Problem:
 first(0(),X) -> nil()
 first(s(X),cons(Y)) -> cons(Y)
 from(X) -> cons(X)

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {6,5}
   transitions:
    cons1(2) -> 6,5
    cons1(4) -> 6,5
    cons1(1) -> 6,5
    cons1(3) -> 6,5
    nil1() -> 5*
    first0(3,1) -> 5*
    first0(3,3) -> 5*
    first0(4,2) -> 5*
    first0(4,4) -> 5*
    first0(1,2) -> 5*
    first0(1,4) -> 5*
    first0(2,1) -> 5*
    first0(2,3) -> 5*
    first0(3,2) -> 5*
    first0(3,4) -> 5*
    first0(4,1) -> 5*
    first0(4,3) -> 5*
    first0(1,1) -> 5*
    first0(1,3) -> 5*
    first0(2,2) -> 5*
    first0(2,4) -> 5*
    00() -> 1*
    nil0() -> 2*
    s0(2) -> 3*
    s0(4) -> 3*
    s0(1) -> 3*
    s0(3) -> 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*
  problem:
   
  Qed