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65 lines
1.9 KiB
Plaintext
65 lines
1.9 KiB
Plaintext
import NNG.Metadata
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import NNG.MyNat.Addition
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Game "NNG"
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World "Function"
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Level 6
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Title "(P → (Q → R)) → ((P → Q) → (P → R))"
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open MyNat
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Introduction
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"
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You can solve this level completely just using `intro`, `apply` and `exact`,
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but if you want to argue forwards instead of backwards then don't forget
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that you can do things like
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`have j : Q → R := f p`
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if `f : P → (Q → R)` and `p : P`. Remember the trick with the colon in `have`:
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we could just write `have j := f p,` but this way we can be sure that `j` is
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what we actually expect it to be.
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I recommend that you start with `intro f` rather than `intro p`
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because even though the goal starts `P → ...`, the brackets mean that
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the goal is not a function from `P` to anything, it's a function from
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`P → (Q → R)` to something. In fact you can save time by starting
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with `intros f h p`, which introduces three variables at once, although you'd
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better then look at your tactic state to check that you called all those new
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terms sensible things.
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After all the intros, you find that your your goal is `Goal: R`…
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"
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Statement
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"Whatever the sets $P$ and $Q$ and $R$ are, we
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make an element of $\\operatorname{Hom}(\\operatorname{Hom}(P,\\operatorname{Hom}(Q,R)),
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\\operatorname{Hom}(\\operatorname{Hom}(P,Q),\\operatorname{Hom}(P,R)))$."
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(P Q R : Type) : (P → (Q → R)) → ((P → Q) → (P → R)) := by
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intro f
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intro h
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intro p
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Hint "
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If you try `have j : {Q} → {R} := {f} {p}`
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now then you can `apply j`.
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Alternatively you can `apply ({f} {p})` directly.
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What happens if you just try `apply {f}`?
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"
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Branch
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apply f
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Hint "Can you figure out what just happened? This is a little
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`apply` easter egg. Why is it mathematically valid?"
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Hint (hidden := true) "Note that there are two goals now, first you need to
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provide an element in ${P}$ which you did not provide before."
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have j : Q → R := f p
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apply j
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apply h
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exact p
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Conclusion
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"
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"
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