import NNG.Metadata
import NNG.MyNat.Addition

Game "NNG"
World "Proposition"
Level 6
Title "(P → (Q → R)) → ((P → Q) → (P → R))"

open MyNat

Introduction
"
You can solve this level completely just using `intro`, `apply` and `exact`,
but if you want to argue forwards instead of backwards then don't forget
that you can do things like `have j : Q → R := f p` if `f : P → (Q → R)`
and `p : P`. I recommend that you start with `intro f` rather than `intro p`
because even though the goal starts `P → ...`, the brackets mean that
the goal is not the statement that `P` implies anything, it's the statement that
$P\\implies (Q\\implies R)$ implies something. In fact I'd recommend that you started
with `intros f h p`, which introduces three variables at once.
You then find that your your goal is `⊢ R`. If you try `have j : Q → R := f p`
now then you can `apply j`. Alternatively you can `apply (f p)` directly.
What happens if you just try `apply f`? Can you figure out what just happened? This is a little
`apply` easter egg. Why is it mathematically valid?
-/

/- Lemma : no-side-bar
If $P$ and $Q$ and $R$ are true/false statements, then
$$(P\\implies(Q\\implies R))\\implies((P\\implies Q)\\implies(P\\implies R)).$$

"

Statement
""
    (P Q R : Prop) : (P → (Q → R)) → ((P → Q) → (P → R)) := by
  intro f
  intro h
  intro p
  have j : Q → R := f p
  apply j
  apply h
  exact p

Conclusion
"

"