Problem:
 norm(nil()) -> 0()
 norm(g(x,y)) -> s(norm(x))
 f(x,nil()) -> g(nil(),x)
 f(x,g(y,z)) -> g(f(x,y),z)
 rem(nil(),y) -> nil()
 rem(g(x,y),0()) -> g(x,y)
 rem(g(x,y),s(z)) -> rem(x,z)

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {7,6,5}
   transitions:
    rem1(3,1) -> 7*
    rem1(3,3) -> 7*
    rem1(4,2) -> 7*
    rem1(4,4) -> 7*
    rem1(1,2) -> 7*
    rem1(1,4) -> 7*
    rem1(2,1) -> 7*
    rem1(2,3) -> 7*
    rem1(3,2) -> 7*
    rem1(3,4) -> 7*
    rem1(4,1) -> 7*
    rem1(4,3) -> 7*
    rem1(1,1) -> 7*
    rem1(1,3) -> 7*
    rem1(2,2) -> 7*
    rem1(2,4) -> 7*
    g1(3,1) -> 7*
    g1(3,3) -> 7*
    g1(4,2) -> 7*
    g1(4,4) -> 7*
    g1(10,1) -> 10,6
    g1(10,3) -> 10,6
    g1(1,2) -> 7*
    g1(1,4) -> 7*
    g1(7,1) -> 10*
    g1(2,1) -> 7*
    g1(7,3) -> 10*
    g1(2,3) -> 7*
    g1(3,2) -> 7*
    g1(3,4) -> 7*
    g1(4,1) -> 7*
    g1(4,3) -> 7*
    g1(10,2) -> 10,6
    g1(10,4) -> 10,6
    g1(1,1) -> 7*
    g1(1,3) -> 7*
    g1(7,2) -> 10*
    g1(2,2) -> 7*
    g1(7,4) -> 10*
    g1(2,4) -> 7*
    nil1() -> 7,10
    f1(3,1) -> 10*
    f1(3,3) -> 10*
    f1(4,2) -> 10*
    f1(4,4) -> 10*
    f1(1,2) -> 10*
    f1(1,4) -> 10*
    f1(2,1) -> 10*
    f1(2,3) -> 10*
    f1(3,2) -> 10*
    f1(3,4) -> 10*
    f1(4,1) -> 10*
    f1(4,3) -> 10*
    f1(1,1) -> 10*
    f1(1,3) -> 10*
    f1(2,2) -> 10*
    f1(2,4) -> 10*
    s1(8) -> 8,5
    norm1(2) -> 8*
    norm1(4) -> 8*
    norm1(1) -> 8*
    norm1(3) -> 8*
    01() -> 8,5
    norm0(2) -> 5*
    norm0(4) -> 5*
    norm0(1) -> 5*
    norm0(3) -> 5*
    nil0() -> 1*
    00() -> 2*
    g0(3,1) -> 3*
    g0(3,3) -> 3*
    g0(4,2) -> 3*
    g0(4,4) -> 3*
    g0(1,2) -> 3*
    g0(1,4) -> 3*
    g0(2,1) -> 3*
    g0(2,3) -> 3*
    g0(3,2) -> 3*
    g0(3,4) -> 3*
    g0(4,1) -> 3*
    g0(4,3) -> 3*
    g0(1,1) -> 3*
    g0(1,3) -> 3*
    g0(2,2) -> 3*
    g0(2,4) -> 3*
    s0(2) -> 4*
    s0(4) -> 4*
    s0(1) -> 4*
    s0(3) -> 4*
    f0(3,1) -> 6*
    f0(3,3) -> 6*
    f0(4,2) -> 6*
    f0(4,4) -> 6*
    f0(1,2) -> 6*
    f0(1,4) -> 6*
    f0(2,1) -> 6*
    f0(2,3) -> 6*
    f0(3,2) -> 6*
    f0(3,4) -> 6*
    f0(4,1) -> 6*
    f0(4,3) -> 6*
    f0(1,1) -> 6*
    f0(1,3) -> 6*
    f0(2,2) -> 6*
    f0(2,4) -> 6*
    rem0(3,1) -> 7*
    rem0(3,3) -> 7*
    rem0(4,2) -> 7*
    rem0(4,4) -> 7*
    rem0(1,2) -> 7*
    rem0(1,4) -> 7*
    rem0(2,1) -> 7*
    rem0(2,3) -> 7*
    rem0(3,2) -> 7*
    rem0(3,4) -> 7*
    rem0(4,1) -> 7*
    rem0(4,3) -> 7*
    rem0(1,1) -> 7*
    rem0(1,3) -> 7*
    rem0(2,2) -> 7*
    rem0(2,4) -> 7*
  problem:
   
  Qed