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:
   [d](x0) = 3x0,
   
   [c](x0) = x0 + 1,
   
   [a](x0) = x0 + 4,
   
   [b](x0) = x0 + 12
  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))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {7,4,1}
    transitions:
     b0(3) -> 1*
     a0(5) -> 6*
     a0(6) -> 4*
     a0(3) -> 7*
     a0(2) -> 5*
     a1(13) -> 14*
     a1(12) -> 13*
     a1(14) -> 15*
     f40() -> 2*
     d0(2) -> 3*
     1 -> 3*
     3 -> 12*
     15 -> 1*
   problem:
    
   Qed