NO

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

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