cleanup

Game/Levels/Addition/L01two_add_one.lean

@@ -5,7 +5,6 @@ World "Addition"
 Level 1
 Title "Two add one is three."
 
---namespace MyNat
 namespace MyNat
 
 Introduction
@@ -28,8 +27,7 @@ Can you prove `2 + 1 = 3`?
 "
 
 /-- $2+1=3$. -/
-Statement
-    : (2 : ℕ) + 1 = 3 := by
+Statement : (2 : ℕ) + 1 = 3 := by
   Hint "Which one of Peano's axioms do we ultimately want to use to rewrite that addition?"
   Hint (hidden := true) "change `1` to `succ 0` with a rewrite, and then
   think about Peano's axioms."
@@ -39,9 +37,11 @@ Statement
   rw [three_eq_succ_two]
   rfl
 
+NewLemma MyNat.two_eq_succ_one MyNat.three_eq_succ_two
+  MyNat.four_eq_succ_three MyNat.five_eq_succ_four
 LemmaTab "Add"
 
 Conclusion
 "
-  Are you up for `2 + 2 = 4`?
+Are you up for `2 + 2 = 4`?
 "

Game/Levels/Addition/L02one_add_one.lean

@@ -5,7 +5,6 @@ World "Addition"
 Level 2
 Title "One add one is two."
 
---namespace MyNat
 namespace MyNat
 
 Introduction
@@ -16,8 +15,7 @@ strategy.
 "
 
 /-- $1+1=2$. -/
-Statement
-    : (1 : ℕ) + 1 = 2 := by
+Statement : (1 : ℕ) + 1 = 2 := by
   Hint "Go ahead and start with `rw [one_eq_succ_zero]`. Do you see what it does?"
   Branch
     rw [one_eq_succ_zero]
@@ -39,5 +37,5 @@ LemmaTab "Add"
 
 Conclusion
 "
-  Are you now up for the first sub-boss `2 + 2 = 4`?
+Are you now up for the first sub-boss `2 + 2 = 4`?
 "

Game/Levels/Addition/L03two_add_two.lean

@@ -5,7 +5,6 @@ World "Addition"
 Level 3
 Title "Two add two is four."
 
---namespace MyNat
 namespace MyNat
 
 Introduction
@@ -15,8 +14,7 @@ about succ 3 and 4, just look at the blah blah I have no idea.
 "
 
 /-- $2+2=4$. -/
-Statement
-    : (2 : ℕ) + 2 = 4 := by
+Statement : (2 : ℕ) + 2 = 4 := by
   nth_rewrite 2 [two_eq_succ_one]
   rw [add_succ, one_eq_succ_zero, add_succ, add_zero, four_eq_succ_three, three_eq_succ_two]
   rfl
@@ -25,7 +23,7 @@ LemmaTab "Add"
 
 Conclusion
 "
-  Nice. In the next level we'll prove `29 + 35 = 64`.
+Nice. In the next level we'll prove `29 + 35 = 64`.
 
-  In one line.
+In one line.
 "

Game/Levels/Addition/L04decide.lean

@@ -18,6 +18,7 @@ on `MyNat` in `Game.MyNat.DecidableEq`. The implementation
 is not at all optimised: I just wanted to get something which could beat
 humans easily.
 "
+
 NewTactic decide
 
 example : (20 : ℕ) + 20 = 40 := by
@@ -25,17 +26,16 @@ example : (20 : ℕ) + 20 = 40 := by
 
 Introduction
 "
-Oh did I mention there was a new tactic? Find it in the blah blah blah.
+Oh did I mention there was a new tactic? Find it highlighted in your inventory.
 "
-/-- $29+35=64. -/
-Statement
-    : (29 : ℕ) + 35 = 64 := by
+
+/-- $29+35=64$. -/
+Statement : (29 : ℕ) + 35 = 64 := by
   decide
 
 LemmaTab "Add"
 
 Conclusion
 "
-  The `decide` tactic destroys sub-bosses such as `2 + 2 = 4`.
-
+The `decide` tactic destroys sub-bosses such as `2 + 2 = 4`.
 "

Game/Levels/Addition/L05zero_add.lean

@@ -47,13 +47,12 @@ because we have formulas for adding `0` and adding successors. See if you
 can do your first induction proof in Lean.
 "
 
-LemmaDoc MyNat.zero_add as "zero_add" in "MyNat" "`zero_add x` is the proof that `0 + x = x`. It's
+LemmaDoc MyNat.zero_add as "zero_add" in "Add" "`zero_add x` is the proof that `0 + x = x`. It's
 a `simp` lemma, because replacing `0 + x` by `x` is almost always what you want
 to do if you're simplifying. "
 
