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

Proof:
 Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {6,5,4}
   transitions:
    31(45) -> 46*
    31(30) -> 31*
    31(52) -> 53*
    31(69) -> 70*
    31(34) -> 35*
    31(9) -> 10*
    41(70) -> 71*
    41(10) -> 11*
    41(46) -> 47*
    51(65) -> 66*
    51(49) -> 50*
    11(50) -> 51*
    11(67) -> 68*
    11(47) -> 48*
    11(44) -> 45*
    11(29) -> 30*
    11(66) -> 67*
    11(33) -> 34*
    21(7) -> 8*
    21(64) -> 65*
    21(19) -> 20*
    21(21) -> 22*
    01(51) -> 52*
    01(31) -> 32*
    01(8) -> 9*
    00(2) -> 4*
    00(1) -> 4*
    00(3) -> 4*
    10(2) -> 1*
    10(1) -> 1*
    10(3) -> 1*
    20(2) -> 2*
    20(1) -> 2*
    20(3) -> 2*
    30(2) -> 3*
    30(1) -> 3*
    30(3) -> 3*
    40(2) -> 5*
    40(1) -> 5*
    40(3) -> 5*
    50(2) -> 6*
    50(1) -> 6*
    50(3) -> 6*
    1 -> 19*
    2 -> 7*
    3 -> 21*
    8 -> 64,44
    9 -> 29*
    11 -> 4*
    20 -> 8*
    22 -> 8*
    30 -> 33*
    32 -> 4*
    35 -> 4*
    45 -> 49*
    48 -> 31*
    50 -> 69*
    53 -> 6*
    68 -> 52*
    71 -> 52*
  problem:
   
  Qed