move all WIP files to level-rewrite branch

Game/Levels/OldAdvProposition/Level_1.lean

@@ -0,0 +1,41 @@
+import Game.Metadata
+import Game.MyNat.Addition
+
+
+World "AdvProposition"
+Level 1
+Title "the `split` tactic"
+
+open MyNat
+
+Introduction
+"
+The logical symbol `∧` means \"and\". If $P$ and $Q$ are propositions, then
+$P\\land Q$ is the proposition \"$P$ and $Q$\".
+"
+namespace MySet
+
+/-- If $P$ and $Q$ are true, then $P\\land Q$ is true. -/
+Statement
+    (P Q : Prop) (p : P) (q : Q) : P ∧ Q := by
+  Hint "If your *goal* is `P ∧ Q` then
+  you can make progress with the `constructor` tactic, which turns one goal `P ∧ Q`
+  into two goals, namely `P` and `Q`."
+  constructor
+  Hint "Now you have two goals. The first one is `P`, which you can proof now. The
+  second one is `Q` and you see it in the queue \"Other Goals\". You will have to prove it afterwards."
+  Hint (hidden := true) "This first goal can be proved with `exact p`."
+  exact p
+  -- Hint "Observe that now the first goal has been proved, so it disappears and you continue
+  -- proving the second goal."
+  -- Hint (hidden := true) "Like the first goal, this is a case for `exact`."
+  -- -- TODO: It looks like these hints get shown above as well, but weirdly the hints from
+  -- -- above to not get shown here.
+  exact q
+
+NewTactic constructor
+NewDefinition And
+
+Conclusion
+"
+"

Game/Levels/OldAdvProposition/Level_10.lean

@@ -0,0 +1,64 @@
+import Game.Metadata
+import Game.MyNat.Addition
+
+
+World "AdvProposition"
+Level 10
+Title "The law of the excluded middle."
+
+open MyNat
+
+Introduction
+"
+We proved earlier that `(P → Q) → (¬ Q → ¬ P)`. The converse,
+that `(¬ Q → ¬ P) → (P → Q)` is certainly true, but trying to prove
+it using what we've learnt so far is impossible (because it is not provable in
+constructive logic).
+
+"
+
+/-- If $P$ and $Q$ are true/false statements, then
+$$(\\lnot Q\\implies \\lnot P)\\implies(P\\implies Q).$$
+ -/
+Statement
+    (P Q : Prop) : (¬ Q → ¬ P) → (P → Q) := by
+  Hint "For example, you could start as always with
+
+  ```
+  intro h p
+  ```
+  "
+  intro h p
+  Hint "From here there is no way to continue with the tactics you've learned so far.
+
+  Instead you can call `by_cases q : Q`. This creates **two goals**, once under the assumption
+  that `Q` is true, once assuming `Q` is false."
+  by_cases q : Q
+  Hint "This first case is trivial."
+  exact q
+  Hint "The second case needs a bit more work, but you can get there with the tactics you've already
+  learned beforehand!"
+  have j := h q
+  exfalso
+  apply j
+  exact p
+
+NewTactic by_cases
+
+Conclusion
+"
+This approach assumed that `P ∨ ¬ P` is true, which is called \"law of excluded middle\".
+It cannot be proven using just tactics like `intro` or `apply`.
+`by_cases p : P` just does `rcases` on this result `P ∨ ¬ P`.
+
+
+OK that's enough logic -- now perhaps it's time to go on to Advanced Addition World!
+Get to it via the main menu.
+"
+
+-- TODO: I cannot really import `tauto` because of the notation `ℕ` that's used
+-- for `MyNat`.
+-- ## Pro tip
+
+-- In fact the tactic `tauto` just kills this goal (and many other logic goals) immediately. You'll be
+-- able to use it from now on.

