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:
 DP Processor:
  DPs:
   a#(a(x1)) -> c#(x1)
   a#(a(x1)) -> b#(c(x1))
   b#(b(x1)) -> d#(x1)
   b#(b(x1)) -> c#(d(x1))
   b#(x1) -> a#(x1)
   c#(c(x1)) -> f#(x1)
   c#(c(x1)) -> d#(f(x1))
   d#(d(x1)) -> f#(x1)
   d#(d(x1)) -> f#(f(x1))
   d#(d(x1)) -> f#(f(f(x1)))
   d#(x1) -> b#(x1)
   f#(f(x1)) -> a#(x1)
   f#(f(x1)) -> g#(a(x1))
   g#(g(x1)) -> a#(x1)
  TRS:
   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)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [g#](x0) = [0 2 0]x0 + [1],
    
    [f#](x0) = [2 0 2]x0 + [3],
    
    [d#](x0) = [2 0 0]x0 + [3],
    
    [b#](x0) = [2 0 0]x0 + [2],
    
    [c#](x0) = [0 0 2]x0,
    
    [a#](x0) = [2 0 0]x0 + [1],
    
              [0 2 0]     [0]
    [g](x0) = [1 1 1]x0 + [1]
              [0 2 0]     [0],
    
              [0 0 0]     [0]
    [f](x0) = [0 0 2]x0 + [3]
              [2 0 1]     [0],
    
              [2 0 1]     [2]
    [d](x0) = [3 0 0]x0 + [0]
              [2 0 1]     [1],
    
              [2 0 1]     [2]
    [b](x0) = [0 0 0]x0 + [0]
              [2 0 1]     [1],
    
              [1 0 1]     [0]
    [c](x0) = [0 0 0]x0 + [0]
              [1 0 1]     [2],
    
              [2 0 1]     [1]
    [a](x0) = [0 0 0]x0 + [0]
              [2 0 1]     [1]
   orientation:
    a#(a(x1)) = [4 0 2]x1 + [3] >= [0 0 2]x1 = c#(x1)
    
    a#(a(x1)) = [4 0 2]x1 + [3] >= [2 0 2]x1 + [2] = b#(c(x1))
    
    b#(b(x1)) = [4 0 2]x1 + [6] >= [2 0 0]x1 + [3] = d#(x1)
    
    b#(b(x1)) = [4 0 2]x1 + [6] >= [4 0 2]x1 + [2] = c#(d(x1))
    
    b#(x1) = [2 0 0]x1 + [2] >= [2 0 0]x1 + [1] = a#(x1)
    
    c#(c(x1)) = [2 0 2]x1 + [4] >= [2 0 2]x1 + [3] = f#(x1)
    
    c#(c(x1)) = [2 0 2]x1 + [4] >= [3] = d#(f(x1))
    
    d#(d(x1)) = [4 0 2]x1 + [7] >= [2 0 2]x1 + [3] = f#(x1)
    
    d#(d(x1)) = [4 0 2]x1 + [7] >= [4 0 2]x1 + [3] = f#(f(x1))
    
    d#(d(x1)) = [4 0 2]x1 + [7] >= [4 0 2]x1 + [3] = f#(f(f(x1)))
    
    d#(x1) = [2 0 0]x1 + [3] >= [2 0 0]x1 + [2] = b#(x1)
    
    f#(f(x1)) = [4 0 2]x1 + [3] >= [2 0 0]x1 + [1] = a#(x1)
    
    f#(f(x1)) = [4 0 2]x1 + [3] >= [1] = g#(a(x1))
    
    g#(g(x1)) = [2 2 2]x1 + [3] >= [2 0 0]x1 + [1] = a#(x1)
    
               [6 0 3]     [4]    [3 0 3]     [4]           
    a(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(c(x1))
               [6 0 3]     [4]    [3 0 3]     [3]           
    
               [6 0 3]     [7]    [4 0 2]     [3]           
    b(b(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = c(d(x1))
               [6 0 3]     [6]    [4 0 2]     [5]           
    
            [2 0 1]     [2]    [2 0 1]     [1]        
    b(x1) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(x1)
            [2 0 1]     [1]    [2 0 1]     [1]        
    
               [2 0 2]     [2]    [2 0 1]     [2]           
    c(c(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = d(f(x1))
               [2 0 2]     [4]    [2 0 1]     [1]           
    
               [6 0 3]     [7]    [0 0 0]     [0]              
    d(d(x1)) = [6 0 3]x1 + [6] >= [4 0 2]x1 + [3] = f(f(f(x1)))
               [6 0 3]     [6]    [2 0 1]     [0]              
    
            [2 0 1]     [2]    [2 0 1]     [2]        
    d(x1) = [3 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(x1)
            [2 0 1]     [1]    [2 0 1]     [1]        
    
               [0 0 0]     [0]    [0 0 0]     [0]           
    f(f(x1)) = [4 0 2]x1 + [3] >= [4 0 2]x1 + [3] = g(a(x1))
               [2 0 1]     [0]    [0 0 0]     [0]           
    
               [2 2 2]     [2]    [2 0 1]     [1]        
    g(g(x1)) = [1 5 1]x1 + [2] >= [0 0 0]x1 + [0] = a(x1)
               [2 2 2]     [2]    [2 0 1]     [1]        
   problem:
    DPs:
     
    TRS:
     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)
   Qed