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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      a__fst#_A(x1,x2) = ((1,0,0),(1,1,0),(0,0,1)) x2
      s_A(x1) = (1,2,2)
      cons_A(x1,x2) = ((1,0,0),(0,0,0),(1,0,0)) x1 + ((0,0,0),(0,0,0),(1,0,0)) x2 + (2,1,5)
      mark#_A(x1) = x1 + (1,3,5)
      len_A(x1) = ((1,0,0),(0,0,0),(1,0,0)) x1 + (0,1,0)
      add_A(x1,x2) = x1 + x2 + (0,1,1)
      a__add#_A(x1,x2) = x2 + (1,3,5)
      mark_A(x1) = x1 + (0,3,4)
      |0|_A() = (1,0,5)
      from_A(x1) = x1 + (3,1,1)
      a__from#_A(x1) = ((1,0,0),(1,0,0),(0,0,0)) x1 + (2,2,5)
      fst_A(x1,x2) = ((1,0,0),(0,0,0),(0,1,0)) x1 + x2 + (0,1,1)
      a__fst_A(x1,x2) = x1 + x2 + (0,2,2)
      nil_A() = (1,0,3)
      a__from_A(x1) = x1 + (3,2,3)
      a__add_A(x1,x2) = x1 + x2 + (0,2,2)
      a__len_A(x1) = ((1,0,0),(0,0,0),(1,0,0)) x1 + (0,1,5)

The next rules are strictly ordered:

  p1, p2, p8, p9, p10, p13

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(len(X)) -> mark#(X)
p2: mark#(add(X1,X2)) -> mark#(X2)
p3: mark#(add(X1,X2)) -> mark#(X1)
p4: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p5: a__add#(|0|(),X) -> mark#(X)
p6: mark#(fst(X1,X2)) -> mark#(X2)
p7: mark#(fst(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}


-- Reduction pair.

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

p1: mark#(len(X)) -> mark#(X)
p2: mark#(fst(X1,X2)) -> mark#(X1)
p3: mark#(fst(X1,X2)) -> mark#(X2)
p4: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p5: a__add#(|0|(),X) -> mark#(X)
p6: mark#(add(X1,X2)) -> mark#(X1)
p7: mark#(add(X1,X2)) -> 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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      mark#_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (0,0,8)
      len_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1 + (2,1,1)
      fst_A(x1,x2) = x1 + x2 + (2,1,1)
      add_A(x1,x2) = ((1,0,0),(1,0,0),(1,1,0)) x1 + ((1,0,0),(1,0,0),(1,1,0)) x2 + (2,4,1)
      a__add#_A(x1,x2) = ((0,0,0),(1,0,0),(0,1,0)) x1 + ((1,0,0),(1,0,0),(0,0,0)) x2 + (1,5,0)
      mark_A(x1) = ((1,0,0),(1,1,0),(1,0,1)) x1 + (0,0,11)
      |0|_A() = (1,7,9)
      a__fst_A(x1,x2) = x1 + x2 + (2,2,2)
      nil_A() = (1,3,3)
      s_A(x1) = (1,4,6)
      cons_A(x1,x2) = (0,0,3)
      a__from_A(x1) = (0,1,4)
      from_A(x1) = (0,1,1)
      a__add_A(x1,x2) = ((1,0,0),(1,0,0),(1,1,0)) x1 + ((1,0,0),(1,0,0),(1,1,0)) x2 + (2,4,2)
      a__len_A(x1) = ((1,0,0),(0,1,0),(0,1,1)) x1 + (2,3,2)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7

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