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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = ((0,0),(1,0)) x1 + (1,1)
            f_A(x1) = ((1,0),(0,0)) x1 + (3,0)
            |0|_A() = (0,0)
            mark#_A(x1) = ((0,0),(1,0)) x1 + (1,1)
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + (2,0)
            s_A(x1) = ((1,0),(0,0)) x1 + (4,9)
            p_A(x1) = ((1,0),(0,0)) x1 + (0,10)
            mark_A(x1) = ((1,0),(0,0)) x1 + (0,10)
            active_A(x1) = ((1,0),(0,0)) x1 + (0,10)
    
      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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = (4,3)
            f_A(x1) = (3,2)
            |0|_A() = (3,2)
            mark#_A(x1) = (4,3)
            cons_A(x1,x2) = (1,1)
            s_A(x1) = (0,0)
            p_A(x1) = (2,0)
            mark_A(x1) = (2,0)
            active_A(x1) = (2,0)
    
      precedence:
      
        p > |0| > active# = mark# = s > f = cons = mark = active
      
      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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,0)) x1 + (1,5)
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (2,15)
            f_A(x1) = ((1,0),(1,1)) x1 + (3,5)
            active#_A(x1) = ((1,0),(1,0)) x1 + (1,5)
            mark_A(x1) = x1 + (0,4)
            s_A(x1) = ((1,0),(0,0)) x1 + (2,11)
            |0|_A() = (1,20)
            p_A(x1) = ((1,0),(0,0)) x1 + (0,6)
            active_A(x1) = x1
    
      precedence:
      
        cons > mark = |0| = active > f > mark# = active# = s = p
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = []
        pi(f) = []
        pi(active#) = []
        pi(mark) = [1]
        pi(s) = []
        pi(|0|) = []
        pi(p) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = (1,2)
            cons_A(x1,x2) = (3,0)
            f_A(x1) = (1,2)
            active#_A(x1) = (1,2)
            mark_A(x1) = (2,4)
            s_A(x1) = (5,5)
            |0|_A() = (4,3)
            p_A(x1) = (3,0)
            active_A(x1) = (0,1)
    
      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) = []
    

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(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#(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)) -> active#(s(mark(X)))
p6: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p7: mark#(f(X)) -> mark#(X)
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,0)) x1 + (0,4)
            cons_A(x1,x2) = ((1,0),(1,0)) x1 + (5,1)
            p_A(x1) = ((1,0),(1,0)) x1 + (0,1)
            active#_A(x1) = x1
            mark_A(x1) = x1 + (0,2)
            s_A(x1) = ((1,0),(0,0)) x1 + (4,7)
            f_A(x1) = ((1,0),(1,0)) x1 + (5,9)
            |0|_A() = (1,5)
            active_A(x1) = x1 + (0,2)
    
      precedence:
      
        f = |0| > s > mark# = cons = p = active# = mark = active
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = []
        pi(p) = []
        pi(active#) = []
        pi(mark) = [1]
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = (6,2)
            cons_A(x1,x2) = (0,3)
            p_A(x1) = (1,4)
            active#_A(x1) = (6,2)
            mark_A(x1) = (4,1)
            s_A(x1) = (5,0)
            f_A(x1) = (3,3)
            |0|_A() = (2,3)
            active_A(x1) = (4,1)
    
      precedence:
      
        mark = s = f = active > |0| > mark# = active# > cons = p
      
      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:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(p(X)) -> mark#(X)
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)) -> mark#(X)
p7: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
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#(p(X)) -> mark#(X)
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#(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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = (2,9)
            p_A(x1) = (8,10)
            f_A(x1) = (8,8)
            active#_A(x1) = ((0,0),(1,0)) x1 + (2,1)
            mark_A(x1) = (9,6)
            s_A(x1) = (3,5)
            |0|_A() = (1,7)
            cons_A(x1,x2) = (0,0)
            active_A(x1) = (9,6)
    
      precedence:
      
        mark = |0| = active > p > mark# = f = active# = s = cons
      
      partial status:
    
