YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = x0 + 1,
   
   [b](x0) = x0 + 12,
   
   [d](x0) = 3x0,
   
   [a](x0) = x0 + 4
  orientation:
   d(a(x1)) = 3x1 + 12 >= 3x1 + 12 = b(d(x1))
   
   b(x1) = x1 + 12 >= x1 + 12 = a(a(a(x1)))
   
   c(d(c(x1))) = 3x1 + 4 >= 3x1 + 4 = a(d(x1))
   
   b(d(d(x1))) = 9x1 + 12 >= 9x1 + 11 = c(c(d(d(c(x1)))))
  problem:
   d(a(x1)) -> b(d(x1))
   b(x1) -> a(a(a(x1)))
   c(d(c(x1))) -> a(d(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [0 0]  
    [c](x0) = [2 2]x0,
    
              [0  0 ]  
    [b](x0) = [-& 3 ]x0,
    
              [0  3 ]  
    [d](x0) = [-& -&]x0,
    
              [0  -&]  
    [a](x0) = [-& 1 ]x0
   orientation:
               [0  4 ]      [0  3 ]             
    d(a(x1)) = [-& -&]x1 >= [-& -&]x1 = b(d(x1))
    
            [0  0 ]      [0  -&]                
    b(x1) = [-& 3 ]x1 >= [-& 3 ]x1 = a(a(a(x1)))
    
                  [5 5]      [0  3 ]             
    c(d(c(x1))) = [7 7]x1 >= [-& -&]x1 = a(d(x1))
   problem:
    d(a(x1)) -> b(d(x1))
    b(x1) -> a(a(a(x1)))
   Bounds Processor:
    bound: 1
    enrichment: match
    automaton:
     final states: {4,1}
     transitions:
      a0(5) -> 6*
      a0(2) -> 5*
      a0(6) -> 4*
      a1(7) -> 8*
      a1(9) -> 10*
      a1(8) -> 9*
      f40() -> 2*
      b0(3) -> 1*
      d0(2) -> 3*
      1 -> 3,7
      3 -> 7*
      10 -> 1*
    problem:
     
    Qed