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:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
             [1 0]     [0]
   [c](x0) = [0 2]x0 + [1],
   
             [1 2]     [2]
   [d](x0) = [0 0]x0 + [0],
   
             [1 0]  
   [a](x0) = [0 2]x0,
   
             [1 0]  
   [b](x0) = [0 0]x0
  orientation:
           [1 0]      [1 0]             
   a(x1) = [0 2]x1 >= [0 0]x1 = b(c(x1))
   
              [1 0]      [1 0]             
   a(b(x1)) = [0 0]x1 >= [0 0]x1 = b(a(x1))
   
              [1 4]     [4]    [1 4]     [2]           
   d(c(x1)) = [0 0]x1 + [0] >= [0 0]x1 + [0] = d(a(x1))
   
              [1 0]     [0]    [1 0]     [0]           
   a(c(x1)) = [0 4]x1 + [2] >= [0 4]x1 + [1] = c(a(x1))
  problem:
   a(x1) -> b(c(x1))
   a(b(x1)) -> b(a(x1))
   a(c(x1)) -> c(a(x1))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {6,4,1}
    transitions:
     b0(3) -> 1*
     b0(5) -> 4*
     a0(2) -> 5*
     c0(2) -> 3*
     c0(5) -> 6*
     f40() -> 2*
     b1(8) -> 9*
     c1(7) -> 8*
     4 -> 5*
     2 -> 7*
     6 -> 5*
     9 -> 5*
   problem:
    
   Qed