YES

Problem:
 a(x1) -> b(c(x1))
 a(b(x1)) -> b(a(x1))
 d(c(x1)) -> d(a(x1))
 a(c(x1)) -> c(a(x1))

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0  0 ]  
   [d](x0) = [-& -&]x0,
   
             [0  -&]  
   [b](x0) = [-& -&]x0,
   
             [0  -&]  
   [c](x0) = [1  1 ]x0,
   
             [0  -&]  
   [a](x0) = [0  0 ]x0
  orientation:
           [0  -&]      [0  -&]             
   a(x1) = [0  0 ]x1 >= [-& -&]x1 = b(c(x1))
   
              [0  -&]      [0  -&]             
   a(b(x1)) = [0  -&]x1 >= [-& -&]x1 = b(a(x1))
   
              [1  1 ]      [0  0 ]             
   d(c(x1)) = [-& -&]x1 >= [-& -&]x1 = d(a(x1))
   
              [0  -&]      [0  -&]             
   a(c(x1)) = [1  1 ]x1 >= [1  1 ]x1 = c(a(x1))
  problem:
   a(x1) -> b(c(x1))
   a(b(x1)) -> b(a(x1))
   a(c(x1)) -> c(a(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0  0 ]  
    [b](x0) = [0  -&]x0,
    
              [1  -&]  
    [c](x0) = [-& 1 ]x0,
    
              [2 2]  
    [a](x0) = [2 0]x0
   orientation:
            [2 2]      [1  1 ]             
    a(x1) = [2 0]x1 >= [1  -&]x1 = b(c(x1))
    
               [2 2]      [2 2]             
    a(b(x1)) = [2 2]x1 >= [2 2]x1 = b(a(x1))
    
               [3 3]      [3 3]             
    a(c(x1)) = [3 1]x1 >= [3 1]x1 = c(a(x1))
   problem:
    a(b(x1)) -> b(a(x1))
    a(c(x1)) -> c(a(x1))
   KBO Processor:
    weight function:
     w0 = 1
     w(b) = w(c) = w(a) = 1
    precedence:
     a > b ~ c
    problem:
     
    Qed