split off test game
still need to adapt the call to the lean binary to provide two arguments
This commit is contained in:
Alexander Bentkamp
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import TestGame.Metadata
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Level 1
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Title "The reflexivity spell"
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Introduction
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"
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Let's learn a first spell: the `rfl` spell. `rfl` stands for \"reflexivity\", which is a fancy
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way of saying that it will prove any goal of the form `A = A`. It doesn't matter how
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complicated `A` is, all that matters is that the left hand side is *exactly equal* to the
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right hand side (a computer scientist would say \"definitionally equal\"). I really mean
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\"press the same buttons on your computer in the same order\" equal.
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For example, `x * y + z = x * y + z` can be proved by `rfl`, but `x + y = y + x` cannot.
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This is a very low level spell, but you need to start somewhere.
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After closing this message, type rfl in the invocation zone and hit Enter or click
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the \"Cast spell\" button.
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"
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Statement (x y z : ℕ) : x * y + z = x * y + z := by
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rfl
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Conclusion "Congratulations for completing your first level! You can now click on the *Go to next level* button."
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Tactics rfl
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import TestGame.Metadata
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Level 2
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Title "The rewriting spell"
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Introduction
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"
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The rewrite spell is the way to \"substitute in\" the value
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of an expression. In general, if you have a hypothesis of the form `A = B`, and your
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goal mentions the left hand side `A` somewhere, then
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the `rewrite` tactic will replace the `A` in your goal with a `B`.
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The documentation for `rewrite` just appeared in your spell book.
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Play around with the menus and see what is there currently.
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More information will appear as you progress.
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Take a look in the top right box at what we have.
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The variables $x$ and $y$ are natural numbers, and we have
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an assumption `h` that $y = x + 7$. Our goal
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is to prove that $2y=2(x+7)$. This goal is obvious -- we just
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substitute in $y = x+7$ and we're done. In Lean, we do
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this substitution using the `rewrite` spell. This spell takes a list of equalities
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or equivalences so you can cast `rewrite [h]`.
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"
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Statement (x y : ℕ) (h : y = x + 7) : 2 * y = 2 * (x + 7) := by
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rewrite [h]
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rfl
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Message (x : ℕ) (y : ℕ) (h : y = x + 7) : 2*(x + 7) = 2*(x + 7) =>
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"Great! Now the goal should be easy to reach using the `rfl` spell."
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Conclusion "Congratulations for completing your second level!"
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Tactics rfl rewrite
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import TestGame.Metadata
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Level 3
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Title "Peano's axioms"
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Introduction
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"
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The team that salvaged the type `ℕ` of natural numbers actually got us three things:
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* a term `0 : ℕ`, interpreted as the number zero.
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* a function `succ : ℕ → ℕ`, with `succ n` interpreted as \"the number after $n$\".
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* The principle of mathematical induction.
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These are essentially the axioms isolated by Peano which uniquely characterise
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the natural numbers (we also need recursion, but we can ignore it for now).
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The first axiom says that $0$ is a natural number. The second says that there
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is a `succ` function which eats a number and spits out the number after it,
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so $\\operatorname{succ}(0)=1$, $\\operatorname{succ}(1)=2$ and so on.
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Peano's last axiom is the principle of mathematical induction. This is a deeper
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fact. It says that if we have infinitely many true/false statements $P(0)$, $P(1)$,
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$P(2)$ and so on, and if $P(0)$ is true, and if for every natural number $d$
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we know that $P(d)$ implies $P(\\operatorname{succ}(d))$, then $P(n)$ must be true for every
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natural number $n$. It's like saying that if you have a long line of dominoes, and if
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you knock the first one down, and if you know that if a domino falls down then the one
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after it will fall down too, then you can deduce that all the dominos will fall down.
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One can also think of it as saying that every natural number
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can be built by starting at `0` and then applying `succ` a finite number of times.
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Peano's insights were firstly that these axioms completely characterise
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the natural numbers, and secondly that these axioms alone can be used to build
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a whole bunch of other structure on the natural numbers, for example
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addition, multiplication and so on.
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This game is all about seeing how far these axioms of Peano can take us.
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Let's practice our use of the `rewrite` tactic in the following example.
