Haskell Demonstration Induction and Recursion
$10-20 USD
Betalt ved levering
1) I need to demonstrate that cut function is fulfilled (∀list:: [a])(∀x :: a) cut x (cut x list) = 0
cut :: Eq a => a -> [a] -> [a]
cut = \a -> \l -> case a of {
y -> case l of {
[] -> [];
(x:xs) -> if x == y then [] else [x] ++ (cut a xs);
};
}
2) I need to demonstrate that (∀list:: [a])(∀x :: a)(∀y :: a) count x (count y list) ≤ count x list.
count :: Eq a => a -> [a] -> N
count = \a -> \l -> case a of {
y -> case l of {
[] -> O;
(x:xs) -> if x == y then S(count a xs) else (count a xs);
};
}
-----------------------------
{- This is the data type of natural numbers which you have to use: -}
{-#LANGUAGE GADTs #-}
{-# OPTIONS_GHC -fno-warn-tabs #-}
{-# OPTIONS_GHC -fno-warn-missing-methods #-}
module Naturales where
import Prelude hiding (Eq, (==), (/=), Ord, (<),(<=),(>),(>=))
import [login to view URL]
data N where { O :: N ;
S :: N -> N
} deriving Show
one :: N
one = S O
two :: N
two = S one
three :: N
three = S two
four :: N
four = S three
five :: N
five = S four
six :: N
six = S five
seven :: N
seven = S six
eight:: N
eight = S seven
nine:: N
nine= S eight
Projekt ID: #28254575
