add NNG
This commit is contained in:
Jon Eugster
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import NNG.Metadata
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import NNG.MyNat.Addition
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Game "NNG"
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World "Addition"
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Level 1
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Title "the induction tactic."
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open MyNat
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set_option tactic.hygienic false
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Introduction
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"
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OK so let's see induction in action. We're going to prove
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```
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zero_add (n : ℕ) : 0 + n = n
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```
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Wait… what is going on here? Didn't we already prove that adding zero to $n$ gave us $n$?
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No we didn't! We proved $n + 0 = n$, and that proof was called `add_zero`. We're now
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trying to establish `zero_add`, the proof that $0 + n = n$.
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But aren't these two theorems the same?
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No they're not! It is *true* that `x + y = y + x`, but we haven't *proved* it yet,
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and in fact we will need both `add_zero` and `zero_add` in order
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to prove this. In fact `x + y = y + x` is the boss level for addition world,
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and `induction` is the only other tactic you'll need to beat it.
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Now `add_zero` is one of Peano's axioms, so we don't need to prove it, we already have it.
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To prove `0 + n = n` we need to use induction on $n$. While we're here,
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note that `zero_add` is about zero add something, and `add_zero` is about something add zero.
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The names of the proofs tell you what the theorems are. Anyway, let's prove `0 + n = n`.
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"
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Statement zero_add
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"For all natural numbers $n$, we have $0 + n = n$."
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(n : ℕ) : 0 + n = n := by
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Hint "You can start a proof by induction over `n` by typing:
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`induction n with d hd`.
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If you use the `with` part, you can name your variable and induction hypothesis, otherwise
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they get default names."
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induction n with n hn
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· Hint "Now you have two goals. Once you proved the first, you will jump to the second one.
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This first goal is the base case $n = 0$.
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Recall that you can use all lemmas that are visible in your inventory."
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Hint (hidden := true) "try using `add_zero`."
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rw [add_zero]
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rfl
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· Hint "Now you jumped to the second goal. Here you have the induction hypothesis
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`{hn} : 0 + {n} = {n}` and you need to prove the statement for `succ {n}`."
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Hint (hidden := true) "look at `add_succ`."
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rw [add_succ]
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Hint (hidden := true) "At this point you see the term `0 + {n}`, so you can use the
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induction hypothesis with `rw [{hn}]`."
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rw [hn]
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rfl
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NewTactic induction
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Conclusion
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"
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## Now venture off on your own.
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Those three tactics:
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* `induction n with d hd`
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* `rw [h]`
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* `rfl`
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will get you quite a long way through this game. Using only these tactics
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you can beat Addition World level 4 (the boss level of Addition World),
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all of Multiplication World including the boss level `a * b = b * a`,
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and even all of Power World including the fiendish final boss. This route will
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give you a good grounding in these three basic tactics; after that, if you
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are still interested, there are other worlds to master, where you can learn
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more tactics.
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But we're getting ahead of ourselves, you still have to beat the rest of Addition World.
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We're going to stop explaining stuff carefully now. If you get stuck or want
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to know more about Lean (e.g. how to do much harder maths in Lean),
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ask in `#new members` at
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[the Lean chat](https://leanprover.zulipchat.com)
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(login required, real name preferred). Kevin or Mohammad or one of the other
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people there might be able to help.
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Good luck! Click on \"Next\" to solve some levels on your own.
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"
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import NNG.Metadata
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import NNG.MyNat.Addition
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Game "NNG"
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World "Addition"
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Level 2
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Title "add_assoc (associativity of addition)"
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open MyNat
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theorem MyNat.zero_add (n : ℕ) : 0 + n = n := by
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induction n with n hn
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· rw [add_zero]
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rfl
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· rw [add_succ]
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rw [hn]
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rfl
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Introduction
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"
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It's well-known that $(1 + 2) + 3 = 1 + (2 + 3)$; if we have three numbers
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to add up, it doesn't matter which of the additions we do first. This fact
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is called *associativity of addition* by mathematicians, and it is *not*
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obvious. For example, subtraction really is not associative: $(6 - 2) - 1$
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is really not equal to $6 - (2 - 1)$. We are going to have to prove
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that addition, as defined the way we've defined it, is associative.
