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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sel#_A(x1,x2) = ((1,0),(1,1)) x1 + (2,1)
            s_A(x1) = ((1,0),(1,1)) x1 + (3,2)
            cons_A(x1,x2) = ((1,0),(0,0)) x2 + (1,14)
            activate_A(x1) = ((1,0),(1,1)) x1 + (6,7)
            from_A(x1) = ((1,0),(0,0)) x1 + (6,15)
            n__from_A(x1) = ((1,0),(0,0)) x1 + (1,16)
            first_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (5,1)
            |0|_A() = (2,2)
            nil_A() = (1,1)
            n__first_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,0)
    
      precedence:
      
        sel# = s = cons = activate = from = n__from = first = |0| = nil = n__first
      
      partial status:
    
        pi(sel#) = []
        pi(s) = []
        pi(cons) = []
        pi(activate) = []
        pi(from) = []
        pi(n__from) = []
        pi(first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(n__first) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            sel#_A(x1,x2) = x1
            s_A(x1) = ((1,0),(1,1)) x1 + (3,3)
            cons_A(x1,x2) = (0,1)
            activate_A(x1) = ((1,0),(0,0)) x1 + (5,4)
            from_A(x1) = (2,2)
            n__from_A(x1) = (1,1)
            first_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (10,5)
            |0|_A() = (1,1)
            nil_A() = (0,0)
            n__first_A(x1,x2) = ((0,0),(1,0)) x2 + (4,4)
    
      precedence:
      
        activate > first > cons = from > n__first > sel# = s = n__from = |0| = nil
      
      partial status:
    
        pi(sel#) = []
        pi(s) = []
        pi(cons) = []
        pi(activate) = []
        pi(from) = []
        pi(n__from) = []
        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)) -> 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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            first#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,1)
            s_A(x1) = ((1,0),(1,1)) x1 + (1,2)
            cons_A(x1,x2) = x1 + x2 + (3,2)
            activate#_A(x1) = ((1,0),(1,1)) x1 + (2,1)
            n__first_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (2,2)
    
      precedence:
      
        n__first > first# = s = cons = activate#
      
      partial status:
    
        pi(first#) = [2]
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(activate#) = [1]
        pi(n__first) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            first#_A(x1,x2) = (0,0)
            s_A(x1) = x1 + (1,1)
            cons_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
            activate#_A(x1) = (0,0)
            n__first_A(x1,x2) = (1,1)
    
      precedence:
      
        n__first > s > cons = activate# > first#
      
      partial status:
    
        pi(first#) = []
        pi(s) = []
        pi(cons) = []
        pi(activate#) = []
        pi(n__first) = []
    

The next rules are strictly ordered:

  p1, p2

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