YES

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

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