 /-- For all natural numbers $n$, we have $0 + n = n$. -/
-Statement zero_add
-    (n : ℕ) : 0 + n = n := by
+Statement zero_add (n : ℕ) : 0 + n = n := by
   Hint "You can start a proof by induction over `n` by typing:
   `induction n with d hd`."
   induction n with d hd

Game/Levels/Addition/L06succ_add.lean

@@ -28,11 +28,14 @@ that `a + succ(b) = succ(a + b)`. The tactic `rw [add_succ]` just says to Lean \
 what the variables are\".
 "
 
-LemmaDoc MyNat.succ_add as "succ_add" in "MyNat" "`succ_add a b` is a proof that `succ a + b = succ (a + b)`."
-/-- For all natural numbers $a, b$, we have
-$ \\operatorname{succ}(a) + b = \\operatorname{succ}(a + b)$. -/
-Statement succ_add
-    (a b : ℕ) : succ a + b = succ (a + b)  := by
+LemmaDoc MyNat.succ_add as "succ_add" in "Add"
+"`succ_add a b` is a proof that `succ a + b = succ (a + b)`."
+
+/--
+For all natural numbers $a, b$, we have
+$ \operatorname{succ}(a) + b = \operatorname{succ}(a + b)$.
+-/
+Statement succ_add (a b : ℕ) : succ a + b = succ (a + b)  := by
   Hint (hidden := true) "You might want to think about whether induction
   on `a` or `b` is the best idea."
   Branch

Game/Levels/Addition/L07add_comm.lean

@@ -15,12 +15,13 @@ Look in your inventory to see the proofs you have available.
 These should be enough.
 "
 
-LemmaDoc MyNat.add_comm as "add_comm" in "MyNat" "`add_comm x y` is a proof of `x + y = y + x`."
+LemmaDoc MyNat.add_comm as "add_comm" in "Add"
+"`add_comm x y` is a proof of `x + y = y + x`."
+
 /-- On the set of natural numbers, addition is commutative.
 In other words, if `a` and `b` are arbitrary natural numbers, then
 $a + b = b + a$. -/
-Statement add_comm
-    (a b : ℕ) : a + b = b + a := by
+Statement add_comm (a b : ℕ) : a + b = b + a := by
   Hint (hidden := true) "Induction on `a` or `b` -- it's all the same in this one."
   Branch
     induction a with d hd
@@ -30,6 +31,9 @@ Statement add_comm
   · simp
   · simp_all [succ_add, add_succ]
 
+-- Adding this instance to make `ac_rfl` work.
+instance : Lean.IsCommutative (α := ℕ) (·+·) := ⟨add_comm⟩
+
 LemmaTab "Add"
 
 Conclusion

Game/Levels/Addition/L08add_assoc.lean

@@ -21,7 +21,7 @@ See if you can prove associativity of addition. Note that now we have
 it probably doesn't matter which variable we induct on. Take your pick!
 "
 
-LemmaDoc MyNat.add_assoc as "add_assoc" in "MyNat" "`add_assoc a b c` is a proof
+LemmaDoc MyNat.add_assoc as "add_assoc" in "Add" "`add_assoc a b c` is a proof
 that `(a + b) + c = a + (b + c)`. Note that in Lean `(a + b) + c` prints
 as `a + b + c`, because the notation for addition is defined to be left
 associative. "
@@ -29,8 +29,7 @@ associative. "
 /-- On the set of natural numbers, addition is associative.
 In other words, if $a, b$ and $c$ are arbitrary natural numbers, we have
 $ (a + b) + c = a + (b + c). $ -/
-Statement add_assoc
-    (a b c : ℕ) : (a + b) + c = a + (b + c) := by
+Statement add_assoc (a b c : ℕ) : (a + b) + c = a + (b + c) := by
   Hint "Note that when Lean writes `a + b + c`, it means `(a + b) + c`. If it wants to talk
   about `a + (b + c)` it will put the brackets in explictly."
   Branch
@@ -45,6 +44,9 @@ Statement add_assoc
   simp
   simp [hc, succ_add, add_succ]
 
+-- Adding this instance to make `ac_rfl` work.
+instance : Lean.IsAssociative (α := ℕ) (·+·) := ⟨add_assoc⟩
+
 NewLemma MyNat.add_assoc
 LemmaTab "Add"
 

Game/Levels/Addition/L09add_left_comm.lean

@@ -21,12 +21,12 @@ like `(d + f) + (h + (a + c)) + (g + e + b) = a + b + c + d + e + f + g + h`.
 Let's start with a simpler one. Can you do it in three rewrites?
 "
 
-LemmaDoc MyNat.add_left_comm as "add_left_comm" in "MyNat" "`add_left_comm a b c` is a proof that `a + (b + c) = b + (a + c)`."
+LemmaDoc MyNat.add_left_comm as "add_left_comm" in "Add"
+"`add_left_comm a b c` is a proof that `a + (b + c) = b + (a + c)`."
 
