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(n__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)
  s(X) -> n__s(X)
  add(X1,X2) -> n__add(X1,X2)
  len(X) -> n__len(X)
  activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
  activate(n__from(X)) -> from(activate(X))
  activate(n__s(X)) -> s(X)
  activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
  activate(n__len(X)) -> len(activate(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) -> s#(n__add(activate(X),Y))
p4: add#(s(X),Y) -> activate#(X)
p5: len#(cons(X,Z)) -> s#(n__len(activate(Z)))
p6: len#(cons(X,Z)) -> activate#(Z)
p7: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))
p8: activate#(n__fst(X1,X2)) -> activate#(X1)
p9: activate#(n__fst(X1,X2)) -> activate#(X2)
p10: activate#(n__from(X)) -> from#(activate(X))
p11: activate#(n__from(X)) -> activate#(X)
p12: activate#(n__s(X)) -> s#(X)
p13: activate#(n__add(X1,X2)) -> add#(activate(X1),activate(X2))
p14: activate#(n__add(X1,X2)) -> activate#(X1)
p15: activate#(n__add(X1,X2)) -> activate#(X2)
p16: activate#(n__len(X)) -> len#(activate(X))
p17: activate#(n__len(X)) -> activate#(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(n__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: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4, p6, p7, p8, p9, p11, p13, p14, p15, p16, p17}


-- 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)) -> activate#(X)
p3: activate#(n__len(X)) -> len#(activate(X))
p4: len#(cons(X,Z)) -> activate#(Z)
p5: activate#(n__add(X1,X2)) -> activate#(X2)
p6: activate#(n__add(X1,X2)) -> activate#(X1)
p7: activate#(n__add(X1,X2)) -> add#(activate(X1),activate(X2))
p8: add#(s(X),Y) -> activate#(X)
p9: activate#(n__from(X)) -> activate#(X)
p10: activate#(n__fst(X1,X2)) -> activate#(X2)
p11: activate#(n__fst(X1,X2)) -> activate#(X1)
p12: activate#(n__fst(X1,X2)) -> fst#(activate(X1),activate(X2))
p13: 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(n__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: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          fst#_A(x1,x2) = max{x1 + 3, x2 + 3}
          s_A(x1) = max{9, x1}
          cons_A(x1,x2) = max{x1 + 10, x2}
          activate#_A(x1) = max{2, x1}
          n__len_A(x1) = x1 + 13
          len#_A(x1) = x1 + 12
          activate_A(x1) = x1
          n__add_A(x1,x2) = max{x1 + 11, x2 + 10}
          add#_A(x1,x2) = max{x1 + 6, x2 + 9}
          n__from_A(x1) = max{21, x1 + 11}
          n__fst_A(x1,x2) = max{x1 + 4, x2 + 7}
          fst_A(x1,x2) = max{x1 + 4, x2 + 7}
          |0|_A = 8
          nil_A = 8
          from_A(x1) = max{21, x1 + 11}
          n__s_A(x1) = max{9, x1}
          add_A(x1,x2) = max{x1 + 11, x2 + 10}
          len_A(x1) = x1 + 13
    
      precedence:
      
        activate = |0| = add = len > from > s = n__from = nil > n__len = n__s > n__add = fst > cons > fst# = activate# = len# = add# > n__fst
      
      partial status:
    
        pi(fst#) = [1, 2]
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(activate#) = [1]
        pi(n__len) = [1]
        pi(len#) = [1]
        pi(activate) = [1]
        pi(n__add) = [1, 2]
        pi(add#) = [1, 2]
        pi(n__from) = [1]
        pi(n__fst) = [1, 2]
        pi(fst) = []
        pi(|0|) = []
        pi(nil) = []
        pi(from) = [1]
        pi(n__s) = [1]
        pi(add) = [1, 2]
        pi(len) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          fst#_A(x1,x2) = max{42, x1 + 2, x2 + 2}
          s_A(x1) = max{46, x1 + 42}
          cons_A(x1,x2) = max{x1 + 43, x2 + 18}
          activate#_A(x1) = max{43, x1 + 3}
          n__len_A(x1) = 17
          len#_A(x1) = 17
          activate_A(x1) = x1 + 37
          n__add_A(x1,x2) = max{x1 + 82, x2 + 81}
          add#_A(x1,x2) = max{82, x1 + 2, x2 + 44}
          n__from_A(x1) = max{7, x1}
          n__fst_A(x1,x2) = max{x1 + 39, x2 + 39}
          fst_A(x1,x2) = 58
          |0|_A = 47
          nil_A = 1
          from_A(x1) = 44
          n__s_A(x1) = max{43, x1 + 6}
          add_A(x1,x2) = max{119, x2 + 81}
          len_A(x1) = 46
    
