NO

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

Proof:
 Matrix Interpretation Processor:
  dimension: 3
  interpretation:
         [1]
   [d] = [0]
         [0],
   
         [0]
   [b] = [0]
         [0],
   
         [0]
   [c] = [1]
         [0],
   
             [1 0 0]     [1]
   [f](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0],
   
         [0]
   [a] = [0]
         [0]
  orientation:
            [1]    [0]      
   f(a()) = [1] >= [1] = c()
            [0]    [0]      
   
            [1]    [1]      
   f(b()) = [1] >= [0] = d()
            [0]    [0]      
   
         [0]    [0]      
   a() = [0] >= [0] = b()
         [0]    [0]      
   
         [0]    [0]      
   b() = [0] >= [0] = a()
         [0]    [0]      
  problem:
   f(b()) -> d()
   a() -> b()
   b() -> a()
  Matrix Interpretation Processor:
   dimension: 3
   interpretation:
          [0]
    [d] = [0]
          [1],
    
          [0]
    [b] = [0]
          [0],
    
              [1 0 0]     [1]
    [f](x0) = [0 0 0]x0 + [0]
              [0 0 0]     [1],
    
          [0]
    [a] = [0]
          [0]
   orientation:
             [1]    [0]      
    f(b()) = [0] >= [0] = d()
             [1]    [1]      
    
          [0]    [0]      
    a() = [0] >= [0] = b()
          [0]    [0]      
    
          [0]    [0]      
    b() = [0] >= [0] = a()
          [0]    [0]      
   problem:
    a() -> b()
    b() -> a()
   Unfolding Processor:
    loop length: 2
    terms:
     a()
     b()
    context: []
    substitution:
     
    Qed