YES

Problem:
 f(a(),x) -> g(a(),x)
 g(a(),x) -> f(b(),x)
 f(a(),x) -> f(b(),x)

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