YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: active#(fst(|0|(),Z)) -> mark#(nil())
p2: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p3: active#(fst(s(X),cons(Y,Z))) -> cons#(Y,fst(X,Z))
p4: active#(fst(s(X),cons(Y,Z))) -> fst#(X,Z)
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: active#(from(X)) -> cons#(X,from(s(X)))
p7: active#(from(X)) -> from#(s(X))
p8: active#(from(X)) -> s#(X)
p9: active#(add(|0|(),X)) -> mark#(X)
p10: active#(add(s(X),Y)) -> mark#(s(add(X,Y)))
p11: active#(add(s(X),Y)) -> s#(add(X,Y))
p12: active#(add(s(X),Y)) -> add#(X,Y)
p13: active#(len(nil())) -> mark#(|0|())
p14: active#(len(cons(X,Z))) -> mark#(s(len(Z)))
p15: active#(len(cons(X,Z))) -> s#(len(Z))
p16: active#(len(cons(X,Z))) -> len#(Z)
p17: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))
p18: mark#(fst(X1,X2)) -> fst#(mark(X1),mark(X2))
p19: mark#(fst(X1,X2)) -> mark#(X1)
p20: mark#(fst(X1,X2)) -> mark#(X2)
p21: mark#(|0|()) -> active#(|0|())
p22: mark#(nil()) -> active#(nil())
p23: mark#(s(X)) -> active#(s(X))
p24: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p25: mark#(cons(X1,X2)) -> cons#(mark(X1),X2)
p26: mark#(cons(X1,X2)) -> mark#(X1)
p27: mark#(from(X)) -> active#(from(mark(X)))
p28: mark#(from(X)) -> from#(mark(X))
p29: mark#(from(X)) -> mark#(X)
p30: mark#(add(X1,X2)) -> active#(add(mark(X1),mark(X2)))
p31: mark#(add(X1,X2)) -> add#(mark(X1),mark(X2))
p32: mark#(add(X1,X2)) -> mark#(X1)
p33: mark#(add(X1,X2)) -> mark#(X2)
p34: mark#(len(X)) -> active#(len(mark(X)))
p35: mark#(len(X)) -> len#(mark(X))
p36: mark#(len(X)) -> mark#(X)
p37: fst#(mark(X1),X2) -> fst#(X1,X2)
p38: fst#(X1,mark(X2)) -> fst#(X1,X2)
p39: fst#(active(X1),X2) -> fst#(X1,X2)
p40: fst#(X1,active(X2)) -> fst#(X1,X2)
p41: s#(mark(X)) -> s#(X)
p42: s#(active(X)) -> s#(X)
p43: cons#(mark(X1),X2) -> cons#(X1,X2)
p44: cons#(X1,mark(X2)) -> cons#(X1,X2)
p45: cons#(active(X1),X2) -> cons#(X1,X2)
p46: cons#(X1,active(X2)) -> cons#(X1,X2)
p47: from#(mark(X)) -> from#(X)
p48: from#(active(X)) -> from#(X)
p49: add#(mark(X1),X2) -> add#(X1,X2)
p50: add#(X1,mark(X2)) -> add#(X1,X2)
p51: add#(active(X1),X2) -> add#(X1,X2)
p52: add#(X1,active(X2)) -> add#(X1,X2)
p53: len#(mark(X)) -> len#(X)
p54: len#(active(X)) -> len#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p2, p5, p9, p10, p14, p17, p19, p20, p23, p24, p26, p27, p29, p30, p32, p33, p34, p36}
  {p43, p44, p45, p46}
  {p37, p38, p39, p40}
  {p47, p48}
  {p41, p42}
  {p49, p50, p51, p52}
  {p53, p54}


-- Reduction pair.

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

p1: mark#(len(X)) -> active#(len(mark(X)))
p2: active#(len(cons(X,Z))) -> mark#(s(len(Z)))
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)) -> active#(add(mark(X1),mark(X2)))
p7: active#(add(s(X),Y)) -> mark#(s(add(X,Y)))
p8: mark#(from(X)) -> mark#(X)
p9: mark#(from(X)) -> active#(from(mark(X)))
p10: active#(add(|0|(),X)) -> mark#(X)
p11: mark#(cons(X1,X2)) -> mark#(X1)
p12: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p13: active#(from(X)) -> mark#(cons(X,from(s(X))))
p14: mark#(s(X)) -> active#(s(X))
p15: active#(fst(s(X),cons(Y,Z))) -> mark#(cons(Y,fst(X,Z)))
p16: mark#(fst(X1,X2)) -> mark#(X2)
p17: mark#(fst(X1,X2)) -> mark#(X1)
p18: mark#(fst(X1,X2)) -> active#(fst(mark(X1),mark(X2)))

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(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, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      len > |0| > nil > fst > s > active# > cons > active > mark# > mark > from > add
    
    argument filter:
  
      pi(mark#) = 1
      pi(len) = [1]
      pi(active#) = 1
      pi(mark) = 1
      pi(cons) = 1
      pi(s) = []
      pi(add) = [1, 2]
      pi(from) = [1]
      pi(|0|) = []
      pi(fst) = [1, 2]
      pi(active) = 1
      pi(nil) = []

The next rules are strictly ordered:

  p2, p3, p4, p5, p7, p8, p10, p13, p15, p16, p17

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The estimated dependency graph contains the following SCCs:

  {p4}


-- Reduction pair.

