nng levels

2023-04-13 16:43:45 +02:00
parent 66fd600462
commit fec4005476
63 changed files with 1119 additions and 184 deletions
@@ -1,22 +1,55 @@
 import NNG.Metadata
-import NNG.MyNat.Addition
+import NNG.MyNat.Multiplication
+import Std.Tactic.RCases
+import Mathlib.Tactic.LeftRight

 Game "NNG"
 World "AdvMultiplication"
 Level 1
-Title ""
+Title "mul_pos"

 open MyNat

 Introduction
 "
+Welcome to Advanced Multiplication World! Before attempting this
+world you should have completed seven other worlds, including
+Multiplication World and Advanced Addition World. There are four
+levels in this world.
+
+Recall that if `b : mynat` is a hypothesis and you do `cases b with n`,
+your one goal will split into two goals, 
+namely the cases `b = 0` and `b = succ(n)`. So `cases` here is like
+a weaker version of induction (you don't get the inductive hypothesis).
+
+## Tricks
+
+1) if your goal is `⊢ X ≠ Y` then `intro h` will give you `h : X = Y` and
+a goal of `⊢ false`. This is because `X ≠ Y` *means* `(X = Y) → false`.
+Conversely if your goal is `false` and you have `h : X ≠ Y` as a hypothesis
+then `apply h` will turn the goal into `X = Y`.
+
+2) if `hab : succ (3 * x + 2 * y + 1) = 0` is a hypothesis and your goal is `⊢ false`,
+then `exact succ_ne_zero _ hab` will solve the goal, because Lean will figure
+out that `_` is supposed to be `3 * x + 2 * y + 1`.

 "

 Statement
-""
-    : true := by
-  trivial
+"The product of two non-zero natural numbers is non-zero."
+    (a b : ℕ) : a ≠ 0 → b ≠ 0 → a * b ≠ 0 := by
+  intro ha hb
+  intro hab
+  induction b with b
+  apply hb
+  rfl
+  rw [mul_succ] at hab
+  apply ha
+  induction a with a
+  rfl
+  rw [add_succ] at hab
+  exfalso
+  exact succ_ne_zero _ hab

 Conclusion
 "
@@ -1,22 +1,34 @@
 import NNG.Metadata
-import NNG.MyNat.Addition
+import NNG.MyNat.Multiplication
+import Std.Tactic.RCases
+import Mathlib.Tactic.LeftRight

 Game "NNG"
 World "AdvMultiplication"
 Level 2
-Title ""
+Title "eq_zero_or_eq_zero_of_mul_eq_zero"

 open MyNat

 Introduction
 "
-
+A variant on the previous level.
 "

-Statement
-""
-    : true := by
-  trivial
+Statement -- eq_zero_or_eq_zero_of_mul_eq_zero
+"If $ab = 0$, then at least one of $a$ or $b$ is equal to zero."
+    (a b : ℕ) (h : a * b = 0) :
+  a = 0 ∨ b = 0 := by
+  induction a with d
+  left
+  rfl
+  induction b with e he
+  right
+  rfl
+  exfalso
+  rw [mul_succ] at h
+  rw [add_succ] at h
+  exact succ_ne_zero _ h

 Conclusion
 "
@@ -1,22 +1,39 @@
 import NNG.Metadata
-import NNG.MyNat.Addition
+import NNG.MyNat.Multiplication
+import Std.Tactic.RCases
+import Mathlib.Tactic.LeftRight

 Game "NNG"
 World "AdvMultiplication"
 Level 3
-Title ""
+Title "mul_eq_zero_iff"

 open MyNat

 Introduction
 "
-
+Now you have `eq_zero_or_eq_zero_of_mul_eq_zero` this is pretty straightforward.
 "

+axiom eq_zero_or_eq_zero_of_mul_eq_zero (a b : ℕ) (h : a * b = 0) : a = 0 ∨ b = 0 
+axiom zero_mul (a : ℕ) : 0 * a = 0
+
 Statement
-""
-    : true := by
-  trivial
+"$ab = 0$, if and only if at least one of $a$ or $b$ is equal to zero.
+"
+    (a b : ℕ): a * b = 0 ↔ a = 0 ∨ b = 0 := by
+  constructor
+  intro h
+  exact eq_zero_or_eq_zero_of_mul_eq_zero a b h
+  intro hab
+  rcases hab with hab | hab
+  rw [hab]
+  rw [zero_mul]
+  rfl
+  rw [hab]
+  rw [mul_zero]
+  rfl
+

 Conclusion
 "
@@ -1,22 +1,63 @@
 import NNG.Metadata
-import NNG.MyNat.Addition
+import NNG.MyNat.Multiplication
+import Std.Tactic.RCases
+import Mathlib.Tactic.LeftRight

 Game "NNG"
 World "AdvMultiplication"
 Level 4
-Title ""
+Title "mul_left_cancel"

 open MyNat

 Introduction
 "
+This is the last of the bonus multiplication levels.
+`mul_left_cancel` will be useful in inequality world.

+People find this level hard. I have probably had more questions about this
+level than all the other levels put together, in fact. Many levels in this
+game can just be solved by \"running at it\" -- do induction on one of the
+variables, keep your head, and you're done. In fact, if you like a challenge,
+it might be instructive if you stop reading after the end of this paragraph and try solving this level now by induction,
+seeing the trouble you run into, and reading the rest of these comments afterwards. This level
+has a sting in the tail. If you are a competent mathematician, try
+and figure out what is going on. Write down a maths proof of the
+theorem in this level. Exactly what statement do you want to prove
+by induction? It is subtle.
+
+Ok so here are some spoilers. The problem with naively running at it, is that if you try induction on,
+say, $c$, then you are imagining a and b as fixed, and your inductive
+hypothesis $P(c)$ is $ab=ac \\implies b=c$. So for your inductive step
+you will be able to assume $ab=ad \\implies b=d$ and your goal will
+be to show $ab=a(d+1) \\implies b=d+1$. When you also assume $ab=a(d+1)$
+you will realise that your inductive hypothesis is *useless*, because
+$ab=ad$ is not true! The statement $P(c)$ (with $a$ and $b$ regarded
+as constants) is not provable by induction.
+
+What you *can* prove by induction is the following *stronger* statement.
+Imagine $a\not=0$ as fixed, and then prove \"for all $b$, if $ab=ac$ then $b=c$\"
+by induction on $c$. This gives us the extra flexibility we require.
+Note that we are quantifying over all $b$ in the inductive hypothesis -- it
+is essential that $b$ is not fixed. 
+
+You can do this in two ways in Lean -- before you start the induction
+you can write `revert b,`. The `revert` tactic is the opposite of the `intro`
+tactic; it replaces the `b` in the hypotheses with \"for all $b$\" in the goal.
+
+Alternatively, you can write `induction c with d hd
+generalizing b` as the first line of the proof. 
+
+If you do not modify your technique in this way, then this level seems
+to be impossible (judging by the comments I've had about it!)
 "

 Statement
-""
-    : true := by
-  trivial
+"If $a \\neq 0$, $b$ and $c$ are natural numbers such that
+$ ab = ac, $
+then $b = c$."
+    (a b c : ℕ) (ha : a ≠ 0) : a * b = a * c → b = c := by
+  sorry

 Conclusion
 "