 /-- If $a, b$ and $c$ are arbitrary natural numbers, we have
 $a + (b + c) = b + (a + c)$. -/
-Statement add_left_comm
-    (a b c : ℕ) : a + (b + c) = b + (a + c) := by
+Statement add_left_comm (a b c : ℕ) : a + (b + c) = b + (a + c) := by
   Hint "Don't use induction; `add_assoc` and `add_comm` are all the tools you need.
     Remember that to rewrite `h : X = Y` backwards (i.e. to change `Y`s to `X`s
     rather than `X`s to `Y`s) use `rw [←h]`"

Game/Levels/Addition/L10annoying.lean

@@ -13,12 +13,11 @@ where the order of the variables needs to be changed
 and the brackets need to be moved around. In this level we learn
 an algorithm which will work for an arbitrary such problem. Let's
 prove `a + b + (c + d) = a + c + d + b`.
-
 "
+
 /-- If $a, b$, $c$ and $d$ are arbitrary natural numbers, we have
 $(a + b) + (c + d) = ((a + c) + d) + b. -/
-Statement
-    (a b c d : ℕ) : a + d + (b + c) = a + b + c + d := by
+Statement (a b c d : ℕ) : a + d + (b + c) = a + b + c + d := by
   Hint "We no longer have to use inducion; `add_assoc` and `add_comm` are
     all the tools we need.
     Start with `repeat rw [add_assoc]` to push all the brackets to the right."
@@ -32,10 +31,11 @@ Statement
   `d` and `b` on the left hand side."
   rw [add_left_comm d b, add_comm d c]
   rfl
+
 LemmaTab "Add"
 
 Conclusion
 "
-  In the last level we use automation to perform this algorithm
-  automatically.
+In the last level we use automation to perform this algorithm
+automatically.
 "

Game/Levels/Addition/L11ac_rfl.lean

@@ -19,15 +19,17 @@ numbers. Let's now write a tactic which automates this.
 
 /-- If $a, b,\\ldots h$ are arbitrary natural numbers, we have
 $(d + f) + (h + (a + c)) + (g + e + b) = a + b + c + d + e + f + g + h$. -/
-Statement
-    (a b c d e f g h : ℕ) : (d + f) + (h + (a + c)) + (g + e + b) = a + b + c + d + e + f + g + h := by
-  simp only [add_assoc, add_left_comm, add_comm]
+Statement (a b c d e f g h : ℕ) :
+    (d + f) + (h + (a + c)) + (g + e + b) = a + b + c + d + e + f + g + h := by
+  ac_rfl
+
+NewTactic ac_rfl
 LemmaTab "Add"
 
 Conclusion
 "
-  Congratulations! You finished addition world. Now go back to the overworld by clicking the
-  home button in the top left. If you want to press on to the final boss
-  of the game then go to Multiplication world next. If you are in no hurry, and would like
-  to learn some more tactics, then you can try Advanced Addition World.
+Congratulations! You finished addition world. Now go back to the overworld by clicking the
+home button in the top left. If you want to press on to the final boss
+of the game then go to Multiplication world next. If you are in no hurry, and would like
+to learn some more tactics, then you can try Advanced Addition World.
 "

