YES

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

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