YES

We show the termination of the TRS R:

  active(from(X)) -> mark(cons(X,from(s(X))))
  active(length(nil())) -> mark(|0|())
  active(length(cons(X,Y))) -> mark(s(length1(Y)))
  active(length1(X)) -> mark(length(X))
  mark(from(X)) -> active(from(mark(X)))
  mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
  mark(s(X)) -> active(s(mark(X)))
  mark(length(X)) -> active(length(X))
  mark(nil()) -> active(nil())
  mark(|0|()) -> active(|0|())
  mark(length1(X)) -> active(length1(X))
  from(mark(X)) -> from(X)
  from(active(X)) -> from(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)
  s(mark(X)) -> s(X)
  s(active(X)) -> s(X)
  length(mark(X)) -> length(X)
  length(active(X)) -> length(X)
  length1(mark(X)) -> length1(X)
  length1(active(X)) -> length1(X)

-- SCC decomposition.

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

p1: active#(from(X)) -> mark#(cons(X,from(s(X))))
p2: active#(from(X)) -> cons#(X,from(s(X)))
p3: active#(from(X)) -> from#(s(X))
p4: active#(from(X)) -> s#(X)
p5: active#(length(nil())) -> mark#(|0|())
p6: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p7: active#(length(cons(X,Y))) -> s#(length1(Y))
p8: active#(length(cons(X,Y))) -> length1#(Y)
p9: active#(length1(X)) -> mark#(length(X))
p10: active#(length1(X)) -> length#(X)
p11: mark#(from(X)) -> active#(from(mark(X)))
p12: mark#(from(X)) -> from#(mark(X))
p13: mark#(from(X)) -> mark#(X)
p14: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p15: mark#(cons(X1,X2)) -> cons#(mark(X1),X2)
p16: mark#(cons(X1,X2)) -> mark#(X1)
p17: mark#(s(X)) -> active#(s(mark(X)))
p18: mark#(s(X)) -> s#(mark(X))
p19: mark#(s(X)) -> mark#(X)
p20: mark#(length(X)) -> active#(length(X))
p21: mark#(nil()) -> active#(nil())
p22: mark#(|0|()) -> active#(|0|())
p23: mark#(length1(X)) -> active#(length1(X))
p24: from#(mark(X)) -> from#(X)
p25: from#(active(X)) -> from#(X)
p26: cons#(mark(X1),X2) -> cons#(X1,X2)
p27: cons#(X1,mark(X2)) -> cons#(X1,X2)
p28: cons#(active(X1),X2) -> cons#(X1,X2)
p29: cons#(X1,active(X2)) -> cons#(X1,X2)
p30: s#(mark(X)) -> s#(X)
p31: s#(active(X)) -> s#(X)
p32: length#(mark(X)) -> length#(X)
p33: length#(active(X)) -> length#(X)
p34: length1#(mark(X)) -> length1#(X)
p35: length1#(active(X)) -> length1#(X)

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(X)

The estimated dependency graph contains the following SCCs:

  {p1, p6, p9, p11, p13, p14, p16, p17, p19, p20, p23}
  {p26, p27, p28, p29}
  {p24, p25}
  {p30, p31}
  {p34, p35}
  {p32, p33}


-- Reduction pair.

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

p1: active#(from(X)) -> mark#(cons(X,from(s(X))))
p2: mark#(length1(X)) -> active#(length1(X))
p3: active#(length1(X)) -> mark#(length(X))
p4: mark#(length(X)) -> active#(length(X))
p5: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p6: mark#(s(X)) -> mark#(X)
p7: mark#(s(X)) -> active#(s(mark(X)))
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p10: mark#(from(X)) -> mark#(X)
p11: mark#(from(X)) -> active#(from(mark(X)))

