nng levels
This commit is contained in:
Jon Eugster
@@ -1,24 +1,42 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.MyNat.AdvAddition
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Game "NNG"
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World "AdvAddition"
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Level 1
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Title ""
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Title "`succ_inj`. A function."
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open MyNat
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Introduction
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"
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Peano's original collection of axioms for the natural numbers contained two further
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assumptions, which have not yet been mentioned in the game:
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```
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succ_inj {a b : mynat} :
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succ(a) = succ(b) → a = b
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zero_ne_succ (a : mynat) :
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zero ≠ succ(a)
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```
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The reason they have not been used yet is that they are both implications,
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that is,
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of the form $P\\implies Q$. This is clear for `succ_inj a b`, which
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says that for all $a$ and $b$ we have $succ(a)=succ(b)\\implies a=b$.
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For `zero_ne_succ` the trick is that $X\ne Y$ is *defined to mean*
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$X = Y\\implies{\\tt false}$. If you have played through Proposition world,
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you now have the required Lean skills (i.e., you know the required
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tactics) to work with these implications.
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Let's finally learn how to use `succ_inj`. You should know a couple
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of ways to prove the below -- one directly using an `exact`,
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and one which uses an `apply` first. But either way you'll need to use `succ_inj`.
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"
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theorem MyNat.succ_inj {a b : ℕ} : succ a = succ b → a = b := by simp only [succ.injEq, imp_self]
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theorem MyNat.zero_ne_succ (a : ℕ) : zero ≠ succ a := by simp only [ne_eq, not_false_iff]
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Statement -- MyNat.succ_inj'
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""
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"For all naturals $a$ and $b$, if we assume $succ(a)=succ(b)$, then we can
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deduce $a=b$."
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{a b : ℕ} (hs : succ a = succ b) : a = b := by
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exact succ_inj hs
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@@ -26,5 +44,11 @@ NewLemma MyNat.succ_inj MyNat.zero_ne_succ
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Conclusion
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"
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## Important thing.
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You can rewrite proofs of *equalities*. If `h : A = B` then `rw h` changes `A`s to `B`s.
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But you *cannot rewrite proofs of implications*. `rw succ_inj` will *never work*
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because `succ_inj` isn't of the form $A = B$, it's of the form $A\\implies B$. This is one
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of the most common mistakes I see from beginners. $\\implies$ and $=$ are *two different things*
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and you need to be clear about which one you are using.
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"
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@@ -1,22 +1,74 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.MyNat.AdvAddition
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import NNG.MyNat.Multiplication
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Game "NNG"
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World "AdvAddition"
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Level 10
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Title ""
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Title "add_left_eq_zero"
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open MyNat
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Introduction
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"
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In Lean, `a ≠ b` is *defined to mean* `(a = b) → false`.
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This means that if you see `a ≠ b` you can *literally treat
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it as saying* `(a = b) → false`. Computer scientists would
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say that these two terms are *definitionally equal*.
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The following lemma, $a+b=0\\implies b=0$, will be useful in inequality world.
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Let me go through the proof, because it introduces several new
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concepts:
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* `cases b`, where `b : mynat`
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* `exfalso`
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* `apply succ_ne_zero`
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We're going to prove $a+b=0\\implies b=0$. Here is the
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strategy. Each natural number is either `0` or `succ(d)` for
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some other natural number `d`. So we can start the proof
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with
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`cases b with d,`
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and then we have two goals, the case `b = 0` (which you can solve easily)
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and the case `b = succ(d)`, which looks like this:
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```
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a d : mynat,
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H : a + succ d = 0
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⊢ succ d = 0
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```
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Our goal is impossible to prove. However our hypothesis `H`
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is also impossible, meaning that we still have a chance!
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First let's see why `H` is impossible. We can
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`rw add_succ at H,`
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to turn `H` into `H : succ (a + d) = 0`. Because
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`succ_ne_zero (a + d)` is a proof that `succ (a + d) ≠ 0`,
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it is also a proof of the implication `succ (a + d) = 0 → false`.
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Hence `succ_ne_zero (a + d) H` is a proof of `false`!
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Unfortunately our goal is not `false`, it's a generic
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false statement.
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Recall however that the `exfalso` command turns any goal into `false`
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(it's logically OK because `false` implies every proposition, true or false).