Game/Levels/Tutorial/L01rfl.lean

@@ -44,14 +44,20 @@ If the goal looks like this:
 ⊢ x^37+691*y^24+1=x^37+691*y^24+1
 ```
 
-then `rfl` will close it. But if it looks like `0 + x = x` then `rfl` won't work, because even though $0+x$ is *equal* to $x$, it is not *exactly the same thing* as *x*. The only thing which is exactly the same as `0 + x` is `0 + x`.
+then `rfl` will close it. But if it looks like `0 + x = x` then `rfl` won't work,
+because even though $0+x$ is *equal* to $x$, it is not *exactly the same thing* as *x*.
+The only thing which is exactly the same as `0 + x` is `0 + x`.
 "
-NewTactic rfl
-DefinitionDoc MyNat as "Nat" "The natural numbers, defined as an inductive type, with two constructors:
 
-* `0 : Nat`
-* `succ (n : Nat) : Nat`
+
+DefinitionDoc MyNat as "ℕ"
+"The natural numbers, defined as an inductive type, with two constructors:
+
+* `0 : ℕ`
+* `succ (n : ℕ) : ℕ`
 "
+
+NewTactic rfl
 NewDefinition MyNat
 
 Conclusion

Game/Levels/Tutorial/L02rw.lean

@@ -24,7 +24,6 @@ by replacing `y` by `x + 7` and then using `rfl`. The tactic which we use to
 do this kind of \"substituting in\" is called the *rewrite* tactic `rw`.
 The tactic `rw [h]` will replace all occurences of the left hand side $y$ of `h`
 in the goal, with the right hand side $x+7$. Try it and see.
-
 "
 
 /-- If $x$ and $y$ are natural numbers, and $y = x + 7$, then $2y = 2(x + 7)$. -/
@@ -108,6 +107,7 @@ h2 : 2 * y = x
 then `rw h1 at h2` will turn `h2` into `h2 : 2 * y = y + 3`.
 -/
 "
+
 NewTactic rw
 
 Conclusion

Game/Levels/Tutorial/L03rw_practice.lean

@@ -18,14 +18,15 @@ Use the `rw` tactic to prove that $ab + 2c^3 + 7 = pq + 2r^3 + 7$. Here are some
 "
 
 /-- If $a=p$, $b=q$ and $c=r$, then $ab+2c^3+7=pq+2r^3+7.$ -/
-Statement
-    (a b c p q r : ℕ) (hap : a = p) (hbq : b = q) (hcr : c = r) : a * b + 2 * c ^ 3 + 7 = p * q + 2 * r ^ 3 + 7 := by
+Statement (a b c p q r : ℕ) (hap : a = p) (hbq : b = q) (hcr : c = r) :
+    a * b + 2 * c ^ 3 + 7 = p * q + 2 * r ^ 3 + 7 := by
   Hint "Switch to editor mode to make it easier to experiment with the `rw` tactic. Note that each tactic needs to start at the beginning of a line."
   rw [hap, hbq, hcr]
   rfl
 
 Conclusion
 "
-In editor mode,you can click around the proof and see the state of Lean's brain at any point in the proof.
+In editor mode,you can click around the proof and see the state of Lean's brain at
+any point in the proof.
 You can also edit your proof and experiment more with it.
 "

Game/Levels/Tutorial/L04onetwothree.lean

@@ -8,7 +8,9 @@ Title "Peano axioms"
 
 namespace MyNat
 
-LemmaDoc MyNat.one_eq_succ_zero as "foobar" in "bazqux" "`one_eq_succ_zero is a proof of `1 = succ 0`."
+LemmaDoc MyNat.one_eq_succ_zero as "one_eq_succ_zero" in "ℕ"
+"`one_eq_succ_zero is a proof of `1 = succ 0`."
+
 NewLemma MyNat.one_eq_succ_zero
 
 Introduction
@@ -35,7 +37,8 @@ function `succ` taking numbers to numbers. We could apply the function to the nu
 We could define `2` to either be `succ 1` or `succ (succ 0)`. Are these two choices definitely
 equal? Let's see if we can prove it.
 "
-/-- $\\operatorname{succ}(1)=\\operatorname{succ}(\\operatorname{succ}(0))$. -/
+
+/-- $\operatorname{succ}(1)=\operatorname{succ}(\operatorname{succ}(0))$. -/
 Statement
     : succ 1 = succ (succ 0) := by
   Hint "You can use `rw` and `one_eq_succ_zero` your assumption `h` to substitute `succ a` with `b`.
@@ -55,6 +58,8 @@ Statement
   Hint (hidden := true) "Now both sides are identical…"
   rfl
 
+NewLemma MyNat.one_eq_succ_zero
+
 Conclusion
 "
 We can now go on to define `2`, `3`, `4` and `5` and create associated

Game/Levels/Tutorial/L05add_zero.lean

@@ -22,37 +22,14 @@ and returns a proof `add_zero a` of `a + 0 = a`.
 Let me show you how to use Lean's simplifier `simp`
 to do a lot of work for us."
 
-LemmaDoc MyNat.add_zero as "add_zero" in "Add"
-"`add_zero n` is a proof that `n + 0 = n`.
-
