YES

Problem:
 0(*(x1)) -> *(1(x1))
 1(*(x1)) -> 0(#(x1))
 #(0(x1)) -> 0(#(x1))
 #(1(x1)) -> 1(#(x1))
 #($(x1)) -> *($(x1))
 #(#(x1)) -> #(x1)
 #(*(x1)) -> *(x1)

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [$](x0) = 8x0,
   
   [#](x0) = 8x0,
   
   [1](x0) = 8x0,
   
   [0](x0) = 8x0,
   
   [*](x0) = 8x0
  orientation:
   0(*(x1)) = 16x1 >= 16x1 = *(1(x1))
   
   1(*(x1)) = 16x1 >= 16x1 = 0(#(x1))
   
   #(0(x1)) = 16x1 >= 16x1 = 0(#(x1))
   
   #(1(x1)) = 16x1 >= 16x1 = 1(#(x1))
   
   #($(x1)) = 16x1 >= 16x1 = *($(x1))
   
   #(#(x1)) = 16x1 >= 8x1 = #(x1)
   
   #(*(x1)) = 16x1 >= 8x1 = *(x1)
  problem:
   0(*(x1)) -> *(1(x1))
   1(*(x1)) -> 0(#(x1))
   #(0(x1)) -> 0(#(x1))
   #(1(x1)) -> 1(#(x1))
   #($(x1)) -> *($(x1))
  KBO Processor:
   weight function:
    w0 = 1
    w($) = w(#) = w(1) = w(0) = w(*) = 1
   precedence:
    # > 1 > 0 > $ ~ *
   problem:
    
   Qed