Added comments

1 files changed, 15 insertions, 3 deletions

diff --git a/week6/camil/9.4.1 b/week6/camil/9.4.1
index 0f279ca..d9b2608 100644
--- a/week6/camil/9.4.1
+++ b/week6/camil/9.4.1
@@ -2,10 +2,22 @@
 
 Induction base:
     Suppose as = []. Then we have:
-    map f (as ++ bs) = map f ([] ++ bs) = map f bs = [] ++ (map f bs) = (map f []) ++ (map f bs) = (map f as) ++ (map f bs).
+
+    map f (as ++ bs)                        // assumption as = []
+    = map f ([] ++ bs)                      // definition of ++, rule 1
+    = map f bs                              // definition of ++, rule 1
+    = [] ++ (map f bs)                      // definition of map, rule 3
+    = (map f []) ++ (map f bs)              // assumption as = []
+    = (map f as) ++ (map f bs).
 
 Induction step:
-    Suppose map f (as ++ bs) = (map f as) ++ (map f bs) for certain as and any bs. Then we have:
-    map f ([a:as] ++ bs) = map f [a:as ++ bs] = [f a : map f (as ++ bs)] = [f a : (map f as) ++ (map f bs)] = [f a : map f as] ++ (map f bs) = (map f [a:as]) ++ (map f bs).
+    Suppose map f (as ++ bs) = (map f as) ++ (map f bs) for certain as and any bs (induction hypothesis). Then we have:
+    
+    map f ([a:as] ++ bs)                    // definition of ++, rule 2
+    = map f [a:as ++ bs]                    // definition of map, rule 4
+    = [f a : map f (as ++ bs)]              // induction hypothesis: assumption map f (as ++ bs) = (map f as) ++ (map f bs)
+    = [f a : (map f as) ++ (map f bs)]      // rewriting list
+    = [f a : map f as] ++ (map f bs)        // definition of map, rule 4
+    = (map f [a:as]) ++ (map f bs).
 
 By the principle of induction we have now proven that map f (as ++ bs) = (map f as) ++ (map f bs) for any finite lists as, bs.
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