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See if you can prove associativity of addition.
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"
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Statement add_assoc
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"On the set of natural numbers, addition is associative.
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In other words, for all natural numbers $a, b$ and $c$, we have
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$ (a + b) + c = a + (b + c). $"
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(a b c : ℕ) : (a + b) + c = a + (b + c) := by
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Hint "Because addition was defined by recursion on the right-most variable,
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use induction on the right-most variable (try other variables at your peril!).
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Note that when Lean writes `a + b + c`, it means `(a + b) + c`. If it wants to talk
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about `a + (b + c)` it will put the brackets in explictly."
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Branch
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induction a
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Hint "Good luck with that…"
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rw [zero_add, zero_add]
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rfl
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Branch
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induction b
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Hint "Good luck with that…"
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induction c with c hc
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Hint (hidden := true) "look at the lemma `add_zero`."
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rw [add_zero]
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Hint "`rw [add_zero]` only rewrites one term of the form `… + 0`, so you might to
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use it multiple times."
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rw [add_zero]
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rfl
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Hint (hidden := true) "`add_succ` might help here."
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rw [add_succ]
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rw [add_succ]
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rw [add_succ]
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Hint (hidden := true) "Now you can use the induction hypothesis."
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rw [hc]
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rfl
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Conclusion
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"
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"
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NewLemma MyNat.zero_add
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Addition.Level_2
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Game "NNG"
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World "Addition"
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Level 3
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Title "succ_add"
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open MyNat
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theorem MyNat.add_assoc (a b c : ℕ) : (a + b) + c = a + (b + c) := by
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induction c with c hc
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· rw [add_zero]
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rw [add_zero]
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rfl
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· rw [add_succ]
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rw [add_succ]
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rw [add_succ]
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rw [hc]
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rfl
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Introduction
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"
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Oh no! On the way to `add_comm`, a wild `succ_add` appears. `succ_add`
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is the proof that `succ(a) + b = succ(a + b)` for `a` and `b` in your
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natural number type. We need to prove this now, because we will need
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to use this result in our proof that `a + b = b + a` in the next level.
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NB: think about why computer scientists called this result `succ_add` .
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There is a logic to all the names.
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Note that if you want to be more precise about exactly where you want
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to rewrite something like `add_succ` (the proof you already have),
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you can do things like `rw [add_succ (succ a)]` or
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`rw [add_succ (succ a) d]`, telling Lean explicitly what to use for
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the input variables for the function `add_succ`. Indeed, `add_succ`
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is a function: it takes as input two variables `a` and `b` and outputs a proof
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that `a + succ(b) = succ(a + b)`. The tactic `rw [add_succ]` just says to Lean \"guess
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what the variables are\".
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"
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Statement succ_add
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"For all natural numbers $a, b$, we have
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$ \\operatorname{succ}(a) + b = \\operatorname{succ}(a + b)$."
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(a b : ℕ) : succ a + b = succ (a + b) := by
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Hint (hidden := true) "You might again want to start by induction
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on the right-most variable."
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Branch
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induction a
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Hint "Induction on `a` will not work."
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induction b with d hd
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· rw [add_zero]
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rfl
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· rw [add_succ]
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rw [hd]
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rw [add_succ]
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rfl
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NewLemma MyNat.add_assoc
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Conclusion
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"
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"
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Addition.Level_3
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Game "NNG"
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World "Addition"
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Level 4
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Title "`add_comm` (boss level)"
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open MyNat
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theorem MyNat.succ_add (a b : ℕ) : succ a + b = succ (a + b) := by
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induction b with d hd
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· rw [add_zero]
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rfl
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· rw [add_succ]
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rw [hd]
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rw [add_succ]
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rfl
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Introduction
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"
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[boss battle music]
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Look in your inventory to see the proofs you have available.
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These should be enough.
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"
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Statement add_comm
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"On the set of natural numbers, addition is commutative.