-This is one of the two axioms for addition."
-
-NewLemma MyNat.add_zero
-
 namespace MyNat
 
-TacticDoc simp "The simplifier. `rw` on steroids.
-
-A bunch of lemmas like `add_zero : ∀ a, a + 0 = a`
-are tagged with the `@[simp]` tag. If the `simp` tactic
-is run by the user, the simplifier will try and rewrite
-as many of the lemmas tagged `@[simp]` as it can.
-
-`simp` is a *finishing tactic*. After you run `simp`,
-the goal should be closed. If it is not, it is best
-practice to write `simp?` instead and then replace the
-output with the appropriate `simp only` call. Inappropriate
-use of `simp` can make for very slow code.
-"
-NewTactic simp -- TODO: Do we want it to be unlocked here?
-
 /-- $(a+(0+0)+(0+0+0)=a.$ -/
-Statement
-  (a : ℕ)  : (a + (0 + 0)) + (0 + 0 + 0) = a := by
+Statement (a : ℕ) : (a + (0 + 0)) + (0 + 0 + 0) = a := by
   Hint "I will walk you through this level so I can show you some
   techniques which will speed up your proving.
 
-  This is an annoying goal. One of rw [add_zero a]` amd `rw [add_zero 0]`
+  This is an annoying goal. One of `rw [add_zero a]` and `rw [add_zero 0]`
   will work, but not the other. Can you figure out which? Try the one
   that works."
   Branch
@@ -79,16 +56,38 @@ Statement
     start once more, and this time just try `simp`."
   simp
 
-DefinitionDoc Add as "Add" "`Add a b`, with notation `a + b`, is
+LemmaDoc MyNat.add_zero as "add_zero" in "Add"
+"`add_zero n` is a proof that `n + 0 = n`.
+
+This is one of the two axioms for addition."
+
+TacticDoc simp "The simplifier. `rw` on steroids.
+
+A bunch of lemmas like `add_zero : ∀ a, a + 0 = a`
+are tagged with the `@[simp]` tag. If the `simp` tactic
+is run by the user, the simplifier will try and rewrite
+as many of the lemmas tagged `@[simp]` as it can.
+
+`simp` is a *finishing tactic*. After you run `simp`,
+the goal should be closed. If it is not, it is best
+practice to write `simp?` instead and then replace the
+output with the appropriate `simp only` call. Inappropriate
+use of `simp` can make for very slow code.
+"
+
+DefinitionDoc Add as "+" "`Add a b`, with notation `a + b`, is
 the usual sum of natural numbers. Internally it is defined by
 induction on one of the variables, but that is an implementation issue;
 All you need to know is that `add_zero` and `zero_add` are both theorems."
 
+NewLemma MyNat.add_zero
+NewTactic simp «repeat» -- TODO: Do we want it to be unlocked here?
 NewDefinition Add
 
 Conclusion
-" The simplifier atempts to solve goals by using *repeated rewriting* of
-  *equalities* and *iff statements*, and then trying `rfl` at the end.
-
-  Let's now learn about Peano's second axiom for addition, `add_succ`.
+"
+The simplifier atempts to solve goals by using *repeated rewriting* of
+*equalities* and *iff statements*, and then trying `rfl` at the end.
+
+Let's now learn about Peano's second axiom for addition, `add_succ`.
 "

Game/Levels/Tutorial/L06add_succ.lean

@@ -24,9 +24,8 @@ LemmaDoc MyNat.add_succ as "add_succ" in "Add"
 
 NewLemma MyNat.add_succ
 
-/-- For all natural numbers $a$, we have $a + \\operatorname{succ}(0) = \\operatorname{succ}(a)$. -/
-Statement
-    : a + succ 0 = succ a := by
+/-- For all natural numbers $a$, we have $a + \operatorname{succ}(0) = \operatorname{succ}(a)$. -/
+Statement : a + succ 0 = succ a := by
   Hint "You find `{a} + succ …` in the goal, so you can use `rw` and `add_succ`
   to make progress."
   Hint (hidden := true) "Explicitely, you can type `rw [add_succ a 0]`

Game/MyNat/Addition.lean

@@ -17,7 +17,8 @@ If `a` is a natural number, then `add_zero a` is the proof that `a + 0 = a`.
 `add_zero` is a `simp` lemma, because if you see `a + 0`
 you usually want to simplify it to `a`.
 -/
-@[simp] theorem add_zero (a : MyNat) : a + 0 = a := by rfl
+@[simp]
+theorem add_zero (a : MyNat) : a + 0 = a := by rfl
 
 /--
 If `a` and `d` are natural numbers, then `add_succ a d` is the proof that