YES

We show the termination of the TRS R:

  active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
  active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
  active(p(s(X))) -> mark(X)
  mark(f(X)) -> active(f(mark(X)))
  mark(|0|()) -> active(|0|())
  mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
  mark(s(X)) -> active(s(mark(X)))
  mark(p(X)) -> active(p(mark(X)))
  f(mark(X)) -> f(X)
  f(active(X)) -> f(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)
  p(mark(X)) -> p(X)
  p(active(X)) -> p(X)

-- SCC decomposition.

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

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

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(X)

The estimated dependency graph contains the following SCCs:

  {p1, p5, p8, p9, p11, p13, p15, p16, p18, p19, p21}
  {p22, p23}
  {p24, p25, p26, p27}
  {p28, p29}
  {p30, p31}


-- Reduction pair.

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

p1: active#(f(|0|())) -> mark#(cons(|0|(),f(s(|0|()))))
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(p(X)) -> mark#(X)
p4: mark#(p(X)) -> active#(p(mark(X)))
p5: active#(p(s(X))) -> mark#(X)
p6: mark#(s(X)) -> mark#(X)
p7: mark#(s(X)) -> active#(s(mark(X)))
p8: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p9: mark#(f(X)) -> mark#(X)
p10: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p11: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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

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{34, x1 - 45}
          f_A(x1) = max{106, x1 + 62}
          |0|_A = 12
          mark#_A(x1) = max{34, x1 - 45}
          cons_A(x1,x2) = max{11, x1 + 6}
          s_A(x1) = max{8, x1 + 4}
          p_A(x1) = x1 + 5
          mark_A(x1) = x1
          active_A(x1) = max{5, x1}
    
      precedence:
      
        active# = f = |0| = mark# = cons = s = p = mark = active
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(|0|) = []
        pi(mark#) = []
        pi(cons) = []
        pi(s) = []
        pi(p) = []
        pi(mark) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 84
          f_A(x1) = 76
          |0|_A = 78
          mark#_A(x1) = 84
          cons_A(x1,x2) = 93
          s_A(x1) = 134
          p_A(x1) = 76
          mark_A(x1) = x1 + 18
          active_A(x1) = 94
    
      precedence:
      
        f > mark = active > s > active# = |0| = mark# = cons > p
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(|0|) = []
        pi(mark#) = []
        pi(cons) = []
        pi(s) = []
        pi(p) = []
        pi(mark) = []
        pi(active) = []
    

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#(cons(X1,X2)) -> mark#(X1)
p2: mark#(p(X)) -> mark#(X)
p3: mark#(p(X)) -> active#(p(mark(X)))
p4: active#(p(s(X))) -> mark#(X)
p5: mark#(s(X)) -> mark#(X)
p6: mark#(s(X)) -> active#(s(mark(X)))
p7: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p8: mark#(f(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p10: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p4: mark#(f(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: active#(p(s(X))) -> mark#(X)
p7: mark#(s(X)) -> active#(s(mark(X)))
p8: mark#(s(X)) -> mark#(X)
p9: mark#(p(X)) -> active#(p(mark(X)))
p10: mark#(p(X)) -> mark#(X)

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 47
          cons_A(x1,x2) = 50
          f_A(x1) = 55
          active#_A(x1) = max{44, x1 - 8}
          mark_A(x1) = 78
          s_A(x1) = 45
          |0|_A = 55
          p_A(x1) = 55
          active_A(x1) = max{56, x1 + 23}
    
      precedence:
      
        |0| > mark# = f = active# = s = p > mark = active > cons
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = []
        pi(f) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(|0|) = []
        pi(p) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 246
          cons_A(x1,x2) = 0
          f_A(x1) = 83
          active#_A(x1) = 246
          mark_A(x1) = 11
          s_A(x1) = 79
          |0|_A = 83
          p_A(x1) = 132
          active_A(x1) = 11
    
      precedence:
      
        f > mark = s = active > cons > mark# = active# = |0| = p
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = []
        pi(f) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(|0|) = []
        pi(p) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p4: mark#(f(X)) -> mark#(X)
p5: active#(p(s(X))) -> mark#(X)
p6: mark#(s(X)) -> active#(s(mark(X)))
p7: mark#(s(X)) -> mark#(X)
p8: mark#(p(X)) -> active#(p(mark(X)))
p9: mark#(p(X)) -> mark#(X)

