YES

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

Proof:
 DP Processor:
  DPs:
   0#(1(2(3(4(x1))))) -> 0#(2(1(3(4(x1)))))
   0#(5(1(2(4(3(x1)))))) -> 0#(5(2(1(4(3(x1))))))
   0#(5(2(4(1(3(x1)))))) -> 0#(1(5(2(4(3(x1))))))
   0#(5(3(1(2(4(x1)))))) -> 0#(1(5(3(2(4(x1))))))
   0#(5(4(1(3(2(x1)))))) -> 0#(5(4(3(1(2(x1))))))
  TRS:
   0(1(2(3(4(x1))))) -> 0(2(1(3(4(x1)))))
   0(5(1(2(4(3(x1)))))) -> 0(5(2(1(4(3(x1))))))
   0(5(2(4(1(3(x1)))))) -> 0(1(5(2(4(3(x1))))))
   0(5(3(1(2(4(x1)))))) -> 0(1(5(3(2(4(x1))))))
   0(5(4(1(3(2(x1)))))) -> 0(5(4(3(1(2(x1))))))
  KBO Processor:
   weight function:
    w0 = 1
    w(0#) = w(5) = w(0) = w(1) = w(2) = w(3) = w(4) = 1
   precedence:
    0# > 5 > 1 > 0 ~ 2 ~ 3 ~ 4
   problem:
    DPs:
     
    TRS:
     
   Qed