YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]     [0]
   [f](x0) = [0 0 0]x0 + [0]
             [0 0 0]     [1],
   
             [1 0 0]     [0]
   [c](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0],
   
             [1 0 0]     [0]
   [d](x0) = [0 0 0]x0 + [1]
             [0 1 0]     [0],
   
             [1 0 1]     [0]
   [g](x0) = [0 0 1]x0 + [0]
             [0 0 0]     [1],
   
             [1 0 0]     [0]
   [a](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0],
   
             [1 0 0]     [0]
   [b](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0]
  orientation:
              [1 0 0]     [0]    [1 0 0]     [0]           
   a(a(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(c(x1))
              [0 0 0]     [0]    [0 0 0]     [0]           
   
              [1 0 0]     [0]    [1 0 0]     [0]           
   b(b(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = c(d(x1))
              [0 0 0]     [0]    [0 0 0]     [0]           
   
           [1 0 0]     [0]    [1 0 0]     [0]        
   b(x1) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(x1)
           [0 0 0]     [0]    [0 0 0]     [0]        
   
              [1 0 0]     [0]    [1 0 0]     [0]           
   c(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = d(f(x1))
              [0 0 0]     [0]    [0 0 0]     [0]           
   
              [1 0 0]     [0]    [1 0 0]     [0]              
   d(d(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [0] = f(f(f(x1)))
              [0 0 0]     [1]    [0 0 0]     [1]              
   
           [1 0 0]     [0]    [1 0 0]     [0]        
   d(x1) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(x1)
           [0 1 0]     [0]    [0 0 0]     [0]        
   
              [1 0 0]     [0]    [1 0 0]     [0]           
   f(f(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = g(a(x1))
              [0 0 0]     [1]    [0 0 0]     [1]           
   
              [1 0 1]     [1]    [1 0 0]     [0]        
   g(g(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(x1)
              [0 0 0]     [1]    [0 0 0]     [0]        
  problem:
   a(a(x1)) -> b(c(x1))
   b(b(x1)) -> c(d(x1))
   b(x1) -> a(x1)
   c(c(x1)) -> d(f(x1))
   d(d(x1)) -> f(f(f(x1)))
   d(x1) -> b(x1)
   f(f(x1)) -> g(a(x1))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [f](x0) = x0 + 2,
    
    [c](x0) = x0 + 3,
    
    [d](x0) = x0 + 4,
    
    [g](x0) = x0,
    
    [a](x0) = x0 + 4,
    
    [b](x0) = x0 + 4
   orientation:
    a(a(x1)) = x1 + 8 >= x1 + 7 = b(c(x1))
    
    b(b(x1)) = x1 + 8 >= x1 + 7 = c(d(x1))
    
    b(x1) = x1 + 4 >= x1 + 4 = a(x1)
    
    c(c(x1)) = x1 + 6 >= x1 + 6 = d(f(x1))
    
    d(d(x1)) = x1 + 8 >= x1 + 6 = f(f(f(x1)))
    
    d(x1) = x1 + 4 >= x1 + 4 = b(x1)
    
    f(f(x1)) = x1 + 4 >= x1 + 4 = g(a(x1))
   problem:
    b(x1) -> a(x1)
    c(c(x1)) -> d(f(x1))
    d(x1) -> b(x1)
    f(f(x1)) -> g(a(x1))
   KBO Processor:
    weight function:
     w0 = 1
     w(g) = w(f) = w(d) = w(b) = w(c) = 1
     w(a) = 0
    precedence:
     a > c > d > g > b > f
    problem:
     
    Qed