YES

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

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
             [1 2]     [1]
   [b](x0) = [0 0]x0 + [1],
   
             [1 2]     [2]
   [c](x0) = [0 0]x0 + [1],
   
             [1 2]     [0]
   [a](x0) = [0 0]x0 + [1],
   
             [1 2]     [2]
   [d](x0) = [0 0]x0 + [1]
  orientation:
                 [1 2]     [4]    [1 2]     [4]           
   a(a(a(x1))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = b(b(x1))
   
                 [1 2]     [7]    [1 2]     [6]           
   b(b(b(x1))) = [0 0]x1 + [1] >= [0 0]x1 + [1] = c(d(x1))
   
           [1 2]     [2]    [1 2]     [2]           
   c(x1) = [0 0]x1 + [1] >= [0 0]x1 + [1] = a(a(x1))
   
           [1 2]     [2]    [1 2]     [2]        
   d(x1) = [0 0]x1 + [1] >= [0 0]x1 + [1] = c(x1)
  problem:
   a(a(a(x1))) -> b(b(x1))
   c(x1) -> a(a(x1))
   d(x1) -> c(x1)
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {6,4,1}
    transitions:
     b0(3) -> 1*
     b0(2) -> 3*
     c0(2) -> 6*
     a1(10) -> 11*
     a1(11) -> 12*
     f40() -> 2*
     a0(2) -> 5*
     a0(5) -> 4*
     12 -> 6*
     2 -> 10*
     1 -> 11,5,12,4,6
   problem:
    
   Qed