prop) : P ∧ (Q ∨ R) ↔ (P ∧ Q) ∨ (P ∧ R) := by
  constructor
@@ -17,7 +17,7 @@ Introduction

 "

-Statement contra
+Statement --contra
 ""
  (P Q : Prop) : (P ∧ ¬ P) → Q := by
  intro h