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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          sel#_A(x1,x2) = x1
          s_A(x1) = x1
          cons_A(x1,x2) = 1
          activate_A(x1) = x1 + 3
          f_A(x1) = 2
          n__f_A(x1) = 2
          n__g_A(x1) = x1 + 4
          g_A(x1) = x1 + 4
          |0|_A() = 1
  
    precedence:
    
      cons = activate = f > n__f = n__g = g > sel# = s = |0|
    
    partial status:
  
      pi(sel#) = [1]
      pi(s) = [1]
      pi(cons) = []
      pi(activate) = [1]
      pi(f) = []
      pi(n__f) = []
      pi(n__g) = []
      pi(g) = [1]
      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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          activate#_A(x1) = x1
          n__g_A(x1) = x1 + 1
          n__f_A(x1) = x1 + 1
  
    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

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- 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(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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          activate#_A(x1) = x1
          n__f_A(x1) = x1 + 1
  
    precedence:
    
      activate# = n__f
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__f) = [1]

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: 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 monotone reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          g#_A(x1) = x1
          s_A(x1) = x1 + 1
  
    precedence:
    
      g# = s
    
    partial status:
  
      pi(g#) = [1]
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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