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(p(X)) -> mark#(X)
p3: mark#(p(X)) -> active#(p(mark(X)))
p4: active#(p(s(X))) -> mark#(X)
p5: mark#(s(X)) -> mark#(X)
p6: mark#(s(X)) -> active#(s(mark(X)))
p7: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p8: mark#(f(X)) -> mark#(X)
p9: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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

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
          cons_A(x1,x2) = max{5, x1}
          p_A(x1) = max{10, x1}
          active#_A(x1) = max{10, x1 - 3}
          mark_A(x1) = max{4, x1}
          s_A(x1) = max{10, x1 + 3}
          f_A(x1) = max{10, x1}
          |0|_A = 5
          active_A(x1) = max{5, x1}
    
      precedence:
      
        s > mark > mark# = cons = p = active# = f = |0| = active
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = [1]
        pi(p) = []
        pi(active#) = []
        pi(mark) = [1]
        pi(s) = []
        pi(f) = [1]
        pi(|0|) = []
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 12
          cons_A(x1,x2) = max{6, x1 + 3}
          p_A(x1) = 13
          active#_A(x1) = 12
          mark_A(x1) = max{2, x1}
          s_A(x1) = 32
          f_A(x1) = max{7, x1 - 45}
          |0|_A = 1
          active_A(x1) = max{5, x1}
    
      precedence:
      
        cons = |0| > mark# = active# > p = s = active > f > mark
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = [1]
        pi(p) = []
        pi(active#) = []
        pi(mark) = [1]
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(p(X)) -> mark#(X)
p3: mark#(p(X)) -> active#(p(mark(X)))
p4: active#(p(s(X))) -> mark#(X)
p5: mark#(s(X)) -> active#(s(mark(X)))
p6: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p7: mark#(f(X)) -> mark#(X)
p8: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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#(cons(X1,X2)) -> mark#(X1)
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p4: mark#(f(X)) -> mark#(X)
p5: mark#(s(X)) -> active#(s(mark(X)))
p6: active#(p(s(X))) -> mark#(X)
p7: mark#(p(X)) -> active#(p(mark(X)))
p8: mark#(p(X)) -> mark#(X)

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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

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{2, x1 - 9}
          cons_A(x1,x2) = max{28, x1 + 8, x2}
          f_A(x1) = max{29, x1 + 13}
          active#_A(x1) = max{1, x1 - 9}
          mark_A(x1) = x1
          s_A(x1) = max{9, x1}
          |0|_A = 17
          p_A(x1) = max{12, x1}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        mark# = cons = active# = s = |0| = p > f = mark = active
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = []
        pi(f) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(|0|) = []
        pi(p) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 54
          cons_A(x1,x2) = 36
          f_A(x1) = 55
          active#_A(x1) = 54
          mark_A(x1) = 37
          s_A(x1) = 96
          |0|_A = 11
          p_A(x1) = 55
          active_A(x1) = 37
    
      precedence:
      
        cons = |0| > f = mark = p = active > mark# = active# = s
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = []
        pi(f) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(|0|) = []
        pi(p) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p4: mark#(s(X)) -> active#(s(mark(X)))
p5: active#(p(s(X))) -> mark#(X)
p6: mark#(p(X)) -> active#(p(mark(X)))
p7: mark#(p(X)) -> mark#(X)

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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#(cons(X1,X2)) -> mark#(X1)
p2: mark#(p(X)) -> mark#(X)
p3: mark#(p(X)) -> active#(p(mark(X)))
p4: active#(p(s(X))) -> mark#(X)
p5: mark#(s(X)) -> active#(s(mark(X)))
p6: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p7: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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

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{7, x1 + 6}
          cons_A(x1,x2) = max{28, x1 + 13}
          p_A(x1) = max{54, x1 + 11}
          active#_A(x1) = x1 + 6
          mark_A(x1) = max{14, x1}
          s_A(x1) = max{42, x1 - 3}
          f_A(x1) = max{54, x1 - 2}
          |0|_A = 29
          active_A(x1) = max{12, x1}
    
      precedence:
      
        s > f > mark# = cons = p = active# = mark = |0| = active
      
      partial status:
    
        pi(mark#) = [1]
        pi(cons) = []
        pi(p) = []
        pi(active#) = [1]
        pi(mark) = [1]
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 45
          cons_A(x1,x2) = 20
          p_A(x1) = 5
          active#_A(x1) = 45
          mark_A(x1) = 68
          s_A(x1) = 50
          f_A(x1) = 47
          |0|_A = 63
          active_A(x1) = max{68, x1 + 4}
    
      precedence:
      
        mark# = cons = p = active# = mark = s = f = |0| = active
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = []
        pi(p) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(p(X)) -> active#(p(mark(X)))
p3: active#(p(s(X))) -> mark#(X)
p4: mark#(s(X)) -> active#(s(mark(X)))
p5: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p6: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p4: active#(p(s(X))) -> mark#(X)
p5: mark#(s(X)) -> active#(s(mark(X)))
p6: mark#(p(X)) -> active#(p(mark(X)))

