YES

We show the termination of the TRS R:

  a__fst(|0|(),Z) -> nil()
  a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
  a__from(X) -> cons(mark(X),from(s(X)))
  a__add(|0|(),X) -> mark(X)
  a__add(s(X),Y) -> s(add(X,Y))
  a__len(nil()) -> |0|()
  a__len(cons(X,Z)) -> s(len(Z))
  mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
  mark(from(X)) -> a__from(mark(X))
  mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
  mark(len(X)) -> a__len(mark(X))
  mark(|0|()) -> |0|()
  mark(s(X)) -> s(X)
  mark(nil()) -> nil()
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  a__fst(X1,X2) -> fst(X1,X2)
  a__from(X) -> from(X)
  a__add(X1,X2) -> add(X1,X2)
  a__len(X) -> len(X)

-- SCC decomposition.

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

p1: a__fst#(s(X),cons(Y,Z)) -> mark#(Y)
p2: a__from#(X) -> mark#(X)
p3: a__add#(|0|(),X) -> mark#(X)
p4: mark#(fst(X1,X2)) -> a__fst#(mark(X1),mark(X2))
p5: mark#(fst(X1,X2)) -> mark#(X1)
p6: mark#(fst(X1,X2)) -> mark#(X2)
p7: mark#(from(X)) -> a__from#(mark(X))
p8: mark#(from(X)) -> mark#(X)
p9: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p10: mark#(add(X1,X2)) -> mark#(X1)
p11: mark#(add(X1,X2)) -> mark#(X2)
p12: mark#(len(X)) -> a__len#(mark(X))
p13: mark#(len(X)) -> mark#(X)
p14: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p13, p14}


-- Reduction pair.

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

p1: a__fst#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(len(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X2)
p5: mark#(add(X1,X2)) -> mark#(X1)
p6: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p7: a__add#(|0|(),X) -> mark#(X)
p8: mark#(from(X)) -> mark#(X)
p9: mark#(from(X)) -> a__from#(mark(X))
p10: a__from#(X) -> mark#(X)
p11: mark#(fst(X1,X2)) -> mark#(X2)
p12: mark#(fst(X1,X2)) -> mark#(X1)
p13: mark#(fst(X1,X2)) -> a__fst#(mark(X1),mark(X2))

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(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, r19

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__fst#_A(x1,x2) = ((0,0),(1,0)) x2 + (7,3)
            s_A(x1) = (5,1)
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + (8,0)
            mark#_A(x1) = ((0,0),(1,0)) x1 + (7,0)
            len_A(x1) = ((1,0),(1,0)) x1 + (1,4)
            add_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (4,4)
            a__add#_A(x1,x2) = ((0,0),(1,0)) x2 + (7,1)
            mark_A(x1) = ((1,0),(1,1)) x1 + (0,3)
            |0|_A() = (1,2)
            from_A(x1) = ((1,0),(0,0)) x1 + (9,1)
            a__from#_A(x1) = ((0,0),(1,0)) x1 + (7,2)
            fst_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (6,2)
            a__fst_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (6,3)
            nil_A() = (2,1)
            a__from_A(x1) = ((1,0),(0,0)) x1 + (9,4)
            a__add_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (4,5)
            a__len_A(x1) = ((1,0),(1,0)) x1 + (1,8)
    
      precedence:
      
        a__fst# = s = cons = mark# = len = add = a__add# = mark = |0| = from = a__from# = fst = a__fst = nil = a__from = a__add = a__len
      
      partial status:
    
        pi(a__fst#) = []
        pi(s) = []
        pi(cons) = []
        pi(mark#) = []
        pi(len) = []
        pi(add) = []
        pi(a__add#) = []
        pi(mark) = []
        pi(|0|) = []
        pi(from) = []
        pi(a__from#) = []
        pi(fst) = []
        pi(a__fst) = []
        pi(nil) = []
        pi(a__from) = []
        pi(a__add) = []
        pi(a__len) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__fst#_A(x1,x2) = (4,7)
            s_A(x1) = (3,0)
            cons_A(x1,x2) = (1,0)
            mark#_A(x1) = (4,7)
            len_A(x1) = (3,0)
            add_A(x1,x2) = (1,1)
            a__add#_A(x1,x2) = (2,8)
            mark_A(x1) = (3,6)
            |0|_A() = (1,0)
            from_A(x1) = (0,2)
            a__from#_A(x1) = (4,7)
            fst_A(x1,x2) = ((0,0),(1,0)) x2
            a__fst_A(x1,x2) = (4,5)
            nil_A() = (1,0)
            a__from_A(x1) = (2,1)
            a__add_A(x1,x2) = ((0,0),(1,0)) x1 + (3,2)
            a__len_A(x1) = (3,1)
    
      precedence:
      
        mark = a__add > add > a__fst = nil > len = |0| = a__len > from = a__from > cons = fst > a__fst# = mark# = a__from# > a__add# > s
      
      partial status:
    
        pi(a__fst#) = []
        pi(s) = []
        pi(cons) = []
        pi(mark#) = []
        pi(len) = []
        pi(add) = []
        pi(a__add#) = []
        pi(mark) = []
        pi(|0|) = []
        pi(from) = []
        pi(a__from#) = []
        pi(fst) = []
        pi(a__fst) = []
        pi(nil) = []
        pi(a__from) = []
        pi(a__add) = []
        pi(a__len) = []
    

