YES

We show the termination of the TRS R:

  active(|2nd|(cons(X,cons(Y,Z)))) -> mark(Y)
  active(from(X)) -> mark(cons(X,from(s(X))))
  mark(|2nd|(X)) -> active(|2nd|(mark(X)))
  mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
  mark(from(X)) -> active(from(mark(X)))
  mark(s(X)) -> active(s(mark(X)))
  |2nd|(mark(X)) -> |2nd|(X)
  |2nd|(active(X)) -> |2nd|(X)
  cons(mark(X1),X2) -> cons(X1,X2)
  cons(X1,mark(X2)) -> cons(X1,X2)
  cons(active(X1),X2) -> cons(X1,X2)
  cons(X1,active(X2)) -> cons(X1,X2)
  from(mark(X)) -> from(X)
  from(active(X)) -> from(X)
  s(mark(X)) -> s(X)
  s(active(X)) -> s(X)

-- SCC decomposition.

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

p1: active#(|2nd|(cons(X,cons(Y,Z)))) -> mark#(Y)
p2: active#(from(X)) -> mark#(cons(X,from(s(X))))
p3: active#(from(X)) -> cons#(X,from(s(X)))
p4: active#(from(X)) -> from#(s(X))
p5: active#(from(X)) -> s#(X)
p6: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p7: mark#(|2nd|(X)) -> |2nd|#(mark(X))
p8: mark#(|2nd|(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p10: mark#(cons(X1,X2)) -> cons#(mark(X1),X2)
p11: mark#(cons(X1,X2)) -> mark#(X1)
p12: mark#(from(X)) -> active#(from(mark(X)))
p13: mark#(from(X)) -> from#(mark(X))
p14: mark#(from(X)) -> mark#(X)
p15: mark#(s(X)) -> active#(s(mark(X)))
p16: mark#(s(X)) -> s#(mark(X))
p17: mark#(s(X)) -> mark#(X)
p18: |2nd|#(mark(X)) -> |2nd|#(X)
p19: |2nd|#(active(X)) -> |2nd|#(X)
p20: cons#(mark(X1),X2) -> cons#(X1,X2)
p21: cons#(X1,mark(X2)) -> cons#(X1,X2)
p22: cons#(active(X1),X2) -> cons#(X1,X2)
p23: cons#(X1,active(X2)) -> cons#(X1,X2)
p24: from#(mark(X)) -> from#(X)
p25: from#(active(X)) -> from#(X)
p26: s#(mark(X)) -> s#(X)
p27: s#(active(X)) -> s#(X)

and R consists of:

r1: active(|2nd|(cons(X,cons(Y,Z)))) -> mark(Y)
r2: active(from(X)) -> mark(cons(X,from(s(X))))
r3: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r4: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(s(X)) -> active(s(mark(X)))
r7: |2nd|(mark(X)) -> |2nd|(X)
r8: |2nd|(active(X)) -> |2nd|(X)
r9: cons(mark(X1),X2) -> cons(X1,X2)
r10: cons(X1,mark(X2)) -> cons(X1,X2)
r11: cons(active(X1),X2) -> cons(X1,X2)
r12: cons(X1,active(X2)) -> cons(X1,X2)
r13: from(mark(X)) -> from(X)
r14: from(active(X)) -> from(X)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p6, p8, p9, p11, p12, p14, p15, p17}
  {p20, p21, p22, p23}
  {p24, p25}
  {p26, p27}
  {p18, p19}


-- Reduction pair.

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

p1: active#(|2nd|(cons(X,cons(Y,Z)))) -> mark#(Y)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(s(X)) -> active#(s(mark(X)))
p4: active#(from(X)) -> mark#(cons(X,from(s(X))))
p5: mark#(from(X)) -> mark#(X)
p6: mark#(from(X)) -> active#(from(mark(X)))
p7: mark#(cons(X1,X2)) -> mark#(X1)
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p9: mark#(|2nd|(X)) -> mark#(X)
p10: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons(X,cons(Y,Z)))) -> mark(Y)
r2: active(from(X)) -> mark(cons(X,from(s(X))))
r3: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r4: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(s(X)) -> active(s(mark(X)))
r7: |2nd|(mark(X)) -> |2nd|(X)
r8: |2nd|(active(X)) -> |2nd|(X)
r9: cons(mark(X1),X2) -> cons(X1,X2)
r10: cons(X1,mark(X2)) -> cons(X1,X2)
r11: cons(active(X1),X2) -> cons(X1,X2)
r12: cons(X1,active(X2)) -> cons(X1,X2)
r13: from(mark(X)) -> from(X)
r14: from(active(X)) -> from(X)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16

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{20, x1 + 6}
          |2nd|_A(x1) = x1 + 19
          cons_A(x1,x2) = max{x1 + 13, x2 - 16}
          mark#_A(x1) = x1 + 20
          s_A(x1) = x1
          mark_A(x1) = x1 + 14
          from_A(x1) = x1 + 27
          active_A(x1) = x1
    
      precedence:
      
        active# = |2nd| = cons = mark# = s = mark = from = active
      
      partial status:
    
        pi(active#) = []
        pi(|2nd|) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(mark) = []
        pi(from) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 33
          |2nd|_A(x1) = 74
          cons_A(x1,x2) = 35
          mark#_A(x1) = max{24, x1 - 2}
          s_A(x1) = max{43, x1}
          mark_A(x1) = 74
          from_A(x1) = max{74, x1}
          active_A(x1) = max{61, x1}
    
      precedence:
      
        cons = mark = active > from > s > active# = mark# > |2nd|
      
      partial status:
    
        pi(active#) = []
        pi(|2nd|) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(mark) = []
        pi(from) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p1, p5, p6, p7, p9