Game/Levels/OldAdvProposition/Level_2.lean

@@ -0,0 +1,48 @@
+import Game.Metadata
+import Game.MyNat.Addition
+
+
+World "AdvProposition"
+Level 2
+Title "the `rcases` tactic"
+
+open MyNat
+
+Introduction
+"
+If `P ∧ Q` is in the goal, then we can make progress with `constructor`.
+But what if `P ∧ Q` is a hypothesis? In this case, the `rcases` tactic will enable
+us to extract proofs of `P` and `Q` from this hypothesis.
+"
+
+/-- If $P$ and $Q$ are true/false statements, then $P\\land Q\\implies Q\\land P$. -/
+Statement -- and_symm
+    (P Q : Prop) : P ∧ Q → Q ∧ P :=  by
+  Hint "The lemma below asks us to prove `P ∧ Q → Q ∧ P`, that is,
+  symmetry of the \"and\" relation. The obvious first move is
+
+  ```
+  intro h
+  ```
+
+  because the goal is an implication and this tactic is guaranteed
+  to make progress."
+  intro h
+  Hint "Now `{h} : P ∧ Q` is a hypothesis, and
+
+  ```
+  rcases {h} with ⟨p, q⟩
+  ```
+
+  will change `{h}`, the proof of `P ∧ Q`, into two proofs `p : P`
+  and `q : Q`.
+
+  You can write `⟨p, q⟩` with `\\<>` or `\\<` and `\\>`. Note that `rcases h` by itself will just
+  automatically name the new assumptions."
+  rcases h with ⟨p, q⟩
+  Hint "Now a combination of `constructor` and `exact` will get you home."
+  constructor
+  exact q
+  exact p
+
+NewTactic rcases

Game/Levels/OldAdvProposition/Level_3.lean

@@ -0,0 +1,46 @@
+import Game.Metadata
+import Game.MyNat.Addition
+
+
+World "AdvProposition"
+Level 3
+Title "and_trans"
+
+open MyNat
+
+Introduction
+"
+Here is another exercise to `rcases` and `constructor`.
+"
+
+/-- If $P$, $Q$ and $R$ are true/false statements, then $P\\land Q$ and
+$Q\\land R$ together imply $P\\land R$. -/
+Statement --and_trans
+    (P Q R : Prop) : P ∧ Q → Q ∧ R → P ∧ R := by
+  Hint "Here's a trick:
+
+  Your first steps would probably be
+  ```
+  intro h
+  rcases h with ⟨p, q⟩
+  ```
+  i.e. introducing a new assumption and then immediately take it appart.
+
+  In that case you could do that in a single step:
+
+  ```
+  intro ⟨p, q⟩
+  ```
+  "
+  intro hpq
+  rcases hpq with ⟨p, q⟩
+  intro hqr
+  rcases hqr with ⟨q', r⟩
+  constructor
+  exact p
+  exact r
+
+Conclusion
+"
+
+"

Game/Levels/OldAdvProposition/Level_4.lean

@@ -0,0 +1,45 @@
+import Game.Metadata
+import Game.MyNat.Addition
+
+
+World "AdvProposition"
+Level 4
+Title ""
+
+open MyNat
+
+Introduction
+"
+The mathematical statement $P\\iff Q$ is equivalent to $(P\\implies Q)\\land(Q\\implies P)$. The `rcases`
+and `constructor` tactics work on hypotheses and goals (respectively) of the form `P ↔ Q`. If you need
+to write an `↔` arrow you can do so by typing `\\iff`, but you shouldn't need to.
+
+"
+
+/-- If $P$, $Q$ and $R$ are true/false statements, then
+$P\\iff Q$ and $Q\\iff R$ together imply $P\\iff R$. -/
+Statement --iff_trans
+    (P Q R : Prop) : (P ↔ Q) → (Q ↔ R) → (P ↔ R) := by
+  Hint "Similar to \"and\", you can use `intro` and `rcases` to add the `P ↔ Q` to your
+  assumptions and split it into its constituent parts."
+  Branch
+    intro hpq
+    intro hqr
+    Hint "Now you want to use `rcases {hpq} with ⟨pq, qp⟩`."
+    rcases hpq with ⟨hpq, hqp⟩
+    rcases hqr with ⟨hqr, hrq⟩
+  intro ⟨pq, qp⟩
+  intro ⟨qr, rq⟩
+  Hint "If you want to prove an iff-statement, you can use `constructor` to split it
+  into its two implications."
+  constructor
+  · intro p
+    apply qr
+    apply pq
+    exact p
+  · intro r
+    apply qp
+    apply rq
+    exact r
+
+NewDefinition Iff

