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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          sel#_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(1,1)) x2 + (1,0)
          s_A(x1) = ((1,0),(1,1)) x1 + (13,4)
          cons_A(x1,x2) = ((1,1),(1,1)) x2 + (3,1)
          activate_A(x1) = ((1,1),(1,1)) x1 + (19,2)
          from_A(x1) = (5,2)
          n__from_A(x1) = (0,1)
          first_A(x1,x2) = ((1,0),(1,0)) x1 + ((0,1),(0,1)) x2 + (4,3)
          |0|_A() = (2,0)
          nil_A() = (1,1)
          n__first_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(0,1)) x2 + (4,3)
  
    precedence:
    
      cons = |0| = nil > sel# = activate > s = from = n__from > first > n__first
    
    partial status:
  
      pi(sel#) = [1]
      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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          first#_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,1)) x2 + (2,2)
          s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
          cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,0)) x2 + (3,2)
          activate#_A(x1) = ((0,1),(1,0)) x1 + (2,2)
          n__first_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(1,1)) x2 + (1,1)
  
    precedence:
    
      s = activate# > first# = n__first > cons
    
    partial status:
  
      pi(first#) = [2]
      pi(s) = []
      pi(cons) = []
      pi(activate#) = []
      pi(n__first) = []

The next rules are strictly ordered:

  p2

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)

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:

  (no SCCs)