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Our hypothesis `h` is a proof that `succ(a) = b` and we want to prove that
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`succ(succ(a))=succ(b)`. In words, we're going to prove that if
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`b` is the number after `a` then `succ(b)` is the number after `succ(a)`.
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Note that the system drops brackets when they're not
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necessary, so `succ b` just means `succ(b)`.
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Now here's a tricky question. Knowing that our goal is `succ (succ a) = succ b`,
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and our assumption is `h : succ a = b`, then what will the goal change
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to when we type
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`rewrite [h]`
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and hit enter? Remember that `rewrite [h]` will
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look for the *left* hand side of `h` in the goal, and will replace it with
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the *right* hand side. Try and figure out how the goal will change, and
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then try it.
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"
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Statement (a b : ℕ) (h : succ a = b) : succ (succ a) = succ b := by
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rewrite [h]
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rfl
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Message (a : ℕ) (b : ℕ) (h : succ a = b) : succ b = succ b =>
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"
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Look: Lean changed `succ a` into `b`, so the goal became `succ b = succ b`.
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That goal is of the form `X = X`, so you know what to do.
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"
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Conclusion "Congratulations for completing the third level!
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You may be wondering whether we could have just substituted in the definition of `b`
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and proved the goal that way. To do that, we would want to replace the right hand
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side of `h` with the left hand side. You do this in Lean by writing `rewrite [<- h]`. You get the
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left-arrow by typing `\\l` and then a space; note that this is a small letter L,
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not a number 1. You can just edit your proof and try it.
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You may also be wondering why we keep writing `succ(b)` instead of `b+1`. This
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is because we haven't defined addition yet! On the next level, the final level
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of the tutorial, we will introduce addition, and then
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we'll be ready to enter Addition World.
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"
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Tactics rfl rewrite
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import TestGame.Metadata
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Level 4
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Title "Addition"
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Introduction
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"
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Peano defined addition `a + b` by induction on `b`, or,
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more precisely, by *recursion* on `b`. He first explained how to add 0 to a number:
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this is the base case.
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* `add_zero (a : ℕ) : a + 0 = a`
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We will call this theorem `add_zero`. It has just appeared in your inventory!
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Mathematicians sometimes call it \"Lemma 2.1\" or \"Hypothesis P6\" or something. But
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computer scientists call it `add_zero` because it tells you
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what the answer to \"$x$ add zero\" is. It's a *much* better name than \"Lemma 2.1\".
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Even better, we can use the rewrite tactic with `add_zero`.
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If you ever see `x + 0` in your goal, `rewrite [add_zero]` will simplify it to `x`.
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This is because `add_zero` is a proof that `x + 0 = x` (more precisely,
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`add_zero x` is a proof that `x + 0 = x` but Lean can figure out the `x` from the context).
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Now here's the inductive step. If you know how to add `d` to `a`, then
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Peano tells you how to add `succ(d)` to `a`. It looks like this:
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* `add_succ (a d : ℕ) : a + succ(d) = succ (a + d)`
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What's going on here is that we assume `a + d` is already
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defined, and we define `a + succ(d)` to be the number after it.
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This is also in your inventory now -- `add_succ` tells you
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how to add a successor to something. If you ever see `... + succ ...`
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in your goal, you should be able to use `rewrite [add_succ]` to make
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progress. Here is a simple example where we shall see both. Let's prove
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that $x$ add the number after $0$ is the number after $x$.
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Observe that the goal mentions `... + succ ...`. So type
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`rewrite [add_succ]`
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and hit enter; see the goal change.
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"
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Statement (a : ℕ ) : a + succ 0 = succ a := by
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rewrite [add_succ]
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rewrite [add_zero]
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rfl
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Message (a : ℕ) : succ (a + 0) = succ a => "
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Do you see that the goal now mentions ` ... + 0 ...`? So type
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`rewrite [add_zero]`
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and try to finish the level alone from there.
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"
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Conclusion "Congratulations for completing your fourth level! This is the end of the tutorial part
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of the game. Serious things start in the next level."
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Tactics rfl rewrite
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Lemmas add_succ add_zero
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import TestGame.Metadata
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import TestGame.Tactics
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Level 5
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Title "The induction_on spell"
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Introduction
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"
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Welcome to Addition World. If you've done all four levels in tutorial world
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and know about `rewrite` and `rfl`, then you're in the right place. Here's
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a reminder of the things you're now equipped with which we'll need in this world.