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{7, x1 - 9}
          from_A(x1) = max{70, x1 + 47}
          mark#_A(x1) = max{7, x1 - 6}
          cons_A(x1,x2) = max{60, x1 + 44}
          s_A(x1) = x1
          length1_A(x1) = 9
          length_A(x1) = 7
          mark_A(x1) = max{9, x1}
          active_A(x1) = max{9, x1}
          nil_A = 10
          |0|_A = 8
    
      precedence:
      
        active# = from = mark# = cons = s = length1 = length = mark = active = nil = |0|
      
      partial status:
    
        pi(active#) = []
        pi(from) = []
        pi(mark#) = []
        pi(cons) = []
        pi(s) = []
        pi(length1) = []
        pi(length) = []
        pi(mark) = []
        pi(active) = []
        pi(nil) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 14
          from_A(x1) = 59
          mark#_A(x1) = 14
          cons_A(x1,x2) = 5
          s_A(x1) = 49
          length1_A(x1) = 0
          length_A(x1) = 95
          mark_A(x1) = 69
          active_A(x1) = 69
          nil_A = 71
          |0|_A = 70
    
      precedence:
      
        cons = length = nil > |0| > from = mark = active > s > active# = mark# > length1
      
      partial status:
    
        pi(active#) = []
        pi(from) = []
        pi(mark#) = []
        pi(cons) = []
        pi(s) = []
        pi(length1) = []
        pi(length) = []
        pi(mark) = []
        pi(active) = []
        pi(nil) = []
        pi(|0|) = []
    

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: active#(from(X)) -> mark#(cons(X,from(s(X))))
p2: mark#(length1(X)) -> active#(length1(X))
p3: active#(length1(X)) -> mark#(length(X))
p4: mark#(length(X)) -> active#(length(X))
p5: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p6: mark#(s(X)) -> mark#(X)
p7: mark#(s(X)) -> active#(s(mark(X)))
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p9: mark#(from(X)) -> mark#(X)
p10: mark#(from(X)) -> active#(from(mark(X)))

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: active#(from(X)) -> mark#(cons(X,from(s(X))))
p2: mark#(from(X)) -> active#(from(mark(X)))
p3: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p4: mark#(from(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: active#(length1(X)) -> mark#(length(X))
p7: mark#(s(X)) -> active#(s(mark(X)))
p8: mark#(s(X)) -> mark#(X)
p9: mark#(length(X)) -> active#(length(X))
p10: mark#(length1(X)) -> active#(length1(X))

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{12, x1 - 20}
          from_A(x1) = x1 + 43
          mark#_A(x1) = max{11, x1 - 12}
          cons_A(x1,x2) = max{25, x2 - 54}
          s_A(x1) = max{24, x1}
          mark_A(x1) = x1
          length_A(x1) = 24
          length1_A(x1) = 24
          active_A(x1) = max{20, x1}
          nil_A = 21
          |0|_A = 24
    
      precedence:
      
        length1 > length > mark = nil > |0| > active# = from = mark# = cons = s = active
      
      partial status:
    
        pi(active#) = []
        pi(from) = []
        pi(mark#) = []
        pi(cons) = []
        pi(s) = []
        pi(mark) = [1]
        pi(length) = []
        pi(length1) = []
        pi(active) = [1]
        pi(nil) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 267
          from_A(x1) = 12
          mark#_A(x1) = 267
          cons_A(x1,x2) = 1
          s_A(x1) = 1
          mark_A(x1) = 0
          length_A(x1) = 143
          length1_A(x1) = 19
          active_A(x1) = max{0, x1 - 13}
          nil_A = 53
          |0|_A = 198
    
      precedence:
      
        cons > s > |0| > mark > active# = mark# = length = length1 = active > from = nil
      
      partial status:
    
        pi(active#) = []
        pi(from) = []
        pi(mark#) = []
        pi(cons) = []
        pi(s) = []
        pi(mark) = []
        pi(length) = []
        pi(length1) = []
        pi(active) = []
        pi(nil) = []
        pi(|0|) = []
    

