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1.

  (APP1) append(Empty, l) = l
  (APP2) append(Cons(x, xs), l) = Cons(x, append(xs, l))

  (REV1) reverse(Empty) = Empty
  (REV2) reverse(Cons(x, xs)) = append(reverse(xs), Cons(x, Empty))

2.

  (M1) mirror(Nil) = Nil
  (M2) mirror(Node(e, l, r)) =  Node(e, mirror(r), mirror(l))

  (F1) flatten(Nil) = Empty
  (F2) flatten(Node(e, l, r)) = Cons(e, append(flatten(l), flatten(r)))

3.

  (UPD1) update(MEmpty, k, v) = Insert((k, v), MEmpty)
  (UPD2) update(Insert((k, v), m), k', v') = if k = k' then Insert((k, v'), m)
                                                       else Insert((k, v), update(m, k', v'))

  (DEL1) delete(MEmpty, k) = MEmpty
  (DEL2) delete(Insert((k, v), m), k') = if k = k' then m
                                                   else Insert((k, v), delete(m, k'))
                                                   

4.

• $ \forall t \in \texttt{BTree}. size(t) = size(mirror(t)) $

  cazul de baza:
  mirror(Nil) = Nil (M1) => size(Nil) = size(mirror(Nil))
  
  ipoteza inductiei: size(l) = size(mirror(l)), size(r) = size(mirror(r))
  size(Node(e, l, r)) = size(l) + size(r) + 1                  (S2)
                      = size(r) + size(l) + 1   
                      = size(mirror(r)) + size(mirror(l)) + 1  (ipoteza inductiei)
                      = size(Node(e, mirror(r), mirror(l)))    (S2)
                      = size(mirror(Node(e, l, r))             (M2)

• $ \forall t \in \texttt{BTree}. size(t) = length(flatten(t)) $

   cazul de baza:
   size(Nil) = (S1) = 0 = (L1) = length(Empty) = (F1) = length(flatten(Nil))
   
   ipoteza inductiei: size(l) = length(flatten(l)), size(r) = length(flatten(r))
   size(Node(e, l, r)) = size(l) + size(r) + 1                           (S2)
                       = length(flatten(l)) + length(flatten(r)) + 1     (ipoteza inductiei)
                       = length(append(flatten(l), flatten(r)) + 1       (vom demonstra)
                       = length(Cons(e, append(flatten(l), flatten(r)))  (L2)
                       = length(flatten(Node(e, l, r)))                  (F2)
                       

mai trebuie sa demonstram pasul intermediar: demonstram ca

   length(append(a, b)) = length(a) + length(b)   (LAPP)
   prin inductie dupa a
   
   cazul de baza:
   length(append(Empty, b)) = length(b)  (APP1)
                            = 0 + length(b)
                            = length(Empty) + length(b)
   
   ipoteza inductiei: length(append(xs, b)) = length(xs) + length(b)
   pasul inductiei:
   length(append(Cons(x, xs), b)) = length(Cons(x, append(xs, b)))  (APP2)
                                  = length(append(xs, b)) + 1       (L2)
                                  = length(xs) + length(b) + 1      (ip. inductiei)
                                  = length(xs) + 1 + length(b)
                                  = length(Cons(x, xs)) + length(b) (L2)

• $ \forall l \in \texttt{List}. append(l, Empty) = l $

    cazul de baza:
    append(Empty, Empty) = Empty  (APP1)
    
    ipoteza inductiei: append(xs, Empty) = s
    append(Cons(x, xs), Empty) = Cons(x, append(xs, Empty))  (APP2)
                               = Cons(x, s)                  (ip. inductiei)

• $ \forall l_1, l_2, l_3 \in \texttt{List}. append(l_1, append(l_2, l_3)) = append(append(l_1, l_2), l_3) $

   facem inductie structurala dupa l_1. Vom nota append(a, b) cu a ++ b ca sa ne fie mai usor
   cazul de baza:
   Empty ++ (l_2 ++ l_3)) = l_2 ++ l_3                  (APP1)
                          = (Empty ++ l_2) ++ l_3       (APP1)
   
   ipoteza inductiei: l_1 ++ (l_2 ++ l_3) = (l_1 ++ l_2) ++ l_3
   pasul inductiei: 
   Cons(x, l_1) ++ (l_2 ++ l_3) = Cons(x, l_1 ++ (l_2 ++ l_3))    (APP2)
                                = Cons(x, (l_1 ++ l_2) ++ l_3)    (ip. inductiei)
                                = Cons(x, l_1 ++ l_2) ++ l_3      (APP2)
                                = (Cons(x, l_1) ++ l_2) ++ l_3    (APP2)

• $ \forall l_1, l_2 \in \texttt{List}. length(append(l_1, l_2)) = length(append(l_2, l_1)) $

   facem inductie structurala dupa l_1
   cazul de baza:
   length(append(Empty, l_2)) = length(l_2)                   (APP1)
                              = length(append(l_2, Empty))    (4.c)
   
   ipoteza inductiei: length(append(l_1, l_2)) = length(append(l_2, l_1))
   pasul inductiei:
   length(append(Cons(x, l_1), l_2)) = length(Cons(x, append(l_1, l_2)))  (APP2)
                                     = length(append(l_1, l_2)) + 1       (L2)
                                     = length(l_1) + length(l_2) + 1      (LAPP)
                                     = length(l_2) + length(l_1) + 1
                                     = length(l_2) + length(Cons(x, l_1)) (L2)
                                     = length(append(l_2, Cons(x, l_1)))  (LAPP)

• $ \forall l_1, l_2 \in \texttt{List}. reverse(append(l_1, l_2)) = append(reverse(l_2), reverse(l_1)) $.