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(s(X)) -> active#(s(mark(X)))
p3: active#(from(X)) -> mark#(cons(X,from(s(X))))
p4: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p5: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons(X,cons(Y,Z)))) -> mark(Y)
r2: active(from(X)) -> mark(cons(X,from(s(X))))
r3: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r4: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(s(X)) -> active(s(mark(X)))
r7: |2nd|(mark(X)) -> |2nd|(X)
r8: |2nd|(active(X)) -> |2nd|(X)
r9: cons(mark(X1),X2) -> cons(X1,X2)
r10: cons(X1,mark(X2)) -> cons(X1,X2)
r11: cons(active(X1),X2) -> cons(X1,X2)
r12: cons(X1,active(X2)) -> cons(X1,X2)
r13: from(mark(X)) -> from(X)
r14: from(active(X)) -> from(X)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(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#(s(X)) -> mark#(X)
p2: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p3: active#(from(X)) -> mark#(cons(X,from(s(X))))
p4: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p5: mark#(s(X)) -> active#(s(mark(X)))

and R consists of:

r1: active(|2nd|(cons(X,cons(Y,Z)))) -> mark(Y)
r2: active(from(X)) -> mark(cons(X,from(s(X))))
r3: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r4: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(s(X)) -> active(s(mark(X)))
r7: |2nd|(mark(X)) -> |2nd|(X)
r8: |2nd|(active(X)) -> |2nd|(X)
r9: cons(mark(X1),X2) -> cons(X1,X2)
r10: cons(X1,mark(X2)) -> cons(X1,X2)
r11: cons(active(X1),X2) -> cons(X1,X2)
r12: cons(X1,active(X2)) -> cons(X1,X2)
r13: from(mark(X)) -> from(X)
r14: from(active(X)) -> from(X)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16

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{15, x1 + 4}
          s_A(x1) = x1
          |2nd|_A(x1) = max{0, x1 - 3}
          active#_A(x1) = max{15, x1}
          mark_A(x1) = x1
          from_A(x1) = max{17, x1 + 14}
          cons_A(x1,x2) = max{12, x1 + 8, x2 - 4}
          active_A(x1) = x1
    
      precedence:
      
        from > mark# = |2nd| > mark > s = active# = active > cons
      
      partial status:
    
        pi(mark#) = [1]
        pi(s) = [1]
        pi(|2nd|) = []
        pi(active#) = [1]
        pi(mark) = [1]
        pi(from) = []
        pi(cons) = []
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{16, x1}
          s_A(x1) = max{247, x1 + 238}
          |2nd|_A(x1) = 25
          active#_A(x1) = max{23, x1 - 8}
          mark_A(x1) = 7
          from_A(x1) = 115
          cons_A(x1,x2) = 23
          active_A(x1) = 6
    
      precedence:
      
        |2nd| > s = from = cons = active > active# > mark# > mark
      
      partial status:
    
        pi(mark#) = [1]
        pi(s) = []
        pi(|2nd|) = []
        pi(active#) = []
        pi(mark) = []
        pi(from) = []
        pi(cons) = []
        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: 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(|2nd|(cons(X,cons(Y,Z)))) -> mark(Y)
r2: active(from(X)) -> mark(cons(X,from(s(X))))
r3: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r4: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(s(X)) -> active(s(mark(X)))
r7: |2nd|(mark(X)) -> |2nd|(X)
r8: |2nd|(active(X)) -> |2nd|(X)
r9: cons(mark(X1),X2) -> cons(X1,X2)
r10: cons(X1,mark(X2)) -> cons(X1,X2)
r11: cons(active(X1),X2) -> cons(X1,X2)
r12: cons(X1,active(X2)) -> cons(X1,X2)
r13: from(mark(X)) -> from(X)
r14: from(active(X)) -> from(X)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2, p3, p4

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

-- Reduction pair.

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

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

and R consists of:

r1: active(|2nd|(cons(X,cons(Y,Z)))) -> mark(Y)
r2: active(from(X)) -> mark(cons(X,from(s(X))))
r3: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r4: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(s(X)) -> active(s(mark(X)))
r7: |2nd|(mark(X)) -> |2nd|(X)
r8: |2nd|(active(X)) -> |2nd|(X)
r9: cons(mark(X1),X2) -> cons(X1,X2)
r10: cons(X1,mark(X2)) -> cons(X1,X2)
r11: cons(active(X1),X2) -> cons(X1,X2)
r12: cons(X1,active(X2)) -> cons(X1,X2)
r13: from(mark(X)) -> from(X)
r14: from(active(X)) -> from(X)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

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

and R consists of:

r1: active(|2nd|(cons(X,cons(Y,Z)))) -> mark(Y)
r2: active(from(X)) -> mark(cons(X,from(s(X))))
r3: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r4: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r5: mark(from(X)) -> active(from(mark(X)))
r6: mark(s(X)) -> active(s(mark(X)))
r7: |2nd|(mark(X)) -> |2nd|(X)
r8: |2nd|(active(X)) -> |2nd|(X)
r9: cons(mark(X1),X2) -> cons(X1,X2)
r10: cons(X1,mark(X2)) -> cons(X1,X2)
r11: cons(active(X1),X2) -> cons(X1,X2)
r12: cons(X1,active(X2)) -> cons(X1,X2)
r13: from(mark(X)) -> from(X)
r14: from(active(X)) -> from(X)
r15: s(mark(X)) -> s(X)
r16: s(active(X)) -> s(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: |2nd|#(mark(X)) -> |2nd|#(X)
p2: |2nd|#(active(X)) -> |2nd|#(X)

and R consists of:

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