YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = x0 + 4,
   
   [d](x0) = x0,
   
   [b](x0) = x0 + 3,
   
   [a](x0) = x0 + 2,
   
   [g](x0) = x0 + 2
  orientation:
   a(x1) = x1 + 2 >= x1 + 2 = g(d(x1))
   
   b(b(b(x1))) = x1 + 9 >= x1 + 8 = c(d(c(x1)))
   
   b(b(x1)) = x1 + 6 >= x1 + 6 = a(g(g(x1)))
   
   c(d(x1)) = x1 + 4 >= x1 + 4 = g(g(x1))
   
   g(g(g(x1))) = x1 + 6 >= x1 + 6 = b(b(x1))
  problem:
   a(x1) -> g(d(x1))
   b(b(x1)) -> a(g(g(x1)))
   c(d(x1)) -> g(g(x1))
   g(g(g(x1))) -> b(b(x1))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [c](x0) = 8x0 + 13,
    
    [d](x0) = x0,
    
    [b](x0) = x0,
    
    [a](x0) = x0,
    
    [g](x0) = x0
   orientation:
    a(x1) = x1 >= x1 = g(d(x1))
    
    b(b(x1)) = x1 >= x1 = a(g(g(x1)))
    
    c(d(x1)) = 8x1 + 13 >= x1 = g(g(x1))
    
    g(g(g(x1))) = x1 >= x1 = b(b(x1))
   problem:
    a(x1) -> g(d(x1))
    b(b(x1)) -> a(g(g(x1)))
    g(g(g(x1))) -> b(b(x1))
   String Reversal Processor:
    a(x1) -> d(g(x1))
    b(b(x1)) -> g(g(a(x1)))
    g(g(g(x1))) -> b(b(x1))
    Bounds Processor:
     bound: 2
     enrichment: match
     automaton:
      final states: {7,4,1}
      transitions:
       f50() -> 2*
       d2(24) -> 25*
       g0(5) -> 6*
       g0(6) -> 4*
       g0(2) -> 3*
       d0(3) -> 1*
       d1(10) -> 11*
       a1(19) -> 20*
       a0(2) -> 5*
       g1(9) -> 10*
       g1(21) -> 22*
       g1(20) -> 21*
       b0(2) -> 8*
       b0(8) -> 7*
       g2(23) -> 24*
       19 -> 23*
       7 -> 24,3,10
       4 -> 8*
       11 -> 5*
       2 -> 19,9
       22 -> 7*
       25 -> 20*
     problem:
      
     Qed