YES

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

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