NO

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

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