Problem:
 0(0(1(2(x1)))) -> 0(2(1(3(3(0(x1))))))
 0(1(0(4(x1)))) -> 0(1(3(0(4(x1)))))
 0(1(0(5(x1)))) -> 0(1(3(5(0(x1)))))
 0(1(5(0(x1)))) -> 0(1(3(5(0(x1)))))
 0(3(1(0(x1)))) -> 0(1(3(5(0(x1)))))
 3(1(0(2(x1)))) -> 1(3(5(0(2(x1)))))
 3(1(0(2(x1)))) -> 3(2(1(3(0(x1)))))
 3(1(0(5(x1)))) -> 1(3(5(3(0(x1)))))
 3(1(2(0(x1)))) -> 2(1(3(3(0(x1)))))
 3(1(2(0(x1)))) -> 3(0(2(1(4(x1)))))
 3(1(2(0(x1)))) -> 3(0(3(2(1(x1)))))
 3(1(2(0(x1)))) -> 3(3(2(1(0(x1)))))
 3(1(2(0(x1)))) -> 3(5(2(1(0(x1)))))
 3(1(2(2(x1)))) -> 3(5(2(2(1(x1)))))
 3(4(2(0(x1)))) -> 3(3(2(4(0(x1)))))
 4(0(1(0(x1)))) -> 1(3(0(4(0(x1)))))
 5(1(0(5(x1)))) -> 1(3(3(5(5(0(x1))))))
 5(2(5(0(x1)))) -> 5(2(3(5(0(x1)))))
 5(3(1(0(x1)))) -> 0(1(3(5(5(x1)))))
 0(0(4(1(0(x1))))) -> 0(4(1(3(0(0(x1))))))
 0(1(2(0(5(x1))))) -> 0(3(0(5(1(2(x1))))))
 0(1(3(1(2(x1))))) -> 3(0(2(2(1(1(x1))))))
 0(1(5(0(4(x1))))) -> 0(4(1(3(5(0(x1))))))
 0(1(5(3(5(x1))))) -> 0(1(3(5(3(5(x1))))))
 0(2(0(3(4(x1))))) -> 0(4(5(2(3(0(x1))))))
 0(2(3(1(5(x1))))) -> 0(1(3(5(5(2(x1))))))
 0(2(4(1(2(x1))))) -> 0(2(1(1(4(2(x1))))))
 0(2(5(0(2(x1))))) -> 0(0(3(5(2(2(x1))))))
 0(2(5(5(0(x1))))) -> 0(0(2(5(1(5(x1))))))
 0(3(1(5(0(x1))))) -> 1(3(0(5(3(0(x1))))))
 0(5(3(2(0(x1))))) -> 0(0(2(1(3(5(x1))))))
 3(1(0(5(2(x1))))) -> 2(4(1(3(5(0(x1))))))
 3(1(3(0(2(x1))))) -> 2(1(3(3(3(0(x1))))))
 3(1(3(2(0(x1))))) -> 3(1(3(0(5(2(x1))))))
 3(1(4(1(2(x1))))) -> 1(4(3(2(2(1(x1))))))
 3(1(4(2(0(x1))))) -> 2(1(3(3(0(4(x1))))))
 3(1(5(0(2(x1))))) -> 2(5(1(3(0(5(x1))))))
 3(3(1(0(0(x1))))) -> 5(1(3(3(0(0(x1))))))
 3(4(0(2(0(x1))))) -> 3(0(0(2(1(4(x1))))))
 3(4(2(3(2(x1))))) -> 3(3(2(5(4(2(x1))))))
 4(3(1(0(2(x1))))) -> 4(2(1(3(0(1(x1))))))
 4(3(1(2(0(x1))))) -> 1(3(0(1(4(2(x1))))))
 4(5(5(1(2(x1))))) -> 5(4(5(2(1(1(x1))))))
 4(5(5(5(0(x1))))) -> 5(5(1(5(0(4(x1))))))
 5(1(0(1(2(x1))))) -> 2(1(1(3(5(0(x1))))))
 5(1(1(0(4(x1))))) -> 5(1(1(4(3(0(x1))))))
 5(1(5(2(0(x1))))) -> 2(1(3(5(0(5(x1))))))
 5(4(1(0(5(x1))))) -> 4(1(3(5(5(0(x1))))))

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {6,5,4,3}
   transitions:
    31(17) -> 18*
    51(16) -> 17*
    21(15) -> 16*
    21(14) -> 15*
    11(19) -> 20*
    11(13) -> 14*
    00(2) -> 3*
    00(1) -> 3*
    10(2) -> 1*
    10(1) -> 1*
    20(2) -> 2*
    20(1) -> 2*
    30(2) -> 4*
    30(1) -> 4*
    40(2) -> 5*
    40(1) -> 5*
    50(2) -> 6*
    50(1) -> 6*
    1 -> 19*
    2 -> 13*
    18 -> 4*
    20 -> 14*
  problem:
   
  Qed