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

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

p1: a__fst#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(len(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X2)
p5: mark#(add(X1,X2)) -> mark#(X1)
p6: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p7: mark#(from(X)) -> mark#(X)
p8: mark#(from(X)) -> a__from#(mark(X))
p9: a__from#(X) -> mark#(X)
p10: mark#(fst(X1,X2)) -> mark#(X2)
p11: mark#(fst(X1,X2)) -> mark#(X1)
p12: mark#(fst(X1,X2)) -> a__fst#(mark(X1),mark(X2))

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

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

p1: a__fst#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(fst(X1,X2)) -> a__fst#(mark(X1),mark(X2))
p3: mark#(fst(X1,X2)) -> mark#(X1)
p4: mark#(fst(X1,X2)) -> mark#(X2)
p5: mark#(from(X)) -> a__from#(mark(X))
p6: a__from#(X) -> mark#(X)
p7: mark#(from(X)) -> mark#(X)
p8: mark#(add(X1,X2)) -> mark#(X1)
p9: mark#(add(X1,X2)) -> mark#(X2)
p10: mark#(len(X)) -> mark#(X)
p11: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__fst(|0|(),Z) -> nil()
r2: a__fst(s(X),cons(Y,Z)) -> cons(mark(Y),fst(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: a__add(|0|(),X) -> mark(X)
r5: a__add(s(X),Y) -> s(add(X,Y))
r6: a__len(nil()) -> |0|()
r7: a__len(cons(X,Z)) -> s(len(Z))
r8: mark(fst(X1,X2)) -> a__fst(mark(X1),mark(X2))
r9: mark(from(X)) -> a__from(mark(X))
r10: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r11: mark(len(X)) -> a__len(mark(X))
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(X)
r14: mark(nil()) -> nil()
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: a__fst(X1,X2) -> fst(X1,X2)
r17: a__from(X) -> from(X)
r18: a__add(X1,X2) -> add(X1,X2)
r19: a__len(X) -> len(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, r19

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__fst#_A(x1,x2) = x1 + x2 + (9,5)
            s_A(x1) = (1,0)
            cons_A(x1,x2) = x1 + (3,12)
            mark#_A(x1) = x1 + (2,17)
            fst_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (7,0)
            mark_A(x1) = ((1,0),(1,1)) x1 + (0,6)
            from_A(x1) = ((1,0),(1,1)) x1 + (4,16)
            a__from#_A(x1) = x1 + (3,18)
            add_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (8,7)
            len_A(x1) = x1 + (4,7)
            a__fst_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (7,1)
            |0|_A() = (1,9)
            nil_A() = (2,1)
            a__from_A(x1) = ((1,0),(1,1)) x1 + (4,19)
            a__add_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (8,8)
            a__len_A(x1) = x1 + (4,8)
    
      precedence:
      
        nil > a__from > from = a__add > s > a__len > len > a__fst# = mark# = a__from# > cons = mark = a__fst > |0| > fst = add
      
      partial status:
    
        pi(a__fst#) = [1, 2]
        pi(s) = []
        pi(cons) = [1]
        pi(mark#) = [1]
        pi(fst) = [1, 2]
        pi(mark) = [1]
        pi(from) = [1]
        pi(a__from#) = [1]
        pi(add) = [1, 2]
        pi(len) = [1]
        pi(a__fst) = [2]
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = [1]
        pi(a__add) = [1, 2]
        pi(a__len) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__fst#_A(x1,x2) = ((1,0),(1,1)) x1 + (1,1)
            s_A(x1) = (3,2)
            cons_A(x1,x2) = ((1,0),(1,0)) x1 + (2,8)
            mark#_A(x1) = ((1,0),(1,1)) x1 + (1,7)
            fst_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (25,1)
            mark_A(x1) = ((1,0),(1,1)) x1 + (12,9)
            from_A(x1) = ((1,0),(1,1)) x1 + (14,21)
            a__from#_A(x1) = ((1,0),(1,0)) x1 + (2,8)
            add_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (11,1)
            len_A(x1) = x1 + (13,6)
            a__fst_A(x1,x2) = ((1,0),(1,0)) x2 + (13,19)
            |0|_A() = (1,2)
            nil_A() = (0,1)
            a__from_A(x1) = ((1,0),(1,1)) x1 + (14,22)
            a__add_A(x1,x2) = ((0,0),(1,0)) x2 + (23,9)
            a__len_A(x1) = x1 + (14,7)
    
      precedence:
      
        a__fst > nil > s = |0| = a__len > from > fst > a__fst# = cons = mark# = mark = a__from# = len = a__add > a__from > add
      
      partial status:
    
        pi(a__fst#) = []
        pi(s) = []
        pi(cons) = []
        pi(mark#) = [1]
        pi(fst) = [1, 2]
        pi(mark) = [1]
        pi(from) = []
        pi(a__from#) = []
        pi(add) = []
        pi(len) = []
        pi(a__fst) = []
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = []
        pi(a__add) = []
        pi(a__len) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11

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