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. lexicographic path order with precedence:
    
      precedence:
      
        f > cons > active# > active > mark > s > |0| > mark# > p
      
      argument filter:
    
        pi(active#) = 1
        pi(f) = [1]
        pi(|0|) = []
        pi(mark#) = 1
        pi(cons) = 1
        pi(s) = [1]
        pi(p) = 1
        pi(mark) = 1
        pi(active) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        active#_A(x1) = 0
        f_A(x1) = 1
        |0|_A() = 1
        mark#_A(x1) = 0
        cons_A(x1,x2) = 2
        s_A(x1) = x1 + 1
        p_A(x1) = x1 + 1
        mark_A(x1) = x1
        active_A(x1) = x1
    

The next rules are strictly ordered:

  p1, p5, p6, p9

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: mark#(s(X)) -> active#(s(mark(X)))
p5: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p6: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
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 estimated dependency graph contains the following SCCs:

  {p1, p2}
  {p5, 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)

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. lexicographic path order with precedence:
    
      precedence:
      
        mark# > p > cons
      
      argument filter:
    
        pi(mark#) = 1
        pi(cons) = 1
        pi(p) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        mark#_A(x1) = x1
        cons_A(x1,x2) = x1 + 1
        p_A(x1) = x1 + 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: mark#(f(X)) -> active#(f(mark(X)))
p2: 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. lexicographic path order with precedence:
    
      precedence:
      
        f > s > |0| > cons > mark > p > active > active# > mark#
      
      argument filter:
    
        pi(mark#) = 1
        pi(f) = [1]
        pi(active#) = 1
        pi(mark) = 1
        pi(s) = [1]
        pi(|0|) = []
        pi(p) = 1
        pi(active) = 1
        pi(cons) = []
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        mark#_A(x1) = x1 + 2
        f_A(x1) = x1 + 1
        active#_A(x1) = x1
        mark_A(x1) = x1 + 1
        s_A(x1) = 5
        |0|_A() = 1
        p_A(x1) = 2
        active_A(x1) = x1
        cons_A(x1,x2) = 2
    

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: 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. lexicographic path order with precedence:
    
      precedence:
      
        f# > active > mark
      
      argument filter:
    
        pi(f#) = 1
        pi(mark) = [1]
        pi(active) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 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

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. lexicographic path order with precedence:
    
      precedence:
      
        cons# > active > mark
      
      argument filter:
    
        pi(cons#) = 1
        pi(mark) = 1
        pi(active) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        cons#_A(x1,x2) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1, p3

We remove them from the problem.

-- SCC decomposition.

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

p1: cons#(X1,active(X2)) -> cons#(X1,X2)
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: cons#(X1,active(X2)) -> cons#(X1,X2)
p2: 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. lexicographic path order with precedence:
    
      precedence:
      
        cons# > mark > active
      
      argument filter:
    
        pi(cons#) = 2
        pi(active) = 1
        pi(mark) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        cons#_A(x1,x2) = x2
        active_A(x1) = x1 + 1
        mark_A(x1) = x1 + 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(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. lexicographic path order with precedence:
    
      precedence:
      
        s# > active > mark
      
      argument filter:
    
        pi(s#) = 1
        pi(mark) = [1]
        pi(active) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        s#_A(x1) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 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

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. lexicographic path order with precedence:
    
      precedence:
      
        p# > active > mark
      
      argument filter:
    
        pi(p#) = 1
        pi(mark) = [1]
        pi(active) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        p#_A(x1) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 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

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