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: active#(from(X)) -> mark#(cons(X,from(s(X))))
p2: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p3: mark#(from(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p5: active#(length1(X)) -> mark#(length(X))
p6: mark#(s(X)) -> active#(s(mark(X)))
p7: mark#(s(X)) -> mark#(X)
p8: mark#(length(X)) -> active#(length(X))
p9: mark#(length1(X)) -> active#(length1(X))

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

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

p1: active#(from(X)) -> mark#(cons(X,from(s(X))))
p2: mark#(length1(X)) -> active#(length1(X))
p3: active#(length1(X)) -> mark#(length(X))
p4: mark#(length(X)) -> active#(length(X))
p5: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p6: mark#(s(X)) -> mark#(X)
p7: mark#(s(X)) -> active#(s(mark(X)))
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p9: mark#(from(X)) -> mark#(X)

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{128, x1 - 23}
          from_A(x1) = 226
          mark#_A(x1) = 128
          cons_A(x1,x2) = 151
          s_A(x1) = 109
          length1_A(x1) = 151
          length_A(x1) = 139
          mark_A(x1) = max{176, x1 + 68}
          active_A(x1) = max{177, x1 + 57}
          nil_A = 120
          |0|_A = 127
    
      precedence:
      
        cons = s = length1 = mark = active = nil > active# = mark# = length > from = |0|
      
      partial status:
    
        pi(active#) = []
        pi(from) = []
        pi(mark#) = []
        pi(cons) = []
        pi(s) = []
        pi(length1) = []
        pi(length) = []
        pi(mark) = [1]
        pi(active) = []
        pi(nil) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 292
          from_A(x1) = 278
          mark#_A(x1) = 292
          cons_A(x1,x2) = 382
          s_A(x1) = 291
          length1_A(x1) = 206
          length_A(x1) = 232
          mark_A(x1) = 193
          active_A(x1) = 193
          nil_A = 192
          |0|_A = 436
    
      precedence:
      
        cons > active# = from = mark# = s = length1 = length = mark = active = nil = |0|
      
      partial status:
    
        pi(active#) = []
        pi(from) = []
        pi(mark#) = []
        pi(cons) = []
        pi(s) = []
        pi(length1) = []
        pi(length) = []
        pi(mark) = []
        pi(active) = []
        pi(nil) = []
        pi(|0|) = []
    

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#(length1(X)) -> active#(length1(X))
p2: active#(length1(X)) -> mark#(length(X))
p3: mark#(length(X)) -> active#(length(X))
p4: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p5: mark#(s(X)) -> mark#(X)
p6: mark#(s(X)) -> active#(s(mark(X)))
p7: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p8: mark#(from(X)) -> mark#(X)

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

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

p1: mark#(length1(X)) -> active#(length1(X))
p2: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p3: mark#(from(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p5: active#(length1(X)) -> mark#(length(X))
p6: mark#(s(X)) -> active#(s(mark(X)))
p7: mark#(s(X)) -> mark#(X)
p8: mark#(length(X)) -> active#(length(X))

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{32, x1 - 2}
          length1_A(x1) = 33
          active#_A(x1) = 32
          length_A(x1) = 23
          cons_A(x1,x2) = 34
          s_A(x1) = max{31, x1}
          from_A(x1) = max{41, x1 + 6}
          mark_A(x1) = max{44, x1 + 9}
          active_A(x1) = max{44, x1}
          nil_A = 43
          |0|_A = 35
    
      precedence:
      
        mark# = length1 = active# = length = cons = s = from = mark = active = nil = |0|
      
      partial status:
    
        pi(mark#) = []
        pi(length1) = []
        pi(active#) = []
        pi(length) = []
        pi(cons) = []
        pi(s) = []
        pi(from) = []
        pi(mark) = []
        pi(active) = []
        pi(nil) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 12
          length1_A(x1) = 16
          active#_A(x1) = 12
          length_A(x1) = 17
          cons_A(x1,x2) = 0
          s_A(x1) = 0
          from_A(x1) = 13
          mark_A(x1) = 13
          active_A(x1) = 13
          nil_A = 14
          |0|_A = 18
    
