fixed test up a bit
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test.hm
56
test.hm
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@ -1,65 +1,53 @@
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-- currently all parameterised types are commented out until kinds are being implemented
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-- no pattern matching, instead a recursor is given based on your inductive definition
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-- for example: the following
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{-
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type List A
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| nil : List A
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| cons : A → List A → List A
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-}
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-- will bring the following into scope
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-- rec[List] : B → (A → B → B) → List A → B
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-- map could then be written as
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{-
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map : (A → B) → List A → List B
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:= rec[List] nil (λx xs. cons (f x) xs)
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-}
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-- one could then define the naturals as follows
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-- one could define the naturals as follows
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type Nat
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| zero : Nat
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| succ : Nat → Nat
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| zero : Nat
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| succ : Nat → Nat
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-- which will bring a recursor for the type into scope
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-- rec[Nat] : A → (A → A) → Nat → A
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-- defining addition as
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add : Nat → Nat → Nat
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:= rec[Nat] (λx. x) (λf n. succ (f n))
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-- since | rec[Nat] : B → (Nat → B → B) → Nat → B
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-- which generalizes to | rec[Nat] : (Nat → Nat) → (Nat → (Nat → Nat) → (Nat → Nat)) → Nat → Nat → Nat
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-- ^ adding 0 ^ recursively adding one more ^ resulting addition type
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-- since | rec[Nat] : A → (A → A) → Nat → A
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-- which generalizes to | rec[Nat] : (Nat → Nat) → ((Nat → Nat) → (Nat → Nat)) → Nat → (Nat → Nat)
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-- ^ adding 0 ^ recursively adding one more ^ resulting addition type
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-- id is used to not add anything, the second function takes the last addition function and adds a layer
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-- of succ onto it, this way it generates a function for adding the right amount.
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-- multiplication is defined similairly
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-- multiplication is defined similairly
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mul : Nat → Nat → Nat
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:= rec[Nat] (λx. zero) (λf n. add n (f n))
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-- here's an example of a simpler type
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-- here's an example of a simpler type
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type Bool
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| true : Bool
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| true : Bool
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| false : Bool
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not : Bool → Bool
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:= rec[Bool] false true
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-- now, let's look at a bit more interesting example
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type Expr
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| ENat : Nat → Expr
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| EAdd : Expr → Expr → Expr
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| EMul : Expr → Expr → Expr
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-- this generates the following recursor
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-- rec[Expr] : (Nat → B) → (B → B → B) → (B → B → B) → Expr → B
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-- rec[Expr] : (Nat → A) → (A → A → A) → (A → A → A) → Expr → A
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eval : Expr → Nat
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:= rec[Expr] (λx. x) add mul
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-- here's some other functions
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isEven : Nat → Bool
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:= rec[Nat] true not
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