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You can probably take it from here.
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"
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Statement
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""
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: true := by
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trivial
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Statement -- add_left_eq_zero
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"If $a$ and $b$ are natural numbers such that
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$$ a + b = 0, $$
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then $b = 0$."
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{{a b : ℕ}} (H : a + b = 0) : b = 0 := by
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induction b with d
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rfl
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rw [add_succ] at H
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exfalso
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apply succ_ne_zero (a + d)
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exact H
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Conclusion
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"
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@@ -1,22 +1,29 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.MyNat.AdvAddition
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Game "NNG"
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World "AdvAddition"
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Level 11
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Title ""
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Title "add_right_eq_zero"
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open MyNat
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theorem MyNat.add_left_eq_zero {{a b : ℕ}} (H : a + b = 0) : b = 0 := by sorry
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Introduction
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"
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We just proved `add_left_eq_zero (a b : mynat) : a + b = 0 → b = 0`.
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Hopefully `add_right_eq_zero` shouldn't be too hard now.
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"
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Statement
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""
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: true := by
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trivial
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"If $a$ and $b$ are natural numbers such that
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$$ a + b = 0, $$
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then $a = 0$."
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{a b : ℕ} : a + b = 0 → a = 0 := by
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intro H
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rw [add_comm] at H
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exact add_left_eq_zero H
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Conclusion
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"
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@@ -1,22 +1,30 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.MyNat.AdvAddition
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Game "NNG"
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World "AdvAddition"
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Level 12
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Title ""
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Title "add_one_eq_succ"
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open MyNat
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Introduction
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"
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We have
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* `succ_eq_add_one (n : mynat) : succ n = n + 1`
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but sometimes the other way is also convenient.
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"
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theorem succ_eq_add_one (d : ℕ) : succ d = d + 1 := by sorry
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Statement
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""
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: true := by
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trivial
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"For any natural number $d$, we have
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$$ d+1 = \\operatorname{succ}(d). $$"
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(d : ℕ) : d + 1 = succ d := by
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rw [succ_eq_add_one]
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rfl
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Conclusion
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"
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@@ -1,22 +1,32 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.MyNat.AdvAddition
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Game "NNG"
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World "AdvAddition"
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Level 13
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Title ""
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Title "ne_succ_self"
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open MyNat
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Introduction
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"
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The last level in Advanced Addition World is the statement
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that $n\\not=\\operatorname{succ}(n)$. When you've done this
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you've completed Advanced Addition World and can move on
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to Advanced Multiplication World (after first doing
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Multiplication World, if you didn't do it already).
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"
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Statement
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""
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: true := by
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trivial
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Statement --ne_succ_self
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"For any natural number $n$, we have
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$$ n \\neq \\operatorname{succ}(n). $$"
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(n : ℕ) : n ≠ succ n := by
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induction n with d hd
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apply zero_ne_succ
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intro hs
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apply hd
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apply succ_inj
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assumption
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Conclusion
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"
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@@ -1,24 +1,53 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.MyNat.AdvAddition
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Game "NNG"
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World "AdvAddition"
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Level 2
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Title ""
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Title "succ_succ_inj"
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open MyNat
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Introduction
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"
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In the below theorem, we need to apply `succ_inj` twice. Once to prove
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$succ(succ(a))=succ(succ(b))\\implies succ(a)=succ(b)$, and then again
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to prove $succ(a)=succ(b)\\implies a=b$. However `succ(a)=succ(b)` is
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nowhere to be found, it's neither an assumption or a goal when we start
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this level. You can make it with `have` or you can use `apply`.
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"
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Statement
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""
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: true := by
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trivial
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"For all naturals $a$ and $b$, if we assume $succ(succ(a))=succ(succ(b))$, then we can
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deduce $a=b$. "
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{a b : ℕ} (h : succ (succ a) = succ ( succ b )) : a = b := by
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apply succ_inj
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apply succ_inj
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assumption
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Conclusion
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"
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## Sample solutions to this level.
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Make sure you understand them all. And remember that `rw` should not be used
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with `succ_inj` -- `rw` works only with equalities or `↔` statements,
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not implications or functions.
