YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [activate](x0) = x0 + 5,
   
   [a] = 0,
   
   [b] = 1,
   
   [f](x0, x1) = 6x0 + 4x1,
   
   [n__b] = 0
  orientation:
   f(X,X) = 10X >= 0 = f(a(),n__b())
   
   b() = 1 >= 0 = a()
   
   b() = 1 >= 0 = n__b()
   
   activate(n__b()) = 5 >= 1 = b()
   
   activate(X) = X + 5 >= X = X
  problem:
   f(X,X) -> f(a(),n__b())
  DP Processor:
   DPs:
    f#(X,X) -> f#(a(),n__b())
   TRS:
    f(X,X) -> f(a(),n__b())
   EDG Processor:
    DPs:
     f#(X,X) -> f#(a(),n__b())
    TRS:
     f(X,X) -> f(a(),n__b())
    graph:
     
    SCC Processor:
     #sccs: 0
     #rules: 0
     #arcs: 0/1