;; To test on issue #143
(rule ((= x 1) (= y x)) ())
(rule ((= x 1) (= y x) (= z y)) ())

(function f (i64) i64)

(set (f 0) 0)

;; a funky id rule
(rule ((f x) (= x y) (= z y)) ((let a (f z)) (set (f (+ z 1)) (+ a 1))))

(run 20)

(print-function f 100)
(check (= (f 10) 10))

(datatype Value (Num i64))
(function fib (Value) Value)

;; a funky fibonacci that test on more complex case and user defined datatype
(rule ((= (Num a) (fib (Num x))) 
		(= (Num b) (fib (Num y)))
		(= x1 x)
		(= y1 y)
		(= a1 a)
		(= b1 b)
		(= x1 x2)
		(= y1 y2)
		(= a1 a2)
		(= b1 b2)
		(= 1 (- x2 y2)))
	((let n (+ x 1)) (let sum (+ a1 b2)) (union (fib (Num n)) (Num sum))))

(union (fib (Num 1)) (Num 1))
(union (fib (Num 2)) (Num 1))

(run 20)

(print-function fib 100)
(check (= (fib (Num 10)) (Num 55)))