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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        sel#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,1,0)) x2
        s_A(x1) = ((1,0,0,0),(0,1,0,0),(1,1,1,0),(0,1,1,0)) x1 + (3,2,1,4)
        cons_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,0,1,0),(0,0,0,1)) x2 + (3,1,1,4)
        activate_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(0,0,0,1)) x1 + (5,0,0,2)
        from_A(x1) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (11,2,1,4)
        n__from_A(x1) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (7,1,0,1)
        n__s_A(x1) = x1 + (3,1,2,1)
        first_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(0,1,0,0)) x1 + (11,2,0,1)
        |0|_A() = (1,1,0,1)
        nil_A() = (0,4,3,0)
        n__first_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,0,0)) x1 + ((0,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,0,0)) x2 + (7,1,0,0)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        cons > from > activate > s > n__s > first > n__first > nil > |0| > n__from > sel#
      
      argument filter:
    
        pi(sel#) = []
        pi(s) = []
        pi(cons) = []
        pi(activate) = [1]
        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. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        first#_A(x1,x2) = x1 + x2 + (1,2,0,0)
        s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,0,1,0)) x1 + (0,1,2,12)
        cons_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,1,1,1)) x2 + (0,1,2,1)
        activate#_A(x1) = ((1,0,0,0),(0,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (0,0,3,0)
        n__first_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(1,0,1,0)) x1 + ((1,0,0,0),(1,0,0,0),(1,1,0,0),(1,0,0,0)) x2 + (2,9,1,1)
        activate_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(0,1,0,0)) x1 + (0,3,1,12)
        n__s_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(0,1,0,0)) x1 + (0,1,1,13)
        n__from_A(x1) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (1,0,1,8)
        from_A(x1) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,0,0)) x1 + (1,2,2,10)
        first_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,1,0,0)) x2 + (2,11,5,17)
        |0|_A() = (1,0,0,1)
        nil_A() = (0,0,0,19)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        nil > activate > first > from > n__from > s > n__first > first# > |0| > n__s > activate# > cons
      
      argument filter:
    
        pi(first#) = [2]
        pi(s) = []
        pi(cons) = 2
        pi(activate#) = 1
        pi(n__first) = []
        pi(activate) = []
        pi(n__s) = []
        pi(n__from) = []
        pi(from) = []
        pi(first) = []
        pi(|0|) = []
        pi(nil) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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