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:
 DP Processor:
  DPs:
   a#(b(x1)) -> a#(x1)
   d#(c(x1)) -> a#(x1)
   d#(c(x1)) -> d#(a(x1))
   a#(c(x1)) -> a#(x1)
  TRS:
   a(x1) -> b(c(x1))
   a(b(x1)) -> b(a(x1))
   d(c(x1)) -> d(a(x1))
   a(c(x1)) -> c(a(x1))
  Usable Rule Processor:
   DPs:
    a#(b(x1)) -> a#(x1)
    d#(c(x1)) -> a#(x1)
    d#(c(x1)) -> d#(a(x1))
    a#(c(x1)) -> a#(x1)
   TRS:
    a(x1) -> b(c(x1))
    a(b(x1)) -> b(a(x1))
    a(c(x1)) -> c(a(x1))
   Matrix Interpretation Processor: dim=2
    
    interpretation:
     [d#](x0) = [2 4]x0,
     
     [a#](x0) = [1 0]x0,
     
               [1 0]     [1]
     [b](x0) = [0 0]x0 + [0],
     
                    [1]
     [c](x0) = x0 + [1],
     
                    [2]
     [a](x0) = x0 + [0]
    orientation:
     a#(b(x1)) = [1 0]x1 + [1] >= [1 0]x1 = a#(x1)
     
     d#(c(x1)) = [2 4]x1 + [6] >= [1 0]x1 = a#(x1)
     
     d#(c(x1)) = [2 4]x1 + [6] >= [2 4]x1 + [4] = d#(a(x1))
     
     a#(c(x1)) = [1 0]x1 + [1] >= [1 0]x1 = a#(x1)
     
                  [2]    [1 0]     [2]           
     a(x1) = x1 + [0] >= [0 0]x1 + [0] = b(c(x1))
     
                [1 0]     [3]    [1 0]     [3]           
     a(b(x1)) = [0 0]x1 + [0] >= [0 0]x1 + [0] = b(a(x1))
     
                     [3]         [3]           
     a(c(x1)) = x1 + [1] >= x1 + [1] = c(a(x1))
    problem:
     DPs:
      
     TRS:
      a(x1) -> b(c(x1))
      a(b(x1)) -> b(a(x1))
      a(c(x1)) -> c(a(x1))
    Qed