Game/Levels/OldAdvProposition/Level_5.lean

@@ -0,0 +1,76 @@
+import Game.Metadata
+import Game.MyNat.Addition
+
+
+World "AdvProposition"
+Level 5
+Title "Easter eggs."
+
+open MyNat
+
+Introduction
+"
+Let's try this again. Try proving it in other ways. (Note that `rcases` is temporarily disabled.)
+
+### A trick.
+
+Instead of using `rcases` on `h : P ↔ Q` you can just access the proofs of `P → Q` and `Q → P`
+directly with `h.1` and `h.2`. So you can solve this level with
+
+```
+intro hpq hqr
+constructor
+intro p
+apply hqr.1
+…
+```
+
+### Another trick
+
+Instead of using `rcases` on `h : P ↔ Q`, you can just `rw [h]`, and this will change all `P`s to `Q`s
+in the goal. You can use this to create a much shorter proof. Note that
+this is an argument for *not* running the `rcases` tactic on an iff statement;
+you cannot rewrite one-way implications, but you can rewrite two-way implications.
+
+
+"
+-- TODO
+-- ### Another trick
+-- `cc` works on this sort of goal too.
+
+/-- If $P$, $Q$ and $R$ are true/false statements, then `P ↔ Q` and `Q ↔ R` together imply `P ↔ R`.
+ -/
+Statement --iff_trans
+    (P Q R : Prop) : (P ↔ Q) → (Q ↔ R) → (P ↔ R) := by
+  intro hpq hqr
+  Hint "Make a choice and continue either with `constructor` or `rw`.
+
+  * if you use `constructor`, you will use `{hqr}.1, {hqr}.2, …` later.
+  * if you use `rw`, you can replace all `P`s with `Q`s using `rw [{hpq}]`"
+  Branch
+    rw [hpq]
+    Branch
+      exact hqr
+    rw [hqr]
+    Hint "Now `rfl` can close this goal.
+
+    TODO: Note that the current modification of `rfl` is too weak to prove this. For now, you can
+    use `simp` instead (which calls the \"real\" `rfl` internally)."
+    simp
+  constructor
+  intro p
+  Hint "Now you can directly `apply {hqr}.1`"
+  apply hqr.1
+  apply hpq.1
+  exact p
+  intro r
+  apply hpq.2
+  apply hqr.2
+  exact r
+
+DisabledTactic rcases
+
+Conclusion
+"
+
+"

Game/Levels/OldAdvProposition/Level_6.lean

@@ -0,0 +1,38 @@
+import Game.Metadata
+import Game.MyNat.Addition
+
+
+World "AdvProposition"
+Level 6
+Title "Or, and the `left` and `right` tactics."
+
+open MyNat
+
+Introduction
+"
+`P ∨ Q` means \"$P$ or $Q$\". So to prove it, you
+need to choose one of `P` or `Q`, and prove that one.
+If `P ∨ Q` is your goal, then `left` changes this
+goal to `P`, and `right` changes it to `Q`.
+"
+-- Note that you can take a wrong turn here. Let's
+-- start with trying to prove $Q\\implies (P\\lor Q)$.
+-- After the `intro`, one of `left` and `right` leads
+-- to an impossible goal, the other to an easy finish.
+
+/-- If $P$ and $Q$ are true/false statements, then $Q\\implies(P\\lor Q)$. -/
+Statement
+    (P Q : Prop) : Q → (P ∨ Q) := by
+  Hint (hidden := true) "Let's start with an initial `intro` again."
+  intro q
+  Hint "Now you need to choose if you want to prove the `left` or `right` side of the goal."
+  Branch
+    left
+    -- TODO: This message is also shown on the correct track. Need strict hints.
+    -- Hint "That was an unfortunate choice, you entered a dead end that cannot be proved anymore.
+    -- Hit \"Undo\"!"
+  right
+  exact q
+
+NewTactic left right
+NewDefinition Or

Game/Levels/OldAdvProposition/Level_7.lean

@@ -0,0 +1,55 @@
+import Game.Metadata
+import Game.MyNat.Addition
+
+
+World "AdvProposition"
+Level 7
+Title "`or_symm`"
+
+open MyNat
+
+Introduction
+"
+Proving that $(P\\lor Q)\\implies(Q\\lor P)$ involves an element of danger.
+"
+
+/-- If $P$ and $Q$ are true/false statements, then
+$$P\\lor Q\\implies Q\\lor P.$$  -/
+Statement --or_symm
+    (P Q : Prop) : P ∨ Q → Q ∨ P := by
+  Hint "`intro h` is the obvious start."
+  intro h
+  Branch
+    left
+    Hint "This is a dead end that is not provable anymore. Hit \"undo\"."
+  Branch
+    right
+    Hint "This is a dead end that is not provable anymore. Hit \"undo\"."
+  Hint "But now, even though the goal is an `∨` statement, both `left` and `right` put
+  you in a situation with an impossible goal. Fortunately,
+  you can do `rcases h with p | q`. (that is a normal vertical slash)
+  "
+  rcases h with p | q
+  Hint " Something new just happened: because
+  there are two ways to prove the assumption `P ∨ Q` (namely, proving `P` or proving `Q`),
+  the `rcases` tactic turns one goal into two, one for each case.
+
+  So now you proof the goal under the assumption that `P` is true, and waiting under \"Other Goals\"
+  there is the same goal but under the assumption that `Q` is true.
+
+  You should be able to make it home from there. "
+  right
+  exact p
+  Hint "Note how now you finished the first goal and jumped to the one, where you assume `Q`."
+  left
+  exact q
+
+Conclusion
+"
+Well done! Note that the syntax for `rcases` is different whether it's an \"or\" or an \"and\".
+
+* `rcases h with ⟨p, q⟩` splits an \"and\" in the assumptions into two parts. You get two assumptions
+but still only one goal.
+* `rcases h with p | q` splits an \"or\" in the assumptions. You get **two goals** which have different
+assumptions, once assumping the lefthand-side of the dismantled \"or\"-assumption, once the righthand-side.
+"

