YES

We show the termination of the TRS R:

  from(X) -> cons(X,n__from(s(X)))
  first(|0|(),Z) -> nil()
  first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
  sel(|0|(),cons(X,Z)) -> X
  sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
  from(X) -> n__from(X)
  first(X1,X2) -> n__first(X1,X2)
  activate(n__from(X)) -> from(X)
  activate(n__first(X1,X2)) -> first(X1,X2)
  activate(X) -> X

-- SCC decomposition.

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

p1: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: sel#(s(X),cons(Y,Z)) -> sel#(X,activate(Z))
p3: sel#(s(X),cons(Y,Z)) -> activate#(Z)
p4: activate#(n__from(X)) -> from#(X)
p5: activate#(n__first(X1,X2)) -> first#(X1,X2)

and R consists of:

r1: from(X) -> cons(X,n__from(s(X)))
r2: first(|0|(),Z) -> nil()
r3: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r4: sel(|0|(),cons(X,Z)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: from(X) -> n__from(X)
r7: first(X1,X2) -> n__first(X1,X2)
r8: activate(n__from(X)) -> from(X)
r9: activate(n__first(X1,X2)) -> first(X1,X2)
r10: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1, p5}


-- 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: from(X) -> cons(X,n__from(s(X)))
r2: first(|0|(),Z) -> nil()
r3: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r4: sel(|0|(),cons(X,Z)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: from(X) -> n__from(X)
r7: first(X1,X2) -> n__first(X1,X2)
r8: activate(n__from(X)) -> from(X)
r9: activate(n__first(X1,X2)) -> first(X1,X2)
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 + 22
          s_A(x1) = x1 + 10
          cons_A(x1,x2) = max{x1 + 5, x2 - 11}
          activate_A(x1) = max{5, x1}
          from_A(x1) = x1 + 13
          n__from_A(x1) = x1 + 13
          first_A(x1,x2) = max{x1 + 2, x2 + 17}
          |0|_A = 2
          nil_A = 1
          n__first_A(x1,x2) = max{x1 + 2, x2 + 17}
    
      precedence:
      
        |0| > nil > sel# = activate = from = n__from > s = first > cons = n__first
      
      partial status:
    
        pi(sel#) = []
        pi(s) = [1]
        pi(cons) = [1]
        pi(activate) = [1]
        pi(from) = [1]
        pi(n__from) = []
        pi(first) = [2]
        pi(|0|) = []
        pi(nil) = []
        pi(n__first) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sel#_A(x1,x2) = 0
          s_A(x1) = max{17, x1 + 12}
          cons_A(x1,x2) = 9
          activate_A(x1) = max{13, x1}
          from_A(x1) = 14
          n__from_A(x1) = 14
          first_A(x1,x2) = x2 + 10
          |0|_A = 8
          nil_A = 9
          n__first_A(x1,x2) = x2 + 7
    
      precedence:
      
        sel# = s = cons = activate = from = n__from = first = |0| > nil = n__first
      
      partial status:
    
        pi(sel#) = []
        pi(s) = []
        pi(cons) = []
        pi(activate) = [1]
        pi(from) = []
        pi(n__from) = []
        pi(first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(n__first) = [2]
    

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: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: activate#(n__first(X1,X2)) -> first#(X1,X2)

and R consists of:

r1: from(X) -> cons(X,n__from(s(X)))
r2: first(|0|(),Z) -> nil()
r3: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r4: sel(|0|(),cons(X,Z)) -> X
r5: sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
r6: from(X) -> n__from(X)
r7: first(X1,X2) -> n__first(X1,X2)
r8: activate(n__from(X)) -> from(X)
r9: activate(n__first(X1,X2)) -> first(X1,X2)
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:
        
          first#_A(x1,x2) = max{x1 + 8, x2 + 6}
          s_A(x1) = max{3, x1 + 1}
          cons_A(x1,x2) = max{x1 + 10, x2 + 5}
          activate#_A(x1) = max{10, x1 + 5}
          n__first_A(x1,x2) = max{x1 + 3, x2 + 1}
    
      precedence:
      
        cons = activate# > first# = n__first > s
      
      partial status:
    
        pi(first#) = [1]
        pi(s) = [1]
        pi(cons) = [2]
        pi(activate#) = []
        pi(n__first) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          first#_A(x1,x2) = x1 + 2
          s_A(x1) = x1
          cons_A(x1,x2) = x2
          activate#_A(x1) = 1
          n__first_A(x1,x2) = x1 + 1
    
      precedence:
      
        s > first# = activate# > cons > n__first
      
      partial status:
    
        pi(first#) = [1]
        pi(s) = [1]
        pi(cons) = [2]
        pi(activate#) = []
        pi(n__first) = []
    

The next rules are strictly ordered:

  p1, p2

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