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))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
    [d#](x0) = [1 1]x0 + [2],
    
    [a#](x0) = [1 0]x0,
    
              [0 0]     [2]
    [d](x0) = [1 1]x0 + [1],
    
              [1 0]     [1]
    [b](x0) = [0 0]x0 + [0],
    
              [1 0]     [3]
    [c](x0) = [1 2]x0 + [2],
    
              [1 0]     [4]
    [a](x0) = [1 2]x0 + [0]
   orientation:
    a#(b(x1)) = [1 0]x1 + [1] >= [1 0]x1 = a#(x1)
    
    d#(c(x1)) = [2 2]x1 + [7] >= [1 0]x1 = a#(x1)
    
    d#(c(x1)) = [2 2]x1 + [7] >= [2 2]x1 + [6] = d#(a(x1))
    
    a#(c(x1)) = [1 0]x1 + [3] >= [1 0]x1 = a#(x1)
    
            [1 0]     [4]    [1 0]     [4]           
    a(x1) = [1 2]x1 + [0] >= [0 0]x1 + [0] = b(c(x1))
    
               [1 0]     [5]    [1 0]     [5]           
    a(b(x1)) = [1 0]x1 + [1] >= [0 0]x1 + [0] = b(a(x1))
    
               [0 0]     [2]    [0 0]     [2]           
    d(c(x1)) = [2 2]x1 + [6] >= [2 2]x1 + [5] = d(a(x1))
    
               [1 0]     [7]    [1 0]     [7]           
    a(c(x1)) = [3 4]x1 + [7] >= [3 4]x1 + [6] = c(a(x1))
   problem:
    DPs:
     
    TRS:
     a(x1) -> b(c(x1))
     a(b(x1)) -> b(a(x1))
     d(c(x1)) -> d(a(x1))
     a(c(x1)) -> c(a(x1))
   Qed