      precedence:
      
        from = |0| > mark# = length1 = active# = length = cons = s = mark = active = nil
      
      partial status:
    
        pi(mark#) = []
        pi(length1) = []
        pi(active#) = []
        pi(length) = []
        pi(cons) = []
        pi(s) = []
        pi(from) = []
        pi(mark) = []
        pi(active) = []
        pi(nil) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(length1(X)) -> active#(length1(X))
p2: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p3: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p4: active#(length1(X)) -> mark#(length(X))
p5: mark#(s(X)) -> active#(s(mark(X)))
p6: mark#(s(X)) -> mark#(X)
p7: mark#(length(X)) -> active#(length(X))

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: mark#(length1(X)) -> active#(length1(X))
p2: active#(length1(X)) -> mark#(length(X))
p3: mark#(length(X)) -> active#(length(X))
p4: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p5: mark#(s(X)) -> mark#(X)
p6: mark#(s(X)) -> active#(s(mark(X)))
p7: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{172, x1 + 2}
          length1_A(x1) = x1 + 186
          active#_A(x1) = max{169, x1 - 14}
          length_A(x1) = x1 + 169
          cons_A(x1,x2) = max{110, x2 + 41}
          s_A(x1) = max{172, x1 + 7}
          mark_A(x1) = max{166, x1 + 7}
          active_A(x1) = max{166, x1}
          from_A(x1) = 117
          nil_A = 176
          |0|_A = 168
    
      precedence:
      
        mark# = length1 = active# = length = cons = s = mark = active = from = nil = |0|
      
      partial status:
    
        pi(mark#) = []
        pi(length1) = []
        pi(active#) = []
        pi(length) = []
        pi(cons) = []
        pi(s) = []
        pi(mark) = []
        pi(active) = []
        pi(from) = []
        pi(nil) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 24
          length1_A(x1) = 46
          active#_A(x1) = 24
          length_A(x1) = 77
          cons_A(x1,x2) = 9
          s_A(x1) = 15
          mark_A(x1) = 15
          active_A(x1) = 15
          from_A(x1) = 15
          nil_A = 1
          |0|_A = 76
    
      precedence:
      
        mark# = active# = length > length1 > mark = active = |0| > cons = s = from = nil
      
      partial status:
    
        pi(mark#) = []
        pi(length1) = []
        pi(active#) = []
        pi(length) = []
        pi(cons) = []
        pi(s) = []
        pi(mark) = []
        pi(active) = []
        pi(from) = []
        pi(nil) = []
        pi(|0|) = []
    

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#(length1(X)) -> active#(length1(X))
p2: active#(length1(X)) -> mark#(length(X))
p3: mark#(length(X)) -> active#(length(X))
p4: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p5: mark#(s(X)) -> active#(s(mark(X)))
p6: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(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#(length1(X)) -> active#(length1(X))
p2: active#(length(cons(X,Y))) -> mark#(s(length1(Y)))
p3: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p4: active#(length1(X)) -> mark#(length(X))
p5: mark#(s(X)) -> active#(s(mark(X)))
p6: mark#(length(X)) -> active#(length(X))

