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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            fst#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (3,4)
            s_A(x1) = ((1,0),(0,0)) x1 + (0,4)
            cons_A(x1,x2) = ((1,0),(0,0)) x2 + (0,1)
            activate#_A(x1) = ((1,0),(1,1)) x1 + (2,13)
            n__len_A(x1) = ((1,0),(0,0)) x1 + (4,5)
            len#_A(x1) = x1 + (3,11)
            activate_A(x1) = x1 + (0,11)
            n__add_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (4,12)
            add#_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (3,11)
            n__from_A(x1) = ((1,0),(1,0)) x1 + (3,2)
            n__fst_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,14)
            fst_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,15)
            |0|_A() = (1,1)
            nil_A() = (0,2)
            from_A(x1) = ((1,0),(1,0)) x1 + (3,2)
            n__s_A(x1) = ((1,0),(0,0)) x1 + (0,3)
            add_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (4,12)
            len_A(x1) = ((1,0),(0,0)) x1 + (4,5)
    
      precedence:
      
        add# > |0| > nil > activate = fst = from = add = len > s = n__add > fst# = activate# = len# = n__from = n__s > n__fst > cons = n__len
      
      partial status:
    
        pi(fst#) = [2]
        pi(s) = []
        pi(cons) = []
        pi(activate#) = [1]
        pi(n__len) = []
        pi(len#) = [1]
        pi(activate) = [1]
        pi(n__add) = [1]
        pi(add#) = []
        pi(n__from) = []
        pi(n__fst) = []
        pi(fst) = []
        pi(|0|) = []
        pi(nil) = []
        pi(from) = []
        pi(n__s) = []
        pi(add) = [1]
        pi(len) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            fst#_A(x1,x2) = x2 + (1,0)
            s_A(x1) = (5,0)
            cons_A(x1,x2) = (4,8)
            activate#_A(x1) = (4,4)
            n__len_A(x1) = (5,5)
            len#_A(x1) = ((1,0),(1,0)) x1
            activate_A(x1) = ((1,0),(1,0)) x1 + (2,2)
            n__add_A(x1,x2) = ((1,0),(0,0)) x1
            add#_A(x1,x2) = (5,5)
            n__from_A(x1) = (5,7)
            n__fst_A(x1,x2) = (6,0)
            fst_A(x1,x2) = (7,7)
            |0|_A() = (0,0)
            nil_A() = (0,0)
            from_A(x1) = (8,8)
            n__s_A(x1) = (0,0)
            add_A(x1,x2) = ((1,0),(0,0)) x1 + (0,1)
            len_A(x1) = (7,6)
    
      precedence:
      
        cons = activate = add > len > fst# = s = activate# = len# = add# = n__fst > n__len > n__add > from > |0| > fst = nil > n__from = n__s
      
      partial status:
    
        pi(fst#) = []
        pi(s) = []
        pi(cons) = []
        pi(activate#) = []
        pi(n__len) = []
        pi(len#) = []
        pi(activate) = []
        pi(n__add) = []
        pi(add#) = []
        pi(n__from) = []
        pi(n__fst) = []
        pi(fst) = []
        pi(|0|) = []
        pi(nil) = []
        pi(from) = []
        pi(n__s) = []
        pi(add) = []
        pi(len) = []
    

The next rules are strictly ordered:

  p3, p7, p8, p10, p12

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)) -> activate#(X)
p3: len#(cons(X,Z)) -> activate#(Z)
p4: activate#(n__add(X1,X2)) -> activate#(X2)
p5: activate#(n__add(X1,X2)) -> activate#(X1)
p6: activate#(n__from(X)) -> activate#(X)
p7: activate#(n__fst(X1,X2)) -> activate#(X1)
p8: 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 estimated dependency graph contains the following SCCs:

  {p2, p4, p5, p6, p7}


-- Reduction pair.

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

p1: activate#(n__len(X)) -> activate#(X)
p2: activate#(n__fst(X1,X2)) -> activate#(X1)
p3: activate#(n__from(X)) -> activate#(X)
p4: activate#(n__add(X1,X2)) -> activate#(X1)
p5: activate#(n__add(X1,X2)) -> activate#(X2)

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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            activate#_A(x1) = ((0,0),(1,0)) x1 + (2,2)
            n__len_A(x1) = x1 + (1,3)
            n__fst_A(x1,x2) = ((1,0),(1,1)) x1 + (3,3)
            n__from_A(x1) = x1 + (3,1)
            n__add_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,1)
    
      precedence:
      
        n__fst > activate# > n__len = n__from = n__add
      
      partial status:
    
        pi(activate#) = []
        pi(n__len) = [1]
        pi(n__fst) = [1]
        pi(n__from) = [1]
        pi(n__add) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            activate#_A(x1) = (0,0)
            n__len_A(x1) = (1,1)
            n__fst_A(x1,x2) = (1,1)
            n__from_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            n__add_A(x1,x2) = (1,1)
    
      precedence:
      
        n__len = n__from > n__add > activate# = n__fst
      
      partial status:
    
        pi(activate#) = []
        pi(n__len) = []
        pi(n__fst) = []
        pi(n__from) = []
        pi(n__add) = []
    

The next rules are strictly ordered:

  p1, p2, p4

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__from(X)) -> activate#(X)
p2: activate#(n__add(X1,X2)) -> activate#(X2)

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__from(X)) -> activate#(X)
p2: activate#(n__add(X1,X2)) -> activate#(X2)

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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            activate#_A(x1) = x1
            n__from_A(x1) = x1 + (1,1)
            n__add_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
    
      precedence:
      
        n__from > activate# = n__add
      
      partial status:
    
        pi(activate#) = []
        pi(n__from) = [1]
        pi(n__add) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            activate#_A(x1) = (0,0)
            n__from_A(x1) = (1,1)
            n__add_A(x1,x2) = x1 + (1,1)
    
      precedence:
      
        n__add > activate# = n__from
      
      partial status:
    
        pi(activate#) = []
        pi(n__from) = []
        pi(n__add) = [1]
    

The next rules are strictly ordered:

  p1, p2

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