        pi(mark#) = []
        pi(p) = []
        pi(f) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(|0|) = []
        pi(cons) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = (4,4)
            p_A(x1) = (1,1)
            f_A(x1) = (5,5)
            active#_A(x1) = (4,4)
            mark_A(x1) = (3,3)
            s_A(x1) = (0,6)
            |0|_A() = (2,2)
            cons_A(x1,x2) = (4,4)
            active_A(x1) = (3,3)
    
      precedence:
      
        mark# = p = f = active# = mark = s = |0| = cons = active
      
      partial status:
    
        pi(mark#) = []
        pi(p) = []
        pi(f) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(|0|) = []
        pi(cons) = []
        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#(p(X)) -> mark#(X)
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#(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 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#(p(X)) -> mark#(X)
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)) -> mark#(X)
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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,0)) x1 + (0,4)
            p_A(x1) = ((1,0),(0,0)) x1 + (0,2)
            active#_A(x1) = ((1,0),(1,1)) x1
            mark_A(x1) = ((1,0),(0,0)) x1 + (0,1)
            s_A(x1) = ((1,0),(0,0)) x1 + (3,3)
            f_A(x1) = ((1,0),(0,0)) x1 + (1,4)
            |0|_A() = (1,10)
            active_A(x1) = ((1,0),(0,0)) x1 + (0,1)
            cons_A(x1,x2) = (0,1)
    
      precedence:
      
        mark# = p = active# = mark = s = f = |0| = active = cons
      
      partial status:
    
        pi(mark#) = []
        pi(p) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(active) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = (0,1)
            p_A(x1) = (2,0)
            active#_A(x1) = (0,1)
            mark_A(x1) = (1,2)
            s_A(x1) = (0,0)
            f_A(x1) = (0,3)
            |0|_A() = (0,0)
            active_A(x1) = (1,2)
            cons_A(x1,x2) = (2,4)
    
      precedence:
      
        mark = s = f = active = cons > |0| > mark# = p = active#
      
      partial status:
    
        pi(mark#) = []
        pi(p) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(f) = []
        pi(|0|) = []
        pi(active) = []
        pi(cons) = []
    

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#(p(X)) -> mark#(X)
p2: mark#(p(X)) -> active#(p(mark(X)))
p3: mark#(s(X)) -> active#(s(mark(X)))
p4: active#(f(s(|0|()))) -> mark#(f(p(s(|0|()))))
p5: mark#(f(X)) -> mark#(X)
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#(p(X)) -> mark#(X)
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: 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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((0,0),(1,0)) x1
            p_A(x1) = ((1,0),(0,0)) x1 + (0,6)
            f_A(x1) = x1 + (2,21)
            active#_A(x1) = ((0,0),(1,0)) x1
            mark_A(x1) = ((1,0),(1,1)) x1 + (0,14)
            s_A(x1) = ((1,0),(1,0)) x1 + (1,24)
            |0|_A() = (2,27)
            active_A(x1) = x1 + (0,1)
            cons_A(x1,x2) = (1,13)
    
      precedence:
      
        p > f = mark > active > |0| > mark# = active# = s > cons
      
      partial status:
    
        pi(mark#) = []
        pi(p) = []
        pi(f) = [1]
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(|0|) = []
        pi(active) = [1]
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = (2,6)
            p_A(x1) = (1,1)
            f_A(x1) = ((1,0),(1,1)) x1 + (0,1)
            active#_A(x1) = (2,6)
            mark_A(x1) = (0,7)
            s_A(x1) = (3,2)
            |0|_A() = (4,8)
            active_A(x1) = (0,7)
            cons_A(x1,x2) = (1,0)
    
      precedence:
      
        mark# = p = f = active# = mark = s = |0| = active = cons
      
      partial status:
    
        pi(mark#) = []
        pi(p) = []
        pi(f) = []
        pi(active#) = []
        pi(mark) = []
        pi(s) = []
        pi(|0|) = []
        pi(active) = []
        pi(cons) = []
    

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#(p(X)) -> mark#(X)
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: 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 estimated dependency graph contains the following SCCs:

  {p1}
  {p2, p3}


-- Reduction pair.

