YES

We show the termination of the TRS R:

  fst(|0|(),Z) -> nil()
  fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
  from(X) -> cons(X,n__from(s(X)))
  add(|0|(),X) -> X
  add(s(X),Y) -> s(n__add(activate(X),Y))
  len(nil()) -> |0|()
  len(cons(X,Z)) -> s(n__len(activate(Z)))
  fst(X1,X2) -> n__fst(X1,X2)
  from(X) -> n__from(X)
  add(X1,X2) -> n__add(X1,X2)
  len(X) -> n__len(X)
  activate(n__fst(X1,X2)) -> fst(X1,X2)
  activate(n__from(X)) -> from(X)
  activate(n__add(X1,X2)) -> add(X1,X2)
  activate(n__len(X)) -> len(X)
  activate(X) -> X

-- SCC decomposition.

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

p1: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: fst#(s(X),cons(Y,Z)) -> activate#(Z)
p3: add#(s(X),Y) -> activate#(X)
p4: len#(cons(X,Z)) -> activate#(Z)
p5: activate#(n__fst(X1,X2)) -> fst#(X1,X2)
p6: activate#(n__from(X)) -> from#(X)
p7: activate#(n__add(X1,X2)) -> add#(X1,X2)
p8: activate#(n__len(X)) -> len#(X)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: add(X1,X2) -> n__add(X1,X2)
r11: len(X) -> n__len(X)
r12: activate(n__fst(X1,X2)) -> fst(X1,X2)
r13: activate(n__from(X)) -> from(X)
r14: activate(n__add(X1,X2)) -> add(X1,X2)
r15: activate(n__len(X)) -> len(X)
r16: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p7, p8}


-- Reduction pair.

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

p1: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__len(X)) -> len#(X)
p3: len#(cons(X,Z)) -> activate#(Z)
p4: activate#(n__add(X1,X2)) -> add#(X1,X2)
p5: add#(s(X),Y) -> activate#(X)
p6: activate#(n__fst(X1,X2)) -> fst#(X1,X2)
p7: fst#(s(X),cons(Y,Z)) -> activate#(Z)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: add(X1,X2) -> n__add(X1,X2)
r11: len(X) -> n__len(X)
r12: activate(n__fst(X1,X2)) -> fst(X1,X2)
r13: activate(n__from(X)) -> from(X)
r14: activate(n__add(X1,X2)) -> add(X1,X2)
r15: activate(n__len(X)) -> len(X)
r16: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      fst#_A(x1,x2) = x1 + x2
      s_A(x1) = x1 + 1
      cons_A(x1,x2) = x2 + 1
      activate#_A(x1) = x1 + 2
      n__len_A(x1) = x1 + 1
      len#_A(x1) = x1 + 3
      n__add_A(x1,x2) = x1 + x2 + 1
      add#_A(x1,x2) = x1 + 1
      n__fst_A(x1,x2) = x1 + x2 + 1

The next rules are strictly ordered:

  p3, p4, p6

We remove them from the problem.

-- SCC decomposition.

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

p1: fst#(s(X),cons(Y,Z)) -> activate#(X)
p2: activate#(n__len(X)) -> len#(X)
p3: add#(s(X),Y) -> activate#(X)
p4: fst#(s(X),cons(Y,Z)) -> activate#(Z)

and R consists of:

r1: fst(|0|(),Z) -> nil()
r2: fst(s(X),cons(Y,Z)) -> cons(Y,n__fst(activate(X),activate(Z)))
r3: from(X) -> cons(X,n__from(s(X)))
r4: add(|0|(),X) -> X
r5: add(s(X),Y) -> s(n__add(activate(X),Y))
r6: len(nil()) -> |0|()
r7: len(cons(X,Z)) -> s(n__len(activate(Z)))
r8: fst(X1,X2) -> n__fst(X1,X2)
r9: from(X) -> n__from(X)
r10: add(X1,X2) -> n__add(X1,X2)
r11: len(X) -> n__len(X)
r12: activate(n__fst(X1,X2)) -> fst(X1,X2)
r13: activate(n__from(X)) -> from(X)
r14: activate(n__add(X1,X2)) -> add(X1,X2)
r15: activate(n__len(X)) -> len(X)
r16: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  (no SCCs)