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example {a b : mynat} (h : succ(succ(a)) = succ(succ(b))) : a = b :=
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begin
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apply succ_inj,
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apply succ_inj,
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exact h
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end
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example {a b : mynat} (h : succ(succ(a)) = succ(succ(b))) : a = b :=
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begin
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apply succ_inj,
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exact succ_inj(h),
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end
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example {a b : mynat} (h : succ(succ(a)) = succ(succ(b))) : a = b :=
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begin
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exact succ_inj(succ_inj(h)),
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end
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"
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@@ -1,22 +1,28 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.MyNat.AdvAddition
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Game "NNG"
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World "AdvAddition"
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Level 3
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Title ""
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Title "succ_eq_succ_of_eq"
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open MyNat
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Introduction
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"
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We are going to prove something completely obvious: if $a=b$ then
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$succ(a)=succ(b)$. This is *not* `succ_inj`!
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This is trivial -- we can just rewrite our proof of `a=b`.
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But how do we get to that proof? Use the `intro` tactic.
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"
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Statement
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""
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: true := by
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trivial
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"For all naturals $a$ and $b$, $a=b\\implies succ(a)=succ(b)$.
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"
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{a b : ℕ} : a = b → succ a = succ b := by
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intro h
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rw [h]
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rfl
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Conclusion
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"
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@@ -1,22 +1,40 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.MyNat.AdvAddition
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Game "NNG"
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World "AdvAddition"
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Level 4
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Title ""
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Title "eq_iff_succ_eq_succ"
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open MyNat
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Introduction
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"
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Here is an `iff` goal. You can split it into two goals (the implications in both
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directions) using the `split` tactic, which is how you're going to have to start.
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`split,`
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Now you have two goals. The first is exactly `succ_inj` so you can close
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it with
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`exact succ_inj,`
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and the second one you could solve by looking up the name of the theorem
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you proved in the last level and doing `exact <that name>`, or alternatively
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you could get some more `intro` practice and seeing if you can prove it
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using `intro`, `rw` and `refl`.
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"
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Statement
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""
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: true := by
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trivial
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"Two natural numbers are equal if and only if their successors are equal.
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"
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(a b : ℕ) : succ a = succ b ↔ a = b := by
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constructor
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exact succ_inj
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intro H
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rw [H]
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rfl
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Conclusion
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"
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@@ -1,22 +1,36 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.MyNat.AdvAddition
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Game "NNG"
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World "AdvAddition"
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Level 5
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Title ""
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Title "add_right_cancel"
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open MyNat
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Introduction
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"
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The theorem `add_right_cancel` is the theorem that you can cancel on the right
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when you're doing addition -- if `a + t = b + t` then `a = b`. After `intro h`
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I'd recommend induction on `t`. Don't forget that `rw add_zero at h` can be used
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to do rewriting of hypotheses rather than the goal.
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"
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Statement
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""
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: true := by
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trivial
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"On the set of natural numbers, addition has the right cancellation property.
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In other words, if there are natural numbers $a, b$ and $c$ such that
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$$ a + t = b + t, $$
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then we have $a = b$."
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(a t b : ℕ) : a + t = b + t → a = b := by
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intro h
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induction t with d hd
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rw [add_zero] at h
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rw [add_zero] at h
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exact h
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apply hd
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rw [add_succ] at h
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rw [add_succ] at h
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exact succ_inj h
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Conclusion
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"
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@@ -1,22 +1,32 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.MyNat.AdvAddition
|
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Game "NNG"
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World "AdvAddition"
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Level 6
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Title ""
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Title "add_left_cancel"
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open MyNat
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Introduction
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"
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The theorem `add_left_cancel` is the theorem that you can cancel on the left
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when you're doing addition -- if `t + a = t + b` then `a = b`.
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There is a three-line proof which ends in `exact add_right_cancel a t b` (or even
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`exact add_right_cancel _ _ _`); this
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strategy involves changing the goal to the statement of `add_right_cancel` somehow.
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"
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Statement
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""
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||||
: true := by
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trivial
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"On the set of natural numbers, addition has the left cancellation property.
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In other words, if there are natural numbers $a, b$ and $t$ such that
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$$ t + a = t + b, $$
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then we have $a = b$."