      precedence:
      
        fst# = n__fst > cons = activate = fst = len > n__add > add > activate# = n__len = len# = add# > s > |0| = nil = from > n__from = n__s
      
      partial status:
    
        pi(fst#) = [2]
        pi(s) = []
        pi(cons) = [2]
        pi(activate#) = [1]
        pi(n__len) = []
        pi(len#) = []
        pi(activate) = [1]
        pi(n__add) = [1, 2]
        pi(add#) = [1]
        pi(n__from) = [1]
        pi(n__fst) = []
        pi(fst) = []
        pi(|0|) = []
        pi(nil) = []
        pi(from) = []
        pi(n__s) = []
        pi(add) = []
        pi(len) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p9, p10, p11, p12, p13

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__add(X1,X2)) -> add#(activate(X1),activate(X2))
p2: add#(s(X),Y) -> activate#(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(n__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: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: activate#(n__add(X1,X2)) -> add#(activate(X1),activate(X2))
p2: add#(s(X),Y) -> activate#(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(n__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: s(X) -> n__s(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: len(X) -> n__len(X)
r13: activate(n__fst(X1,X2)) -> fst(activate(X1),activate(X2))
r14: activate(n__from(X)) -> from(activate(X))
r15: activate(n__s(X)) -> s(X)
r16: activate(n__add(X1,X2)) -> add(activate(X1),activate(X2))
r17: activate(n__len(X)) -> len(activate(X))
r18: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{23, x1}
          n__add_A(x1,x2) = max{50, x1 + 25, x2 + 43}
          add#_A(x1,x2) = max{26, x1 + 25, x2 + 15}
          activate_A(x1) = max{7, x1}
          s_A(x1) = max{24, x1}
          fst_A(x1,x2) = max{31, x2 + 23}
          |0|_A = 0
          nil_A = 1
          cons_A(x1,x2) = max{5, x2 - 15}
          n__fst_A(x1,x2) = max{31, x2 + 23}
          from_A(x1) = 5
          n__from_A(x1) = 4
          n__s_A(x1) = max{24, x1}
          add_A(x1,x2) = max{50, x1 + 25, x2 + 43}
          len_A(x1) = 25
          n__len_A(x1) = 25
    
      precedence:
      
        |0| = nil > activate# = n__add = add > add# > activate = s = from = len > fst = n__from > n__fst = n__s > cons = n__len
      
      partial status:
    
        pi(activate#) = [1]
        pi(n__add) = []
        pi(add#) = [1, 2]
        pi(activate) = [1]
        pi(s) = []
        pi(fst) = [2]
        pi(|0|) = []
        pi(nil) = []
        pi(cons) = []
        pi(n__fst) = [2]
        pi(from) = []
        pi(n__from) = []
        pi(n__s) = []
        pi(add) = []
        pi(len) = []
        pi(n__len) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{63, x1 + 43}
          n__add_A(x1,x2) = 11
          add#_A(x1,x2) = max{62, x1 + 12, x2 + 7}
          activate_A(x1) = x1 + 17
          s_A(x1) = 15
          fst_A(x1,x2) = 36
          |0|_A = 17
          nil_A = 1
          cons_A(x1,x2) = 4
          n__fst_A(x1,x2) = 1
          from_A(x1) = 11
          n__from_A(x1) = 11
          n__s_A(x1) = 14
          add_A(x1,x2) = 36
          len_A(x1) = 16
          n__len_A(x1) = 0
    
      precedence:
      
        |0| = n__s > activate# = n__add = add# = activate = fst = nil = n__fst = add = len = n__len > s = cons = from > n__from
      
      partial status:
    
        pi(activate#) = []
        pi(n__add) = []
        pi(add#) = [2]
        pi(activate) = [1]
        pi(s) = []
        pi(fst) = []
        pi(|0|) = []
        pi(nil) = []
        pi(cons) = []
        pi(n__fst) = []
        pi(from) = []
        pi(n__from) = []
        pi(n__s) = []
        pi(add) = []
        pi(len) = []
        pi(n__len) = []
    

The next rules are strictly ordered:

  p1, p2

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