YES

We show the termination of the TRS R:

  from(X) -> cons(X,n__from(n__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)
  s(X) -> n__s(X)
  first(X1,X2) -> n__first(X1,X2)
  activate(n__from(X)) -> from(activate(X))
  activate(n__s(X)) -> s(activate(X))
  activate(n__first(X1,X2)) -> first(activate(X1),activate(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#(activate(X))
p5: activate#(n__from(X)) -> activate#(X)
p6: activate#(n__s(X)) -> s#(activate(X))
p7: activate#(n__s(X)) -> activate#(X)
p8: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p9: activate#(n__first(X1,X2)) -> activate#(X1)
p10: activate#(n__first(X1,X2)) -> activate#(X2)

and R consists of:

r1: from(X) -> cons(X,n__from(n__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: s(X) -> n__s(X)
r8: first(X1,X2) -> n__first(X1,X2)
r9: activate(n__from(X)) -> from(activate(X))
r10: activate(n__s(X)) -> s(activate(X))
r11: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r12: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1, p5, p7, p8, p9, p10}


-- 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(n__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: s(X) -> n__s(X)
r8: first(X1,X2) -> n__first(X1,X2)
r9: activate(n__from(X)) -> from(activate(X))
r10: activate(n__s(X)) -> s(activate(X))
r11: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r12: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8, r9, r10, r11, r12

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) = max{x1 + 3, x2 + 8}
          s_A(x1) = max{19, x1 + 12}
          cons_A(x1,x2) = max{6, x2 + 5}
          activate_A(x1) = max{6, x1}
          from_A(x1) = 6
          n__from_A(x1) = 0
          n__s_A(x1) = max{19, x1 + 12}
          first_A(x1,x2) = max{20, x1 + 14, x2 + 7}
          |0|_A = 21
          nil_A = 20
          n__first_A(x1,x2) = max{20, x1 + 14, x2 + 7}
    
      precedence:
      
        sel# > first = |0| = nil = n__first > activate > from > cons > n__from > s > n__s
      
      partial status:
    
        pi(sel#) = [1]
        pi(s) = [1]
        pi(cons) = []
        pi(activate) = [1]
        pi(from) = []
        pi(n__from) = []
        pi(n__s) = [1]
        pi(first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(n__first) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          sel#_A(x1,x2) = max{22, x1 + 14}
          s_A(x1) = max{26, x1 + 9}
          cons_A(x1,x2) = 30
          activate_A(x1) = x1 + 24
          from_A(x1) = 31
          n__from_A(x1) = 7
          n__s_A(x1) = max{26, x1 + 9}
          first_A(x1,x2) = 23
          |0|_A = 0
          nil_A = 6
          n__first_A(x1,x2) = 0
    
      precedence:
      
        s = activate = n__s > sel# = from = n__from = |0| = nil = n__first > first > cons
      
      partial status:
    
        pi(sel#) = [1]
        pi(s) = []
        pi(cons) = []
        pi(activate) = []
        pi(from) = []
        pi(n__from) = []
        pi(n__s) = []
        pi(first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(n__first) = []
    

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)) -> activate#(X2)
p3: activate#(n__first(X1,X2)) -> activate#(X1)
p4: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p5: activate#(n__s(X)) -> activate#(X)
p6: activate#(n__from(X)) -> activate#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(n__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: s(X) -> n__s(X)
r8: first(X1,X2) -> n__first(X1,X2)
r9: activate(n__from(X)) -> from(activate(X))
r10: activate(n__s(X)) -> s(activate(X))
r11: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r12: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r6, r7, r8, r9, r10, r11, r12

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{1, x1 - 1, x2}
          s_A(x1) = max{2, x1}
          cons_A(x1,x2) = max{1, x1, x2}
          activate#_A(x1) = x1
          n__first_A(x1,x2) = max{10, x1, x2 + 5}
          activate_A(x1) = max{4, x1}
          n__s_A(x1) = max{2, x1}
          n__from_A(x1) = max{3, x1}
          from_A(x1) = max{3, x1}
          first_A(x1,x2) = max{10, x1, x2 + 5}
          |0|_A = 2
          nil_A = 1
    
      precedence:
      
        first# > activate > s = from = first = |0| = nil > n__first > cons = n__s > n__from > activate#
      
      partial status:
    
        pi(first#) = [2]
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(activate#) = [1]
        pi(n__first) = [1, 2]
        pi(activate) = [1]
        pi(n__s) = [1]
        pi(n__from) = [1]
        pi(from) = [1]
        pi(first) = [1, 2]
        pi(|0|) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          first#_A(x1,x2) = max{5, x2}
          s_A(x1) = max{8, x1}
          cons_A(x1,x2) = max{x1 + 5, x2}
          activate#_A(x1) = x1
          n__first_A(x1,x2) = max{x1 + 4, x2}
          activate_A(x1) = x1 + 4
          n__s_A(x1) = max{4, x1}
          n__from_A(x1) = x1 + 6
          from_A(x1) = max{10, x1 + 6}
          first_A(x1,x2) = 3
          |0|_A = 0
          nil_A = 4
    
      precedence:
      
        first# = n__first > |0| > activate > s > n__from = from > cons > activate# > n__s = first = nil
      
      partial status:
    
        pi(first#) = [2]
        pi(s) = [1]
        pi(cons) = [2]
        pi(activate#) = [1]
        pi(n__first) = [1, 2]
        pi(activate) = [1]
        pi(n__s) = [1]
        pi(n__from) = []
        pi(from) = [1]
        pi(first) = []
        pi(|0|) = []
        pi(nil) = []
    

The next rules are strictly ordered:

  p4, p5

We remove them from the problem.

-- SCC decomposition.

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)) -> activate#(X2)
p3: activate#(n__first(X1,X2)) -> activate#(X1)
p4: activate#(n__from(X)) -> activate#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(n__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: s(X) -> n__s(X)
r8: first(X1,X2) -> n__first(X1,X2)
r9: activate(n__from(X)) -> from(activate(X))
r10: activate(n__s(X)) -> s(activate(X))
r11: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r12: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4}


-- Reduction pair.

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

p1: activate#(n__first(X1,X2)) -> activate#(X2)
p2: activate#(n__from(X)) -> activate#(X)
p3: activate#(n__first(X1,X2)) -> activate#(X1)

and R consists of:

r1: from(X) -> cons(X,n__from(n__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: s(X) -> n__s(X)
r8: first(X1,X2) -> n__first(X1,X2)
r9: activate(n__from(X)) -> from(activate(X))
r10: activate(n__s(X)) -> s(activate(X))
r11: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r12: 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{3, x1 + 2}
          n__first_A(x1,x2) = max{x1 + 1, x2}
          n__from_A(x1) = max{1, x1}
    
      precedence:
      
        activate# = n__first = n__from
      
      partial status:
    
        pi(activate#) = [1]
        pi(n__first) = [1, 2]
        pi(n__from) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = x1 + 2
          n__first_A(x1,x2) = max{x1 + 1, x2 + 1}
          n__from_A(x1) = x1 + 1
    
      precedence:
      
        activate# = n__first = n__from
      
      partial status:
    
        pi(activate#) = []
        pi(n__first) = [2]
        pi(n__from) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3

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