Game/Levels/OldAdvProposition/Level_8.lean

@@ -0,0 +1,53 @@
+import Game.Metadata
+import Game.MyNat.Addition
+
+
+World "AdvProposition"
+Level 8
+Title "`and_or_distrib_left`"
+
+open MyNat
+
+Introduction
+"
+We know that $x(y+z)=xy+xz$ for numbers, and this
+is called distributivity of multiplication over addition.
+The same is true for `∧` and `∨` -- in fact `∧` distributes
+over `∨` and `∨` distributes over `∧`. Let's prove one of these.
+"
+
+/-- If $P$. $Q$ and $R$ are true/false statements, then
+$$P\\land(Q\\lor R)\\iff(P\\land Q)\\lor (P\\land R).$$  -/
+Statement --and_or_distrib_left
+    (P Q R : Prop) : P ∧ (Q ∨ R) ↔ (P ∧ Q) ∨ (P ∧ R) := by
+  constructor
+  intro h
+  rcases h with ⟨hp, hqr⟩
+  rcases hqr with q | r
+  left
+  constructor
+  exact hp
+  exact q
+  right
+  constructor
+  exact hp
+  exact r
+  intro h
+  rcases h with hpq | hpr
+  rcases hpq with ⟨p, q⟩
+  constructor
+  exact p
+  left
+  exact q
+  rcases hpr with ⟨hp, hr⟩
+  constructor
+  exact hp
+  right
+  exact hr
+
+
+Conclusion
+"
+You already know enough to embark on advanced addition world. But the next two levels comprise
+just a couple more things.
+"

Game/Levels/OldAdvProposition/Level_9.lean

@@ -0,0 +1,52 @@
+import Game.Metadata
+import Game.MyNat.Addition
+
+
+World "AdvProposition"
+Level 9
+Title "`exfalso` and proof by contradiction. "
+
+open MyNat
+
+Introduction
+"
+It's certainly true that $P\\land(\\lnot P)\\implies Q$ for any propositions $P$
+and $Q$, because the left hand side of the implication is false. But how do
+we prove that `False` implies any proposition $Q$? A cheap way of doing it in
+Lean is using the `exfalso` tactic, which changes any goal at all to `False`.
+You might think this is a step backwards, but if you have a hypothesis `h : ¬ P`
+then after `rw not_iff_imp_false at h,` you can `apply h,` to make progress.
+Try solving this level without using `cc` or `tauto`, but using `exfalso` instead.
+"
+
+/-- If $P$ and $Q$ are true/false statements, then
+$$(P\\land(\\lnot P))\\implies Q.$$ -/
+Statement --contra
+  (P Q : Prop) : (P ∧ ¬ P) → Q := by
+  Hint "Start as usual with `intro ⟨p, np⟩`."
+  Branch
+    exfalso
+    -- TODO: This hint needs to be strict
+    -- Hint "Not so quick! Now you just threw everything away."
+  intro h
+  Hint "You should also call `rcases` on your assumption `{h}`."
+  rcases h with ⟨p, np ⟩
+  -- TODO: This hint should before the last `exact p` step again.
+  Hint "Now you can call `exfalso` to throw away your goal `Q`. It will be replaced with `False` and
+  which means you will have to prove a contradiction."
+  Branch
+    -- TODO: Would `contradiction` not be more useful to introduce than `exfalso`?
+    contradiction
+  exfalso
+  Hint "Recall that `{np} : ¬ P` means `np : P → False`, which means you can simply `apply {np}` now.
+
+  You can also first call `rw [Not] at {np}` to make this step more explicit."
+  Branch
+    rw [Not] at np
+  apply np
+  exact p
+
+-- TODO: `contradiction`?
+NewTactic exfalso
+-- DisabledTactic cc
+LemmaTab "Prop"