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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__fst#_A(x1,x2) = ((0,1),(0,0)) x1 + x2 + (16,0)
          s_A(x1) = (11,1)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + (9,3)
          mark#_A(x1) = ((1,0),(0,0)) x1 + (8,3)
          len_A(x1) = ((1,0),(0,0)) x1 + (10,0)
          add_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (0,2)
          a__add#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (1,3)
          mark_A(x1) = ((1,0),(0,0)) x1 + (0,3)
          |0|_A() = (9,2)
          from_A(x1) = ((1,0),(0,0)) x1 + (10,0)
          a__from#_A(x1) = ((1,0),(0,0)) x1 + (11,3)
          fst_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (12,3)
          a__fst_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (12,3)
          nil_A() = (0,1)
          a__from_A(x1) = ((1,0),(0,0)) x1 + (10,3)
          a__add_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (0,3)
          a__len_A(x1) = ((1,0),(0,0)) x1 + (10,3)
  
    precedence:
    
      mark# = a__add# = a__from# > s = add = mark = fst = a__fst = a__add = a__len > cons = from = a__from > |0| > a__fst# > len = nil
    
    partial status:
  
      pi(a__fst#) = [2]
      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:

  p13

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: 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)

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:

  {p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

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

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

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,1),(1,1)) x1
          cons_A(x1,x2) = ((0,1),(1,0)) x1 + (2,7)
          fst_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(1,1)) x2 + (4,6)
          from_A(x1) = ((1,1),(1,1)) x1 + (9,5)
          a__from#_A(x1) = ((1,1),(1,1)) x1 + (1,11)
          mark_A(x1) = ((1,1),(1,1)) x1 + (0,3)
          add_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(1,1)) x2 + (0,8)
          a__add#_A(x1,x2) = ((1,1),(1,1)) x2 + (4,0)
          |0|_A() = (2,2)
          len_A(x1) = x1 + (1,8)
          a__fst_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(1,1)) x2 + (4,6)
          nil_A() = (1,1)
          s_A(x1) = (3,6)
          a__from_A(x1) = ((1,1),(1,1)) x1 + (10,14)
          a__add_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(1,1)) x2 + (1,8)
          a__len_A(x1) = x1 + (9,8)
  
    precedence:
    
      mark = add = s = a__add = a__len > a__fst > mark# = fst = a__add# = len > a__from# > from = a__from > cons > |0| > nil
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = []
      pi(fst) = [2]
      pi(from) = []
      pi(a__from#) = []
      pi(mark) = []
      pi(add) = []
      pi(a__add#) = []
      pi(|0|) = []
      pi(len) = [1]
      pi(a__fst) = [2]
      pi(nil) = []
      pi(s) = []
      pi(a__from) = [1]
      pi(a__add) = []
      pi(a__len) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

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

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, p5, p6, p7, p8, p9, p10}


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(len(X)) -> mark#(X)
p3: mark#(add(X1,X2)) -> mark#(X2)
p4: mark#(add(X1,X2)) -> mark#(X1)
p5: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p6: a__add#(|0|(),X) -> mark#(X)
p7: mark#(from(X)) -> mark#(X)
p8: mark#(fst(X1,X2)) -> mark#(X2)
p9: 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 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:

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

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(len(X)) -> mark#(X)
p3: mark#(add(X1,X2)) -> mark#(X2)
p4: mark#(add(X1,X2)) -> mark#(X1)
p5: a__add#(|0|(),X) -> mark#(X)
p6: mark#(from(X)) -> mark#(X)
p7: mark#(fst(X1,X2)) -> mark#(X2)
p8: 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, p6, p7, p8}


-- Reduction pair.

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

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

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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,0),(0,0)) x1 + (2,2)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,1),(1,1)) x2 + (1,1)
          fst_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,0),(0,0)) x2 + (3,2)
          from_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          add_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,2)
          len_A(x1) = ((1,0),(0,0)) x1 + (3,2)
  
    precedence:
    
      cons > from > fst > mark# = add > len
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = []
      pi(fst) = []
      pi(from) = []
      pi(add) = []
      pi(len) = []

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(fst(X1,X2)) -> mark#(X1)
p3: mark#(fst(X1,X2)) -> mark#(X2)
p4: mark#(from(X)) -> mark#(X)
p5: mark#(add(X1,X2)) -> mark#(X2)
p6: mark#(len(X)) -> mark#(X)

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}


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(len(X)) -> mark#(X)
p3: mark#(add(X1,X2)) -> mark#(X2)
p4: mark#(from(X)) -> mark#(X)
p5: mark#(fst(X1,X2)) -> mark#(X2)
p6: 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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,1),(0,0)) x1 + (2,2)
          cons_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (3,1)
          len_A(x1) = ((1,1),(0,0)) x1 + (3,2)
          add_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,0)) x2 + (1,1)
          from_A(x1) = ((0,1),(1,0)) x1 + (1,1)
          fst_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(1,0)) x2 + (3,1)
  
    precedence:
    
      mark# = cons > len > add = from = fst
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]
      pi(len) = []
      pi(add) = []
      pi(from) = []
      pi(fst) = []

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(len(X)) -> mark#(X)
p3: mark#(add(X1,X2)) -> mark#(X2)
p4: mark#(from(X)) -> mark#(X)
p5: 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}


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(fst(X1,X2)) -> mark#(X1)
p3: mark#(from(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X2)
p5: mark#(len(X)) -> mark#(X)

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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,1),(0,0)) x1 + (2,2)
          cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (3,2)
          fst_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,1)
          from_A(x1) = ((0,1),(1,0)) x1 + (1,1)
          add_A(x1,x2) = ((1,1),(1,1)) x1 + x2 + (1,1)
          len_A(x1) = ((0,0),(1,1)) x1 + (1,1)
  
    precedence:
    
      cons > from = len > mark# > add > fst
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]
      pi(fst) = []
      pi(from) = []
      pi(add) = [2]
      pi(len) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(fst(X1,X2)) -> mark#(X1)
p3: mark#(from(X)) -> mark#(X)
p4: mark#(len(X)) -> mark#(X)

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}


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(len(X)) -> mark#(X)
p3: mark#(from(X)) -> mark#(X)
p4: 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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((0,1),(0,1)) x1 + (2,2)
          cons_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (0,1)
          len_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          from_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          fst_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
  
    precedence:
    
      len = from > mark# > cons = fst
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]
      pi(len) = []
      pi(from) = []
      pi(fst) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(from(X)) -> mark#(X)
p3: 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}


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(fst(X1,X2)) -> mark#(X1)
p3: mark#(from(X)) -> mark#(X)

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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((0,1),(0,0)) x1
          cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
          fst_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
          from_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      mark# = cons = fst > from
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]
      pi(fst) = [2]
      pi(from) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(from(X)) -> mark#(X)

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}


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(from(X)) -> mark#(X)

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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((0,1),(0,1)) x1 + (2,2)
          cons_A(x1,x2) = x1 + ((1,1),(0,1)) x2 + (1,1)
          from_A(x1) = x1 + (1,1)
  
    precedence:
    
      mark# = cons = from
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]
      pi(from) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(from(X)) -> mark#(X)

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}


-- Reduction pair.

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

p1: mark#(from(X)) -> mark#(X)

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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          mark#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          from_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      mark# = from
    
    partial status:
  
      pi(mark#) = []
      pi(from) = []

The next rules are strictly ordered:

  p1

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