YES

We show the termination of the TRS R:

  f(f(X)) -> c(n__f(g(n__f(X))))
  c(X) -> d(activate(X))
  h(X) -> c(n__d(X))
  f(X) -> n__f(X)
  d(X) -> n__d(X)
  activate(n__f(X)) -> f(X)
  activate(n__d(X)) -> d(X)
  activate(X) -> X

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(f(X)) -> c#(n__f(g(n__f(X))))
p2: c#(X) -> d#(activate(X))
p3: c#(X) -> activate#(X)
p4: h#(X) -> c#(n__d(X))
p5: activate#(n__f(X)) -> f#(X)
p6: activate#(n__d(X)) -> d#(X)

and R consists of:

r1: f(f(X)) -> c(n__f(g(n__f(X))))
r2: c(X) -> d(activate(X))
r3: h(X) -> c(n__d(X))
r4: f(X) -> n__f(X)
r5: d(X) -> n__d(X)
r6: activate(n__f(X)) -> f(X)
r7: activate(n__d(X)) -> d(X)
r8: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p3, p5}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(f(X)) -> c#(n__f(g(n__f(X))))
p2: c#(X) -> activate#(X)
p3: activate#(n__f(X)) -> f#(X)

and R consists of:

r1: f(f(X)) -> c(n__f(g(n__f(X))))
r2: c(X) -> d(activate(X))
r3: h(X) -> c(n__d(X))
r4: f(X) -> n__f(X)
r5: d(X) -> n__d(X)
r6: activate(n__f(X)) -> f(X)
r7: activate(n__d(X)) -> d(X)
r8: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{4, x1}
          f_A(x1) = max{12, x1 + 11}
          c#_A(x1) = x1 + 3
          n__f_A(x1) = max{6, x1 + 2}
          g_A(x1) = 0
          activate#_A(x1) = max{3, x1 + 1}
    
      precedence:
      
        f# = f = c# = n__f = activate# > g
      
      partial status:
    
        pi(f#) = [1]
        pi(f) = [1]
        pi(c#) = [1]
        pi(n__f) = [1]
        pi(g) = []
        pi(activate#) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 20
          f_A(x1) = max{12, x1 + 8}
          c#_A(x1) = max{13, x1 + 12}
          n__f_A(x1) = x1 + 9
          g_A(x1) = 2
          activate#_A(x1) = x1 + 11
    
      precedence:
      
        f# = f = c# = n__f = g = activate#
      
      partial status:
    
        pi(f#) = [1]
        pi(f) = [1]
        pi(c#) = [1]
        pi(n__f) = [1]
        pi(g) = []
        pi(activate#) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3

We remove them from the problem.  Then no dependency pair remains.