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^2
      order: standard order
      interpretations:
        f#_A(x1) = x1
        f_A(x1) = ((1,0),(1,1)) x1 + (10,3)
        c#_A(x1) = ((0,1),(0,0)) x1 + (3,0)
        n__f_A(x1) = ((1,0),(1,1)) x1 + (10,1)
        n__g_A(x1) = (1,1)
        activate#_A(x1) = ((0,1),(0,0)) x1 + (2,0)
        activate_A(x1) = ((1,0),(1,1)) x1 + (2,1)
        c_A(x1) = ((1,1),(1,1)) x1 + (5,2)
        d_A(x1) = ((1,0),(1,0)) x1 + (2,0)
        g_A(x1) = (2,1)
        n__d_A(x1) = ((1,0),(1,0)) x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = (0,0)
        f_A(x1) = (1,2)
        c#_A(x1) = (2,1)
        n__f_A(x1) = (1,2)
        n__g_A(x1) = (4,7)
        activate#_A(x1) = (1,0)
        activate_A(x1) = ((0,0),(1,0)) x1 + (2,1)
        c_A(x1) = (0,3)
        d_A(x1) = ((1,0),(1,1)) x1 + (3,6)
        g_A(x1) = (3,6)
        n__d_A(x1) = ((0,0),(1,0)) x1 + (4,7)
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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