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## Data:
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* a type called `ℕ`
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* a term `0 : ℕ`, interpreted as the number zero.
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* a function `succ : ℕ → ℕ`, with `succ n` interpreted as \"the number after `n`\".
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* Usual numerical notation 0,1,2 etc (although 2 onwards will be of no use to us until much later ;-) ).
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* Addition (with notation `a + b`).
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## Theorems:
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* `add_zero (a : ℕ) : a + 0 = a`. Use with `rewrite [add_zero]`.
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* `add_succ (a b : ℕ) : a + succ(b) = succ(a + b)`. Use with `rewrite [add_succ]`.
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* The principle of mathematical induction. Use with `induction_on` (see below)
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## Spells:
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* `rfl` : proves goals of the form `X = X`
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* `rewrite [h]` : if h is a proof of `A = B`, changes all A's in the goal to B's.
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* `induction_on n with d hd` : we're going to learn this right now.
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# Important thing:
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This is a *really* good time to check you understand about the spell book and the inventory on
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the left. Eveything you need is collected in those lists. They
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will prove invaluable as the number of theorems we prove gets bigger. On the other hand,
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we only need to learn one more spell to really start going places, so let's learn about
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that spell right now.
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OK so let's see induction in action. We're going to prove
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`zero_add (n : ℕ) : 0 + n = n`.
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That is: for all natural numbers $n$, $0+n=n$. Wait $-$ what is going on here?
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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 $0 + n = n$. But aren't these two theorems
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the same? No they're not! It is *true* that `x + y = y + x`, but we haven't
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*proved* it yet, 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_on` is the only other spell 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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(indeed, if you've opened the Addition World theorem statements on the left, you can even see 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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Start by casting `induction_on n`.
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"
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Statement (n : ℕ) : 0 + n = n := by
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induction_on n
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rewrite [add_zero]
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rfl
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rewrite [add_succ]
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rewrite [ind_hyp]
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rfl
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Message : (0 : ℕ) + 0 = 0 => "
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We now have *two goals!* The
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induction spell has generated for us a base case with `n = 0` (the goal at the top)
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and an inductive step (the goal underneath). The golden rule: **spells operate on the current goal** --
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the goal at the top. So let's just worry about that top goal now, the base case `0 + 0 = 0`.
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Remember that `add_zero` (the proof we have already) is the proof of `x + 0 = x`
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(for any $x$) so we can try
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`rewrite [add_zero]`
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What do you think the goal will change to? Remember to just keep
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focussing on the top goal, ignore the other one for now, it's not changing
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and we're not working on it.
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"
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Message (n : ℕ) (ind_hyp: 0 + n = n) : 0 + succ n = succ n =>
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"
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Great! You solved the base case. We are now be back down
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to one goal -- the inductive step.
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We have a fixed natural number `n`, and the inductive hypothesis `ind_hyp : 0 + n = n`
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saying that we have a proof of `0 + n = n`.
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Our goal is to prove `0 + succ n = succ n`. In words, we're showing that
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if the lemma is true for `n`, then it's also true for the number after `n`.
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That's the inductive step. Once we've proved this inductive step, we will have proved
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`zero_add` by the principle of mathematical induction.
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To prove our goal, we need to use `add_succ`. We know that `add_succ 0 n`
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is the result that `0 + succ n = succ (0 + n)`, so the first thing
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we need to do is to replace the left hand side `0 + succ n` of our
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goal with the right hand side. We do this with the `rewrite` spell. You can write
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`rewrite [add_succ]`
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(or even `rewrite [add_succ 0 n]` if you want to give Lean all the inputs instead of making it
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figure them out itself).
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"
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Message (n : ℕ) (ind_hyp: 0 + n = n) : succ (0 + n) = succ n =>
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"Well-done! We're almost there. It's time to use our induction hypothesis.
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Cast
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`rewrite [ind_hyp]`
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and finish by yourself.
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"
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Conclusion "Congratulations for completing your first inductive proof!"
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Tactics rfl rewrite induction_on
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Lemmas add_succ add_zero
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Reference in New Issue
Block a user