Code Question: Algebraic types in haskell

How can I simplify an expression using basic arithmetic?

From stackoverflow

You can use the technique described here: http://augustss.blogspot.com/2007/04/overloading-haskell-numbers-part-2.html . Make your type be of the necassary type-classes (Num, Fractional, Floating) so that -, +, * and so on works for your type. Then if the expression tree is finally built, you can operate on it to see what you can simplify.
I'm not sure what you mean, but if you have an expression datatype you can define a recursive eval-function. In this case eval means simplify.

For example,
```
data Exp = Lit Int
         | Plus Exp Exp
         | Times Exp Exp

eval :: Exp -> Int
eval (Lit x)     = x
eval (Plus x y)  = eval x + eval y
eval (Times x y) = eval x * eval y
```
It gets really interesting once you add variables to the language, but this is the most basic form of an expression-evaluator.
module Expr where

-- Variables are named by strings, assumed to be identifiers. type Variable = String

-- Representation of expressions. data Expr = Const Integer | Var Variable | Plus Expr Expr | Minus Expr Expr | Mult Expr Expr deriving (Eq, Show)

Simplifications such as 0*e=e*0=0 and 1*e=e*1=0+e=e+0=e-0=e and simplifying constant subexpressions, e.g. Plus (Const 1) (Const 2) would become Const 3. I would not expect variables (or variables and constants) to be concatenated: Var "st" is a distinct variable from Var "s".

they need to be written like the following simplify (Plus (Var'x') (Const 0)) = Var"x"

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