Problem:
 fst(0(),Z) -> nil()
 fst(s(),cons(Y)) -> cons(Y)
 from(X) -> cons(X)
 add(0(),X) -> X
 add(s(),Y) -> s()
 len(nil()) -> 0()
 len(cons(X)) -> s()

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {8,7,6,5}
   transitions:
    s1() -> 8,7
    01() -> 8*
    cons1(2) -> 6,5
    cons1(4) -> 6,5
    cons1(1) -> 6,5
    cons1(3) -> 6,5
    nil1() -> 5*
    fst0(3,1) -> 5*
    fst0(3,3) -> 5*
    fst0(4,2) -> 5*
    fst0(4,4) -> 5*
    fst0(1,2) -> 5*
    fst0(1,4) -> 5*
    fst0(2,1) -> 5*
    fst0(2,3) -> 5*
    fst0(3,2) -> 5*
    fst0(3,4) -> 5*
    fst0(4,1) -> 5*
    fst0(4,3) -> 5*
    fst0(1,1) -> 5*
    fst0(1,3) -> 5*
    fst0(2,2) -> 5*
    fst0(2,4) -> 5*
    00() -> 1*
    nil0() -> 2*
    s0() -> 3*
    cons0(2) -> 4*
    cons0(4) -> 4*
    cons0(1) -> 4*
    cons0(3) -> 4*
    from0(2) -> 6*
    from0(4) -> 6*
    from0(1) -> 6*
    from0(3) -> 6*
    add0(3,1) -> 7*
    add0(3,3) -> 7*
    add0(4,2) -> 7*
    add0(4,4) -> 7*
    add0(1,2) -> 7*
    add0(1,4) -> 7*
    add0(2,1) -> 7*
    add0(2,3) -> 7*
    add0(3,2) -> 7*
    add0(3,4) -> 7*
    add0(4,1) -> 7*
    add0(4,3) -> 7*
    add0(1,1) -> 7*
    add0(1,3) -> 7*
    add0(2,2) -> 7*
    add0(2,4) -> 7*
    len0(2) -> 8*
    len0(4) -> 8*
    len0(1) -> 8*
    len0(3) -> 8*
    1 -> 7*
    2 -> 7*
    3 -> 7*
    4 -> 7*
  problem:
   
  Qed