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(t a b : ℕ) : t + a = t + b → a = b := by
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rw [add_comm]
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rw [add_comm t]
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exact add_right_cancel a t b
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Conclusion
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"
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@@ -1,23 +1,34 @@
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import NNG.Metadata
|
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import NNG.MyNat.Addition
|
||||
import NNG.MyNat.AdvAddition
|
||||
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Game "NNG"
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World "AdvAddition"
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Level 7
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Title ""
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Title "add_right_cancel_iff"
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open MyNat
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Introduction
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"
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It's sometimes convenient to have the \"if and only if\" version
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of theorems like `add_right_cancel`. Remember that you can use `split`
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to split an `↔` goal into the `→` goal and the `←` goal.
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## Pro tip:
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`exact add_right_cancel _ _ _` means \"let Lean figure out the missing inputs\"
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"
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||||
|
||||
Statement
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||||
""
|
||||
: true := by
|
||||
trivial
|
||||
|
||||
Statement --add_right_cancel_iff
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||||
"For all naturals $a$, $b$ and $t$,
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$$ a + t = b + t\\iff a=b. $$
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||||
"
|
||||
(t a b : ℕ) : a + t = b + t ↔ a = b := by
|
||||
constructor
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||||
exact add_right_cancel _ _ _ -- done that way already,
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||||
intro H -- H : a = b,
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||||
rw [H]
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||||
rfl
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||||
Conclusion
|
||||
"
|
||||
|
||||
|
||||
@@ -1,22 +1,29 @@
|
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import NNG.Metadata
|
||||
import NNG.MyNat.Addition
|
||||
import NNG.MyNat.AdvAddition
|
||||
|
||||
Game "NNG"
|
||||
World "AdvAddition"
|
||||
Level 8
|
||||
Title ""
|
||||
Title "eq_zero_of_add_right_eq_self"
|
||||
|
||||
open MyNat
|
||||
|
||||
Introduction
|
||||
"
|
||||
|
||||
The lemma you're about to prove will be useful when we want to prove that $\\leq$ is antisymmetric.
|
||||
There are some wrong paths that you can take with this one.
|
||||
"
|
||||
|
||||
Statement
|
||||
""
|
||||
: true := by
|
||||
trivial
|
||||
Statement --eq_zero_of_add_right_eq_self
|
||||
"If $a$ and $b$ are natural numbers such that
|
||||
$$ a + b = a, $$
|
||||
then $b = 0$."
|
||||
{a b : ℕ} : a + b = a → b = 0 := by
|
||||
intro h
|
||||
apply add_left_cancel a
|
||||
rw [h]
|
||||
rw [add_zero]
|
||||
rfl
|
||||
|
||||
Conclusion
|
||||
"
|
||||
|
||||
@@ -1,22 +1,32 @@
|
||||
import NNG.Metadata
|
||||
import NNG.MyNat.Addition
|
||||
import NNG.MyNat.AdvAddition
|
||||
|
||||
Game "NNG"
|
||||
World "AdvAddition"
|
||||
Level 9
|
||||
Title ""
|
||||
Title "succ_ne_zero"
|
||||
|
||||
open MyNat
|
||||
|
||||
Introduction
|
||||
"
|
||||
Levels 9 to 13 introduce the last axiom of Peano, namely
|
||||
that $0\\not=\\operatorname{succ}(a)$. The proof of this is called `zero_ne_succ a`.
|
||||
|
||||
`zero_ne_succ (a : mynat) : 0 ≠ succ(a)`
|
||||
|
||||
The `symmetry` tactic will turn any goal of the form `R x y` into `R y x`,
|
||||
if `R` is a symmetric binary relation (for example `=` or `≠`).
|
||||
In particular, you can prove `succ_ne_zero` below by first using
|
||||
`symmetry` and then `exact zero_ne_succ a`.
|
||||
"
|
||||
|
||||
Statement
|
||||
""
|
||||
: true := by
|
||||
trivial
|
||||
Statement -- succ_ne_zero
|
||||
"Zero is not the successor of any natural number."
|
||||
(a : ℕ) : succ a ≠ 0 := by
|
||||
apply Ne.symm
|
||||
exact zero_ne_succ a
|
||||
|
||||
|
||||
Conclusion
|
||||
"
|
||||
|
||||
Reference in New Issue
Block a user