NO

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

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