YES

We show the termination of the TRS R:

  f(X) -> cons(X,n__f(n__g(X)))
  g(|0|()) -> s(|0|())
  g(s(X)) -> s(s(g(X)))
  sel(|0|(),cons(X,Y)) -> X
  sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
  f(X) -> n__f(X)
  g(X) -> n__g(X)
  activate(n__f(X)) -> f(activate(X))
  activate(n__g(X)) -> g(activate(X))
  activate(X) -> X

-- SCC decomposition.

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

p1: g#(s(X)) -> g#(X)
p2: sel#(s(X),cons(Y,Z)) -> sel#(X,activate(Z))
p3: sel#(s(X),cons(Y,Z)) -> activate#(Z)
p4: activate#(n__f(X)) -> f#(activate(X))
p5: activate#(n__f(X)) -> activate#(X)
p6: activate#(n__g(X)) -> g#(activate(X))
p7: activate#(n__g(X)) -> activate#(X)

and R consists of:

r1: f(X) -> cons(X,n__f(n__g(X)))
r2: g(|0|()) -> s(|0|())
r3: g(s(X)) -> s(s(g(X)))
r4: sel(|0|(),cons(X,Y)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: f(X) -> n__f(X)
r7: g(X) -> n__g(X)
r8: activate(n__f(X)) -> f(activate(X))
r9: activate(n__g(X)) -> g(activate(X))
r10: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p2}
  {p5, p7}
  {p1}


-- Reduction pair.

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

p1: sel#(s(X),cons(Y,Z)) -> sel#(X,activate(Z))

and R consists of:

r1: f(X) -> cons(X,n__f(n__g(X)))
r2: g(|0|()) -> s(|0|())
r3: g(s(X)) -> s(s(g(X)))
r4: sel(|0|(),cons(X,Y)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: f(X) -> n__f(X)
r7: g(X) -> n__g(X)
r8: activate(n__f(X)) -> f(activate(X))
r9: activate(n__g(X)) -> g(activate(X))
r10: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, 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:
        
          sel#_A(x1,x2) = x1 + 7
          s_A(x1) = max{7, x1}
          cons_A(x1,x2) = max{x1 + 3, x2}
          activate_A(x1) = x1 + 7
          f_A(x1) = max{15, x1 + 8}
          n__f_A(x1) = x1 + 8
          n__g_A(x1) = max{5, x1}
          g_A(x1) = max{6, x1}
          |0|_A = 8
    
      precedence:
      
        sel# = activate = f > cons = n__f = g > s = n__g = |0|
      
      partial status:
    
        pi(sel#) = [1]
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(activate) = [1]
        pi(f) = [1]
        pi(n__f) = [1]
        pi(n__g) = [1]
        pi(g) = [1]
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sel#_A(x1,x2) = max{0, x1 - 9}
          s_A(x1) = x1 + 8
          cons_A(x1,x2) = max{x1 + 13, x2 + 10}
          activate_A(x1) = max{9, x1 + 5}
          f_A(x1) = max{10, x1 + 3}
          n__f_A(x1) = x1 + 2
          n__g_A(x1) = x1 + 4
          g_A(x1) = max{10, x1 + 9}
          |0|_A = 6
    
      precedence:
      
        sel# = s = cons = activate = f = n__f = n__g = g = |0|
      
      partial status:
    
        pi(sel#) = []
        pi(s) = [1]
        pi(cons) = [2]
        pi(activate) = [1]
        pi(f) = []
        pi(n__f) = [1]
        pi(n__g) = [1]
        pi(g) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: activate#(n__g(X)) -> activate#(X)
p2: activate#(n__f(X)) -> activate#(X)

and R consists of:

r1: f(X) -> cons(X,n__f(n__g(X)))
r2: g(|0|()) -> s(|0|())
r3: g(s(X)) -> s(s(g(X)))
r4: sel(|0|(),cons(X,Y)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: f(X) -> n__f(X)
r7: g(X) -> n__g(X)
r8: activate(n__f(X)) -> f(activate(X))
r9: activate(n__g(X)) -> g(activate(X))
r10: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: g#(s(X)) -> g#(X)

and R consists of:

r1: f(X) -> cons(X,n__f(n__g(X)))
r2: g(|0|()) -> s(|0|())
r3: g(s(X)) -> s(s(g(X)))
r4: sel(|0|(),cons(X,Y)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: f(X) -> n__f(X)
r7: g(X) -> n__g(X)
r8: activate(n__f(X)) -> f(activate(X))
r9: activate(n__g(X)) -> g(activate(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:
        
          g#_A(x1) = max{4, x1 + 3}
          s_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        g# = s
      
      partial status:
    
        pi(g#) = [1]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{1, x1 - 1}
          s_A(x1) = x1
    
      precedence:
      
        g# = s
      
      partial status:
    
        pi(g#) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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