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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

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{2, x1 - 3}
          cons_A(x1,x2) = max{2, x1, x2}
          f_A(x1) = max{2, x1 - 12}
          active#_A(x1) = max{2, x1 - 15}
          mark_A(x1) = x1 + 11
          s_A(x1) = x1 + 6
          |0|_A = 2
          p_A(x1) = x1 + 6
          active_A(x1) = max{13, x1}
    
      precedence:
      
        mark# = cons = f = active# = mark = s = |0| = p = active
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = []
        pi(f) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(|0|) = []
        pi(p) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 1
          cons_A(x1,x2) = 8
          f_A(x1) = 92
          active#_A(x1) = 1
          mark_A(x1) = 7
          s_A(x1) = 13
          |0|_A = 52
          p_A(x1) = 18
          active_A(x1) = 7
    
      precedence:
      
        mark# = cons = f = active# = mark = s = p = active > |0|
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = []
        pi(f) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(|0|) = []
        pi(p) = []
        pi(active) = []
    

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: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p4: active#(p(s(X))) -> mark#(X)
p5: mark#(s(X)) -> active#(s(mark(X)))

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(s(X)) -> active#(s(mark(X)))
p3: active#(p(s(X))) -> mark#(X)
p4: mark#(f(X)) -> active#(f(mark(X)))
p5: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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

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{112, x1 + 88}
          cons_A(x1,x2) = x1 + 30
          s_A(x1) = x1 + 175
          active#_A(x1) = max{111, x1 + 75}
          mark_A(x1) = x1 + 11
          p_A(x1) = max{31, x1 - 135}
          f_A(x1) = max{75, x1 - 9}
          |0|_A = 30
          active_A(x1) = max{41, x1}
    
      precedence:
      
        mark = |0| > mark# = cons = s > active# = p = f = active
      
      partial status:
    
        pi(mark#) = [1]
        pi(cons) = [1]
        pi(s) = [1]
        pi(active#) = [1]
        pi(mark) = []
        pi(p) = []
        pi(f) = []
        pi(|0|) = []
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{328, x1 + 324}
          cons_A(x1,x2) = x1 + 26
          s_A(x1) = max{123, x1 + 41}
          active#_A(x1) = x1 + 171
          mark_A(x1) = 25
          p_A(x1) = 8
          f_A(x1) = 18
          |0|_A = 16
          active_A(x1) = max{9, x1 + 7}
    
      precedence:
      
        mark# > s = mark = f = active > cons = active# = p = |0|
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = [1]
        pi(s) = [1]
        pi(active#) = [1]
        pi(mark) = []
        pi(p) = []
        pi(f) = []
        pi(|0|) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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: f#(mark(X)) -> f#(X)
p2: f#(active(X)) -> f#(X)

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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:
        
          f#_A(x1) = x1 + 2
          mark_A(x1) = x1
          active_A(x1) = x1 + 2
    
      precedence:
      
        mark > active > f#
      
      partial status:
    
        pi(f#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        f# = mark = active
      
      partial status:
    
        pi(f#) = [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: 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(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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: s#(mark(X)) -> s#(X)
p2: s#(active(X)) -> s#(X)

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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: p#(mark(X)) -> p#(X)
p2: p#(active(X)) -> p#(X)

and R consists of:

r1: active(f(|0|())) -> mark(cons(|0|(),f(s(|0|()))))
r2: active(f(s(|0|()))) -> mark(f(p(s(|0|()))))
r3: active(p(s(X))) -> mark(X)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(|0|()) -> active(|0|())
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(p(X)) -> active(p(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: cons(mark(X1),X2) -> cons(X1,X2)
r12: cons(X1,mark(X2)) -> cons(X1,X2)
r13: cons(active(X1),X2) -> cons(X1,X2)
r14: cons(X1,active(X2)) -> cons(X1,X2)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)
r17: p(mark(X)) -> p(X)
r18: p(active(X)) -> p(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:
        
          p#_A(x1) = max{6, x1 + 4}
          mark_A(x1) = max{4, x1 + 3}
          active_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        p# = mark = active
      
      partial status:
    
        pi(p#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          p#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        p# = mark = active
      
      partial status:
    
        pi(p#) = [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.