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^1
        order: lexicographic order
        interpretations:
          a__fst#_A(x1,x2) = x1 + x2
          s_A(x1) = 3
          cons_A(x1,x2) = x1 + 3
          mark#_A(x1) = x1 + 2
          len_A(x1) = x1
          add_A(x1,x2) = x1 + x2 + 4
          a__add#_A(x1,x2) = x1 + x2 + 1
          mark_A(x1) = x1
          |0|_A() = 3
          from_A(x1) = x1 + 4
          a__from#_A(x1) = x1 + 3
          fst_A(x1,x2) = x1 + x2 + 1
          a__fst_A(x1,x2) = x1 + x2 + 1
          nil_A() = 3
          a__from_A(x1) = x1 + 4
          a__add_A(x1,x2) = x1 + x2 + 4
          a__len_A(x1) = x1
  
    precedence:
    
      a__from# > s = cons = mark = fst = a__fst = nil = a__from = a__len > a__add# > a__fst# = mark# = from > len > add = a__add > |0|
    
    partial status:
  
      pi(a__fst#) = []
      pi(s) = []
      pi(cons) = []
      pi(mark#) = []
      pi(len) = []
      pi(add) = []
      pi(a__add#) = [1]
      pi(mark) = []
      pi(|0|) = []
      pi(from) = []
      pi(a__from#) = []
      pi(fst) = []
      pi(a__fst) = []
      pi(nil) = []
      pi(a__from) = []
      pi(a__add) = [1, 2]
      pi(a__len) = []

The next rules are strictly ordered:

  p6

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: a__add#(|0|(),X) -> mark#(X)
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:

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

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#(fst(X1,X2)) -> a__fst#(mark(X1),mark(X2))
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)) -> mark#(X1)
p8: mark#(add(X1,X2)) -> mark#(X2)
p9: mark#(len(X)) -> mark#(X)
p10: 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:

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


-- Reduction pair.

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

p1: mark#(fst(X1,X2)) -> mark#(X1)
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#(from(X)) -> mark#(X)
p7: mark#(from(X)) -> a__from#(mark(X))
p8: a__from#(X) -> mark#(X)
p9: mark#(fst(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:

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

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(fst(X1,X2)) -> mark#(X1)
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#(from(X)) -> mark#(X)
p7: mark#(from(X)) -> a__from#(mark(X))
p8: mark#(fst(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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p8}


-- Reduction pair.

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

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

  (no rules)

Take the monotone reduction pair:

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

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

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          fst_A(x1,x2) = x1 + x2 + 1
          len_A(x1) = x1 + 1
          add_A(x1,x2) = x1 + x2 + 1
          from_A(x1) = x1 + 1
  
    precedence:
    
      mark# = fst = len = add = from
    
    partial status:
  
      pi(mark#) = [1]
      pi(fst) = [1, 2]
      pi(len) = [1]
      pi(add) = [1, 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#(len(X)) -> mark#(X)
p2: mark#(add(X1,X2)) -> mark#(X2)
p3: mark#(add(X1,X2)) -> mark#(X1)
p4: mark#(from(X)) -> mark#(X)
p5: mark#(fst(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 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#(len(X)) -> mark#(X)
p2: mark#(fst(X1,X2)) -> mark#(X2)
p3: mark#(from(X)) -> mark#(X)
p4: mark#(add(X1,X2)) -> mark#(X1)
p5: 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

  (no rules)

Take the reduction pair:

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

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#(fst(X1,X2)) -> mark#(X2)
p2: mark#(from(X)) -> mark#(X)
p3: mark#(add(X1,X2)) -> mark#(X1)
p4: 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 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#(fst(X1,X2)) -> mark#(X2)
p2: mark#(add(X1,X2)) -> mark#(X2)
p3: mark#(add(X1,X2)) -> mark#(X1)
p4: 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^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          fst_A(x1,x2) = x1 + x2 + 1
          add_A(x1,x2) = x1 + x2 + 1
          from_A(x1) = x1 + 1
  
    precedence:
    
      mark# = fst = add = from
    
    partial status:
  
      pi(mark#) = []
      pi(fst) = [2]
      pi(add) = [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#(add(X1,X2)) -> mark#(X2)
p2: mark#(add(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 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#(add(X1,X2)) -> mark#(X2)
p2: mark#(from(X)) -> mark#(X)
p3: mark#(add(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^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          add_A(x1,x2) = x1 + x2 + 1
          from_A(x1) = x1 + 1
  
    precedence:
    
      mark# = add = from
    
    partial status:
  
      pi(mark#) = []
      pi(add) = [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)
p2: mark#(add(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}


-- Reduction pair.

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

p1: mark#(from(X)) -> mark#(X)
p2: mark#(add(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^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          from_A(x1) = x1 + 1
          add_A(x1,x2) = x1 + x2 + 1
  
    precedence:
    
      mark# > from = add
    
    partial status:
  
      pi(mark#) = []
      pi(from) = [1]
      pi(add) = [2]

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#(add(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}


-- Reduction pair.

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

p1: mark#(add(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^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          add_A(x1,x2) = x1 + x2 + 1
  
    precedence:
    
      mark# = add
    
    partial status:
  
      pi(mark#) = []
      pi(add) = [2]

The next rules are strictly ordered:

  p1

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