YES

We show the termination of the TRS R:

  f(f(X)) -> c(n__f(n__g(n__f(X))))
  c(X) -> d(activate(X))
  h(X) -> c(n__d(X))
  f(X) -> n__f(X)
  g(X) -> n__g(X)
  d(X) -> n__d(X)
  activate(n__f(X)) -> f(activate(X))
  activate(n__g(X)) -> g(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(n__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#(activate(X))
p6: activate#(n__f(X)) -> activate#(X)
p7: activate#(n__g(X)) -> g#(X)
p8: activate#(n__d(X)) -> d#(X)

and R consists of:

r1: f(f(X)) -> c(n__f(n__g(n__f(X))))
r2: c(X) -> d(activate(X))
r3: h(X) -> c(n__d(X))
r4: f(X) -> n__f(X)
r5: g(X) -> n__g(X)
r6: d(X) -> n__d(X)
r7: activate(n__f(X)) -> f(activate(X))
r8: activate(n__g(X)) -> g(X)
r9: activate(n__d(X)) -> d(X)
r10: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p3, p5, p6}


-- Reduction pair.

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

p1: f#(f(X)) -> c#(n__f(n__g(n__f(X))))
p2: c#(X) -> activate#(X)
p3: activate#(n__f(X)) -> activate#(X)
p4: activate#(n__f(X)) -> f#(activate(X))

and R consists of:

r1: f(f(X)) -> c(n__f(n__g(n__f(X))))
r2: c(X) -> d(activate(X))
r3: h(X) -> c(n__d(X))
r4: f(X) -> n__f(X)
r5: g(X) -> n__g(X)
r6: d(X) -> n__d(X)
r7: activate(n__f(X)) -> f(activate(X))
r8: activate(n__g(X)) -> g(X)
r9: activate(n__d(X)) -> d(X)
r10: activate(X) -> X

The set of usable rules consists of

  r1, r2, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1) = x1 + 3
        f_A(x1) = x1 + 6
        c#_A(x1) = x1 + 1
        n__f_A(x1) = x1 + 6
        n__g_A(x1) = 1
        activate#_A(x1) = x1
        activate_A(x1) = x1 + 2
        c_A(x1) = x1 + 2
        d_A(x1) = 1
        g_A(x1) = 2
        n__d_A(x1) = 0
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        n__d > g > activate > f > c > d > c# > activate# > f# > n__g > n__f
      
      argument filter:
    
        pi(f#) = []
        pi(f) = []
        pi(c#) = 1
        pi(n__f) = []
        pi(n__g) = []
        pi(activate#) = []
        pi(activate) = [1]
        pi(c) = 1
        pi(d) = []
        pi(g) = []
        pi(n__d) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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