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 13
          length1_A(x1) = x1 + 15
          active#_A(x1) = x1 + 13
          length_A(x1) = 14
          cons_A(x1,x2) = max{2, x2}
          s_A(x1) = 0
          mark_A(x1) = max{6, x1}
          active_A(x1) = max{4, x1}
          from_A(x1) = max{10, x1 + 3}
          nil_A = 10
          |0|_A = 5
    
      precedence:
      
        mark# > from > active# = |0| > length1 > mark = nil > active > length = cons = s
      
      partial status:
    
        pi(mark#) = [1]
        pi(length1) = []
        pi(active#) = [1]
        pi(length) = []
        pi(cons) = []
        pi(s) = []
        pi(mark) = [1]
        pi(active) = [1]
        pi(from) = []
        pi(nil) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{34, x1 - 27}
          length1_A(x1) = 18
          active#_A(x1) = max{33, x1 + 30}
          length_A(x1) = 1
          cons_A(x1,x2) = 4
          s_A(x1) = 32
          mark_A(x1) = 4
          active_A(x1) = max{4, x1 - 5}
          from_A(x1) = 15
          nil_A = 10
          |0|_A = 26
    
      precedence:
      
        mark# = s = from = |0| > cons = nil > length1 = active# = mark = active > length
      
      partial status:
    
        pi(mark#) = []
        pi(length1) = []
        pi(active#) = []
        pi(length) = []
        pi(cons) = []
        pi(s) = []
        pi(mark) = []
        pi(active) = []
        pi(from) = []
        pi(nil) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cons#_A(x1,x2) = max{2, x1 + 1, x2 + 1}
          mark_A(x1) = max{1, x1}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        cons# = mark = active
      
      partial status:
    
        pi(cons#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cons#_A(x1,x2) = max{x1 + 1, x2 - 1}
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        cons# = mark = active
      
      partial status:
    
        pi(cons#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          from#_A(x1) = x1 + 2
          mark_A(x1) = x1
          active_A(x1) = x1 + 2
    
      precedence:
      
        mark > active > from#
      
      partial status:
    
        pi(from#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          from#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        from# = mark = active
      
      partial status:
    
        pi(from#) = [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: s#(mark(X)) -> s#(X)
p2: s#(active(X)) -> s#(X)

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          s#_A(x1) = max{6, x1 + 4}
          mark_A(x1) = max{4, x1 + 3}
          active_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        s# = mark = active
      
      partial status:
    
        pi(s#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          s#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        s# = mark = active
      
      partial status:
    
        pi(s#) = [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: length1#(mark(X)) -> length1#(X)
p2: length1#(active(X)) -> length1#(X)

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          length1#_A(x1) = x1 + 2
          mark_A(x1) = x1
          active_A(x1) = x1 + 2
    
      precedence:
      
        mark > active > length1#
      
      partial status:
    
        pi(length1#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          length1#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        length1# = mark = active
      
      partial status:
    
        pi(length1#) = [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: length#(mark(X)) -> length#(X)
p2: length#(active(X)) -> length#(X)

and R consists of:

r1: active(from(X)) -> mark(cons(X,from(s(X))))
r2: active(length(nil())) -> mark(|0|())
r3: active(length(cons(X,Y))) -> mark(s(length1(Y)))
r4: active(length1(X)) -> mark(length(X))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(length(X)) -> active(length(X))
r9: mark(nil()) -> active(nil())
r10: mark(|0|()) -> active(|0|())
r11: mark(length1(X)) -> active(length1(X))
r12: from(mark(X)) -> from(X)
r13: from(active(X)) -> from(X)
r14: cons(mark(X1),X2) -> cons(X1,X2)
r15: cons(X1,mark(X2)) -> cons(X1,X2)
r16: cons(active(X1),X2) -> cons(X1,X2)
r17: cons(X1,active(X2)) -> cons(X1,X2)
r18: s(mark(X)) -> s(X)
r19: s(active(X)) -> s(X)
r20: length(mark(X)) -> length(X)
r21: length(active(X)) -> length(X)
r22: length1(mark(X)) -> length1(X)
r23: length1(active(X)) -> length1(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          length#_A(x1) = x1 + 2
          mark_A(x1) = x1
          active_A(x1) = x1 + 2
    
      precedence:
      
        mark > active > length#
      
      partial status:
    
        pi(length#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          length#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        length# = mark = active
      
      partial status:
    
        pi(length#) = [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.