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 monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      f# > activate# > c# > g > n__f > f
    
    argument filter:
  
      pi(f#) = 1
      pi(f) = [1]
      pi(c#) = 1
      pi(n__f) = 1
      pi(g) = 1
      pi(activate#) = 1

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

  (no rules)

The estimated dependency graph contains the following SCCs:

  (no SCCs)