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In other words, for all natural numbers $a$ and $b$, we have
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$a + b = b + a$."
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(a b : ℕ) : a + b = b + a := by
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Branch
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induction a with d hd
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· rw [zero_add]
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rw [add_zero]
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rfl
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· rw [succ_add]
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rw [hd]
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rw [add_succ]
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rfl
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induction b with d hd
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· rw [zero_add]
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rw [add_zero]
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rfl
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· rw [add_succ]
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rw [hd]
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rw [succ_add]
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rfl
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NewLemma MyNat.succ_add
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Conclusion
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"
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If you got this far -- nice! You're nearly ready to make a choice:
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Multiplication World or Function World. But there are just a couple
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more useful lemmas in Addition World which you should prove first.
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Press on to level 5.
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"
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Addition.Level_4
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Game "NNG"
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World "Addition"
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Level 5
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Title "succ_eq_add_one"
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open MyNat
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theorem MyNat.add_comm (a b : ℕ) : a + b = b + a := by
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induction b with d hd
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· rw [zero_add]
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rw [add_zero]
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rfl
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· rw [add_succ]
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rw [hd]
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rw [succ_add]
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rfl
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theorem MyNat.one_eq_succ_zero : (1 : ℕ) = succ 0 := by
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rfl
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NewLemma MyNat.add_comm MyNat.one_eq_succ_zero
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Introduction
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"
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I've just added `one_eq_succ_zero` (a proof of $1 = \\operatorname{succ}(0)$)
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to your list of theorems; this is true
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by definition of $1$, but we didn't need it until now.
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Levels 5 and 6 are the two last levels in Addition World.
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Level 5 involves the number $1$. When you see a $1$ in your goal,
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you can write `rw [one_eq_succ_zero]` to get back
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to something which only mentions `0`. This is a good move because $0$ is easier for us to
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manipulate than $1$ right now, because we have
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some theorems about $0$ (`zero_add`, `add_zero`), but, other than `1 = succ 0`,
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no theorems at all which mention $1$. Let's prove one now.
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"
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Statement succ_eq_add_one
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"For any natural number $n$, we have
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$ \\operatorname{succ}(n) = n+1$ ."
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(n : ℕ) : succ n = n + 1 := by
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rw [one_eq_succ_zero]
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rw [add_succ]
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rw [add_zero]
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rfl
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Conclusion
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"
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Well done! On to the last level!
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"
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Addition.Level_5
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Game "NNG"
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World "Addition"
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Level 6
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Title "add_right_comm"
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open MyNat
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Introduction
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"
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Lean sometimes writes `a + b + c`. What does it mean? The convention is
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that if there are no brackets displayed in an addition formula, the brackets
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are around the left most `+` (Lean's addition is \"left associative\").
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So the goal in this level is `(a + b) + c = (a + c) + b`. This isn't
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quite `add_assoc` or `add_comm`, it's something you'll have to prove
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by putting these two theorems together.
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If you hadn't picked up on this already, `rw [add_assoc]` will
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change `(x + y) + z` to `x + (y + z)`, but to change it back
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you will need `rw [← add_assoc]`. Get the left arrow by typing `\\l`
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then the space bar (note that this is L for left, not a number 1).
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Similarly, if `h : a = b` then `rw [h]` will change `a`'s to `b`'s
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and `rw [← h]` will change `b`'s to `a`'s.
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Also, you can be (and will need to be, in this level) more precise
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about where to rewrite theorems. `rw add_comm,` will just find the
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first `? + ?` it sees and swap it around. You can target more specific
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additions like this: `rw add_comm a` will swap around
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additions of the form `a + ?`, and `rw add_comm a b,` will only
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swap additions of the form `a + b`.
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"
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Statement add_right_comm
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"For all natural numbers $a, b$ and $c$, we have
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$a + b + c = a + c + b$."
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(a b c : ℕ) : a + b + c = a + c + b := by
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rw [add_assoc]
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rw [add_comm b c]
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rw [←add_assoc]
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rfl
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Conclusion
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"
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"
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Reference in New Issue
Block a user