hello, below is the code that i wrote as an excercise for myself (I am still learning haskell). it implements a straighforward way to simplify boolean expressions, and should be self-explanatory. my question is, if i have an expression such as ((Const False) :&: <subexp>), will <subexp> be reduced first (according to the definition 'simplify (x :&: y) = simplify' ((simplify x) :&: (simplify y))') or will laziness do the right thing, and emit (Const False) without looking into <exp>? i think the latter, but would appreciate a word from an expert. thanks konst PS: any coding suggestions, etc. are also welcome infixr 3 :&: infixr 2 :|: data Exp = Const Bool | Sym String | Not Exp | Exp :&: Exp | Exp :|: Exp instance Eq Exp where (Const x) == (Const y) = x==y (Sym x) == (Sym y) = x==y (Not x) == (Not y) = x==y (x :&: y) == (u :&: v) = x==u && y==v || x==v && y==u (x :|: y) == (u :|: v) = x==u && y==v || x==v && y==u _ == _ = False simplify (x :&: y) = simplify' ((simplify x) :&: (simplify y)) simplify (x :|: y) = simplify' ((simplify x) :|: (simplify y)) simplify (Not x) = simplify' (Not (simplify x)) simplify x = x simplify' (Not (Const True)) = Const False simplify' (Not (Const False)) = Const True simplify' (Not (Not x)) = x simplify' ((Not x) :&: y) | x==y = Const False simplify' (x :&: (Not y)) | x==y = Const False simplify' ((Not x) :|: y) | x==y = Const True simplify' (x :|: (Not y)) | x==y = Const True simplify' ((Const False) :&: _) = Const False simplify' (_ :&: (Const False)) = Const False simplify' ((Const True) :&: x) = x simplify' (x :&: (Const True)) = x simplify' ((Const True) :|: _) = Const True simplify' (_ :|: (Const True)) = Const True simplify' ((Const False) :|: x) = x simplify' (x :|: (Const False)) = x simplify' (x :&: y) | x==y = x simplify' (x :|: y) | x==y = x simplify' x = x