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

p1: mark#(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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            p_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        mark# = p
      
      partial status:
    
        pi(mark#) = []
        pi(p) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(0,0)) x1
            p_A(x1) = (1,1)
    
      precedence:
      
        p > mark#
      
      partial status:
    
        pi(mark#) = []
        pi(p) = []
    

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: 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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = x1 + (0,1)
            f_A(x1) = ((1,0),(1,1)) x1
            active#_A(x1) = x1
            mark_A(x1) = x1
            s_A(x1) = ((1,0),(0,0)) x1 + (1,3)
            |0|_A() = (1,6)
            p_A(x1) = ((1,0),(0,0)) x1
            active_A(x1) = x1
            cons_A(x1,x2) = x1
    
      precedence:
      
        s = |0| > p > mark# > mark > active = cons > f = active#
      
      partial status:
    
        pi(mark#) = [1]
        pi(f) = [1]
        pi(active#) = [1]
        pi(mark) = [1]
        pi(s) = []
        pi(|0|) = []
        pi(p) = []
        pi(active) = [1]
        pi(cons) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(0,0)) x1 + (2,3)
            f_A(x1) = ((1,0),(1,1)) x1 + (1,0)
            active#_A(x1) = ((1,0),(1,1)) x1 + (1,1)
            mark_A(x1) = (0,0)
            s_A(x1) = (7,5)
            |0|_A() = (0,0)
            p_A(x1) = (3,1)
            active_A(x1) = x1 + (2,0)
            cons_A(x1,x2) = ((0,0),(1,0)) x1
    
      precedence:
      
        |0| > p > mark# = active# > f = mark = s = active = cons
      
      partial status:
    
        pi(mark#) = []
        pi(f) = [1]
        pi(active#) = [1]
        pi(mark) = []
        pi(s) = []
        pi(|0|) = []
        pi(p) = []
        pi(active) = [1]
        pi(cons) = []
    

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#(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:

  (no SCCs)

-- 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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = x1 + (1,2)
            mark_A(x1) = x1 + (2,1)
            active_A(x1) = ((1,0),(1,1)) x1 + (2,1)
    
      precedence:
      
        f# = mark = active
      
      partial status:
    
        pi(f#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = ((1,0),(1,0)) x1 + (0,2)
            mark_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            active_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        f# = mark = active
      
      partial status:
    
        pi(f#) = []
        pi(mark) = []
        pi(active) = []
    

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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            cons#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (2,2)
            mark_A(x1) = ((1,0),(0,0)) x1 + (1,3)
            active_A(x1) = ((1,0),(0,0)) x1 + (3,1)
    
      precedence:
      
        cons# = mark = active
      
      partial status:
    
        pi(cons#) = []
        pi(mark) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            cons#_A(x1,x2) = (0,0)
            mark_A(x1) = (1,1)
            active_A(x1) = (1,1)
    
      precedence:
      
        mark > cons# = active
      
      partial status:
    
        pi(cons#) = []
        pi(mark) = []
        pi(active) = []
    

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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            s#_A(x1) = x1 + (1,2)
            mark_A(x1) = x1 + (2,1)
            active_A(x1) = ((1,0),(1,1)) x1 + (2,1)
    
      precedence:
      
        s# = mark > active
      
      partial status:
    
        pi(s#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            s#_A(x1) = ((1,0),(1,0)) x1 + (0,2)
            mark_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            active_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        s# = mark = active
      
      partial status:
    
        pi(s#) = []
        pi(mark) = []
        pi(active) = []
    

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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            p#_A(x1) = x1 + (1,2)
            mark_A(x1) = x1 + (2,1)
            active_A(x1) = ((1,0),(1,1)) x1 + (2,1)
    
      precedence:
      
        p# = mark = active
      
      partial status:
    
        pi(p#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            p#_A(x1) = ((1,0),(1,0)) x1 + (0,2)
            mark_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            active_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        p# = mark = active
      
      partial status:
    
        pi(p#) = []
        pi(mark) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p1, p2

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