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

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

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      cons > mark#
    
    argument filter:
  
      pi(mark#) = [1]
      pi(cons) = [1]

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: cons#(mark(X1),X2) -> cons#(X1,X2)
p2: cons#(X1,active(X2)) -> cons#(X1,X2)
p3: cons#(active(X1),X2) -> cons#(X1,X2)
p4: cons#(X1,mark(X2)) -> cons#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      cons# > active > mark
    
    argument filter:
  
      pi(cons#) = 2
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p2, p4

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(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: cons#(mark(X1),X2) -> cons#(X1,X2)
p2: cons#(active(X1),X2) -> cons#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      cons# > active > mark
    
    argument filter:
  
      pi(cons#) = 1
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: fst#(mark(X1),X2) -> fst#(X1,X2)
p2: fst#(X1,active(X2)) -> fst#(X1,X2)
p3: fst#(active(X1),X2) -> fst#(X1,X2)
p4: fst#(X1,mark(X2)) -> fst#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      fst# > active > mark
    
    argument filter:
  
      pi(fst#) = 2
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p2, p4

We remove them from the problem.

-- SCC decomposition.

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

p1: fst#(mark(X1),X2) -> fst#(X1,X2)
p2: fst#(active(X1),X2) -> fst#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(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: fst#(mark(X1),X2) -> fst#(X1,X2)
p2: fst#(active(X1),X2) -> fst#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      fst# > active > mark
    
    argument filter:
  
      pi(fst#) = 1
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: from#(mark(X)) -> from#(X)
p2: from#(active(X)) -> from#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      from# > active > mark
    
    argument filter:
  
      pi(from#) = 1
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

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

-- Reduction pair.

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

p1: s#(mark(X)) -> s#(X)
p2: s#(active(X)) -> s#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      s# > active > mark
    
    argument filter:
  
      pi(s#) = 1
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

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

-- Reduction pair.

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

p1: add#(mark(X1),X2) -> add#(X1,X2)
p2: add#(X1,active(X2)) -> add#(X1,X2)
p3: add#(active(X1),X2) -> add#(X1,X2)
p4: add#(X1,mark(X2)) -> add#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      add# > active > mark
    
    argument filter:
  
      pi(add#) = 2
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p2, p4

We remove them from the problem.

-- SCC decomposition.

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

p1: add#(mark(X1),X2) -> add#(X1,X2)
p2: add#(active(X1),X2) -> add#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(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: add#(mark(X1),X2) -> add#(X1,X2)
p2: add#(active(X1),X2) -> add#(X1,X2)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      add# > active > mark
    
    argument filter:
  
      pi(add#) = 1
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: len#(mark(X)) -> len#(X)
p2: len#(active(X)) -> len#(X)

and R consists of:

r1: active(fst(|0|(),Z)) -> mark(nil())
r2: active(fst(s(X),cons(Y,Z))) -> mark(cons(Y,fst(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: active(add(|0|(),X)) -> mark(X)
r5: active(add(s(X),Y)) -> mark(s(add(X,Y)))
r6: active(len(nil())) -> mark(|0|())
r7: active(len(cons(X,Z))) -> mark(s(len(Z)))
r8: mark(fst(X1,X2)) -> active(fst(mark(X1),mark(X2)))
r9: mark(|0|()) -> active(|0|())
r10: mark(nil()) -> active(nil())
r11: mark(s(X)) -> active(s(X))
r12: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r13: mark(from(X)) -> active(from(mark(X)))
r14: mark(add(X1,X2)) -> active(add(mark(X1),mark(X2)))
r15: mark(len(X)) -> active(len(mark(X)))
r16: fst(mark(X1),X2) -> fst(X1,X2)
r17: fst(X1,mark(X2)) -> fst(X1,X2)
r18: fst(active(X1),X2) -> fst(X1,X2)
r19: fst(X1,active(X2)) -> fst(X1,X2)
r20: s(mark(X)) -> s(X)
r21: s(active(X)) -> s(X)
r22: cons(mark(X1),X2) -> cons(X1,X2)
r23: cons(X1,mark(X2)) -> cons(X1,X2)
r24: cons(active(X1),X2) -> cons(X1,X2)
r25: cons(X1,active(X2)) -> cons(X1,X2)
r26: from(mark(X)) -> from(X)
r27: from(active(X)) -> from(X)
r28: add(mark(X1),X2) -> add(X1,X2)
r29: add(X1,mark(X2)) -> add(X1,X2)
r30: add(active(X1),X2) -> add(X1,X2)
r31: add(X1,active(X2)) -> add(X1,X2)
r32: len(mark(X)) -> len(X)
r33: len(active(X)) -> len(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      len# > active > mark
    
    argument filter:
  
      pi(len#) = 1
      pi(mark) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33

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