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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 9
          f_A(x1) = max{17, x1 + 9}
          c#_A(x1) = x1 + 9
          n__f_A(x1) = x1 + 9
          n__g_A(x1) = max{1, x1 - 27}
          activate#_A(x1) = max{9, x1 + 8}
          activate_A(x1) = x1 + 8
          c_A(x1) = x1 + 6
          d_A(x1) = max{5, x1 - 2}
          g_A(x1) = max{9, x1 - 20}
          n__d_A(x1) = max{1, x1 - 2}
    
      precedence:
      
        n__g = activate = g > f# = f = c# = n__f = activate# = c = d = n__d
      
      partial status:
    
        pi(f#) = [1]
        pi(f) = [1]
        pi(c#) = [1]
        pi(n__f) = []
        pi(n__g) = []
        pi(activate#) = [1]
        pi(activate) = [1]
        pi(c) = []
        pi(d) = []
        pi(g) = []
        pi(n__d) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{17, x1 + 9}
          f_A(x1) = x1 + 20
          c#_A(x1) = 19
          n__f_A(x1) = 9
          n__g_A(x1) = 7
          activate#_A(x1) = 18
          activate_A(x1) = 8
          c_A(x1) = 41
          d_A(x1) = 41
          g_A(x1) = 8
          n__d_A(x1) = 33
    
      precedence:
      
        n__d > activate > n__g = c = g > d > f# > c# > f > n__f = activate#
      
      partial status:
    
        pi(f#) = [1]
        pi(f) = []
        pi(c#) = []
        pi(n__f) = []
        pi(n__g) = []
        pi(activate#) = []
        pi(activate) = []
        pi(c) = []
        pi(d) = []
        pi(g) = []
        pi(n__d) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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: activate#(n__f(X)) -> activate#(X)
p3: 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 estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: activate#(n__f(X)) -> 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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{4, x1 + 3}
          n__f_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        activate# = n__f
      
      partial status:
    
        pi(activate#) = [1]
        pi(n__f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{1, x1 - 1}
          n__f_A(x1) = x1
    
      precedence:
      
        activate# = n__f
      
      partial status:
    
        pi(activate#) = []
        pi(n__f) = [1]
    

The next rules are strictly ordered:

  p1

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