YES

We show the termination of the TRS R:

  first(|0|(),X) -> nil()
  first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
  from(X) -> cons(X,n__from(n__s(X)))
  first(X1,X2) -> n__first(X1,X2)
  from(X) -> n__from(X)
  s(X) -> n__s(X)
  activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
  activate(n__from(X)) -> from(activate(X))
  activate(n__s(X)) -> s(activate(X))
  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: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p3: activate#(n__first(X1,X2)) -> activate#(X1)
p4: activate#(n__first(X1,X2)) -> activate#(X2)
p5: activate#(n__from(X)) -> from#(activate(X))
p6: activate#(n__from(X)) -> activate#(X)
p7: activate#(n__s(X)) -> s#(activate(X))
p8: activate#(n__s(X)) -> activate#(X)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p6, p8}


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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    first#_A(x1,x2) = max{x1 + 1, x2 + 2}
    s_A(x1) = x1
    cons_A(x1,x2) = max{x1 + 2, x2}
    activate#_A(x1) = x1 + 2
    n__s_A(x1) = x1
    n__from_A(x1) = x1 + 3
    n__first_A(x1,x2) = max{x1, x2}
    activate_A(x1) = x1
    first_A(x1,x2) = max{x1, x2}
    |0|_A = 0
    nil_A = 0
    from_A(x1) = x1 + 3

The next rules are strictly ordered:

  p3

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    first#_A(x1,x2) = max{x1 - 5, x2 + 1}
    s_A(x1) = x1 + 4
    cons_A(x1,x2) = max{3, x2}
    activate#_A(x1) = x1 + 1
    n__first_A(x1,x2) = max{x1 + 3, x2 + 3}
    activate_A(x1) = x1
    n__s_A(x1) = x1 + 4
    first_A(x1,x2) = max{x1 + 3, x2 + 3}
    |0|_A = 0
    nil_A = 3
    from_A(x1) = 3
    n__from_A(x1) = 3

The next rules are strictly ordered:

  p2, p3, 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)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  (no SCCs)