<!--
-eql [] l2 = empty l2
-eql (l1h:l1t) l2 = and (not (empty l2))
- (l1h == head l2)
- (eql l1t (tail l2))
-
-Simple rearangement:
-
-eql [] = \l2 -> empty l2
-eql (l1h:l1t) = \l2 -> and (not (empty l2))
- (l1h == head l2)
- (eql l1t (tail l2))
-
-and another one rearrangement:
-
-eql [] = \l2 -> empty l2
-eql (l1h:l1t) = let prev = eql l1t
- in \l2 -> and (not (empty l2))
- (l1h == head l2)
- (prev (tail l2))
-
-Now it fits the pattern of foldr
-
-So, the end result:
-
-eql = foldr f z
- where
- z = empty
- f h z = \l2 -> and (not (empty l2))
- (h == head l2)
- (z (tail l2))
--->
-
-<!--
let list_equal =
\left right. left
; here's our f
(\might_be_equal right_tail. and might_be_equal (isempty right_tail))
-->
+<!--
+list_equal [] right = isempty right
+list_equal (hd:tl) right = and (not (isempty right))
+ (hd == head right)
+ (list_equal tl (tail right))
+
+Rearangement:
+
+list_equal [] = \right -> isempty right
+list_equal (hd:tl) = let prev = list_equal tl
+ in \right -> and (not (isempty right))
+ (hd == head right)
+ (prev (tail right))
+
+Now it fits the pattern of fold_right
+
+let list_equal = \left. left (\hd sofar. \right. and (and (not (isempty right)) (eq hd (head right))) (sofar (tail right))) isempty
+
+-->
+