NO

Problem:
 f(n__b(),X,n__c()) -> f(X,c(),X)
 c() -> b()
 b() -> n__b()
 c() -> n__c()
 activate(n__b()) -> b()
 activate(n__c()) -> c()
 activate(X) -> X

Proof:
 Matrix Interpretation Processor:
  dimension: 1
  interpretation:
   [activate](x0) = x0 + 6,
   
   [b] = 4,
   
   [c] = 4,
   
   [f](x0, x1, x2) = 2x0 + 3x1 + x2,
   
   [n__c] = 4,
   
   [n__b] = 4
  orientation:
   f(n__b(),X,n__c()) = 3X + 12 >= 3X + 12 = f(X,c(),X)
   
   c() = 4 >= 4 = b()
   
   b() = 4 >= 4 = n__b()
   
   c() = 4 >= 4 = n__c()
   
   activate(n__b()) = 10 >= 4 = b()
   
   activate(n__c()) = 10 >= 4 = c()
   
   activate(X) = X + 6 >= X = X
  problem:
   f(n__b(),X,n__c()) -> f(X,c(),X)
   c() -> b()
   b() -> n__b()
   c() -> n__c()
  Unfolding Processor:
   loop length: 4
   terms:
    f(c(),c(),c())
    f(c(),c(),n__c())
    f(b(),c(),n__c())
    f(n__b(),c(),n__c())
   context: []
   substitution:
    
   Qed