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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = x1
        |2nd|_A(x1) = x1
        cons_A(x1,x2) = max{5, x1 + 4, x2 - 3}
        mark#_A(x1) = x1
        s_A(x1) = x1 + 2
        mark_A(x1) = x1
        from_A(x1) = x1 + 9
        active_A(x1) = x1
  
    precedence:
    
      s > active# = |2nd| = cons = mark# = mark = from = active
    
    partial status:
  
      pi(active#) = []
      pi(|2nd|) = []
      pi(cons) = []
      pi(mark#) = []
      pi(s) = []
      pi(mark) = [1]
      pi(from) = []
      pi(active) = [1]

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

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)) -> active#(cons(mark(X1),X2))
p8: mark#(|2nd|(X)) -> mark#(X)
p9: 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, p6, p7, p8, p9}


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

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{24, x1 - 2}
        |2nd|_A(x1) = max{91, x1 + 90}
        cons_A(x1,x2) = max{x1 + 20, x2 - 33}
        mark#_A(x1) = x1 + 21
        mark_A(x1) = x1
        from_A(x1) = x1 + 43
        s_A(x1) = x1 + 3
        active_A(x1) = max{1, x1}
  
    precedence:
    
      active# = |2nd| = cons = mark# = mark = from = s = active
    
    partial status:
  
      pi(active#) = []
      pi(|2nd|) = []
      pi(cons) = []
      pi(mark#) = []
      pi(mark) = [1]
      pi(from) = []
      pi(s) = []
      pi(active) = [1]

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

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

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


-- 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)) -> active#(from(mark(X)))
p6: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p7: mark#(|2nd|(X)) -> mark#(X)
p8: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{5, x1}
        |2nd|_A(x1) = x1 + 3
        cons_A(x1,x2) = max{x1 + 29, x2 - 21}
        mark#_A(x1) = x1 + 9
        s_A(x1) = max{11, x1 + 10}
        mark_A(x1) = x1
        from_A(x1) = x1 + 39
        active_A(x1) = max{1, x1}
  
    precedence:
    
      active# = |2nd| = cons = mark# = s = mark = from = active
    
    partial status:
  
      pi(active#) = []
      pi(|2nd|) = []
      pi(cons) = []
      pi(mark#) = [1]
      pi(s) = []
      pi(mark) = [1]
      pi(from) = []
      pi(active) = [1]

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: 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#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: mark#(|2nd|(X)) -> mark#(X)
p7: 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, p6, p7}


-- 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#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p3: active#(from(X)) -> mark#(cons(X,from(s(X))))
p4: mark#(|2nd|(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: mark#(s(X)) -> active#(s(mark(X)))
p7: mark#(s(X)) -> 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = max{18, x1 - 3}
        |2nd|_A(x1) = max{63, x1 + 59}
        cons_A(x1,x2) = max{35, x1 + 31, x2 - 25}
        mark#_A(x1) = max{5, x1 - 2}
        mark_A(x1) = max{4, x1}
        from_A(x1) = max{44, x1 + 39}
        s_A(x1) = max{21, x1}
        active_A(x1) = x1
  
    precedence:
    
      active# = |2nd| = cons = mark# = mark = from = s = active
    
    partial status:
  
      pi(active#) = []
      pi(|2nd|) = []
      pi(cons) = []
      pi(mark#) = []
      pi(mark) = [1]
      pi(from) = []
      pi(s) = []
      pi(active) = [1]

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#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p2: active#(from(X)) -> mark#(cons(X,from(s(X))))
p3: mark#(|2nd|(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p5: mark#(s(X)) -> active#(s(mark(X)))
p6: mark#(s(X)) -> 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, p6}


-- Reduction pair.

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

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

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{11, x1}
        |2nd|_A(x1) = x1 + 13
        active#_A(x1) = max{13, x1 - 1}
        mark_A(x1) = x1
        from_A(x1) = x1 + 19
        cons_A(x1,x2) = max{x1 + 17, x2 - 27}
        s_A(x1) = max{20, x1 + 9}
        active_A(x1) = max{13, x1}
  
    precedence:
    
      mark# = |2nd| = active# = mark = from = cons = s = active
    
    partial status:
  
      pi(mark#) = [1]
      pi(|2nd|) = []
      pi(active#) = []
      pi(mark) = [1]
      pi(from) = []
      pi(cons) = []
      pi(s) = []
      pi(active) = [1]

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

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 3
        |2nd|_A(x1) = x1 + 17
        active#_A(x1) = max{1, x1}
        mark_A(x1) = x1
        from_A(x1) = x1 + 14
        cons_A(x1,x2) = max{x1 + 11, x2 - 8}
        s_A(x1) = x1 + 5
        active_A(x1) = max{1, x1}
  
    precedence:
    
      from = s > mark# = active# = mark = cons = active > |2nd|
    
    partial status:
  
      pi(mark#) = [1]
      pi(|2nd|) = []
      pi(active#) = [1]
      pi(mark) = [1]
      pi(from) = []
      pi(cons) = []
      pi(s) = []
      pi(active) = [1]

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#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p2: mark#(|2nd|(X)) -> mark#(X)
p3: mark#(s(X)) -> active#(s(mark(X)))
p4: mark#(s(X)) -> 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:

  {p2, p4}


-- Reduction pair.

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

p1: mark#(|2nd|(X)) -> mark#(X)
p2: mark#(s(X)) -> 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

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 1
        |2nd|_A(x1) = x1
        s_A(x1) = x1
  
    precedence:
    
      mark# = |2nd| = s
    
    partial status:
  
      pi(mark#) = [1]
      pi(|2nd|) = [1]
      pi(s) = [1]

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#(s(X)) -> 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}


-- Reduction pair.

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

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

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 2
        s_A(x1) = x1 + 2
  
    precedence:
    
      mark# = s
    
    partial status:
  
      pi(mark#) = [1]
      pi(s) = [1]

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: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons#_A(x1,x2) = max{0, x2 - 2}
        mark_A(x1) = x1
        active_A(x1) = max{3, x1 + 1}
  
    precedence:
    
      cons# = mark = active
    
    partial status:
  
      pi(cons#) = []
      pi(mark) = [1]
      pi(active) = [1]

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: cons#(mark(X1),X2) -> cons#(X1,X2)
p2: cons#(active(X1),X2) -> cons#(X1,X2)
p3: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: cons#(mark(X1),X2) -> cons#(X1,X2)
p2: cons#(X1,mark(X2)) -> cons#(X1,X2)
p3: cons#(active(X1),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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons#_A(x1,x2) = max{0, x1 - 2}
        mark_A(x1) = max{3, x1 + 1}
        active_A(x1) = x1
  
    precedence:
    
      cons# = mark = active
    
    partial status:
  
      pi(cons#) = []
      pi(mark) = [1]
      pi(active) = [1]

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: cons#(X1,mark(X2)) -> cons#(X1,X2)
p2: cons#(active(X1),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 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,mark(X2)) -> cons#(X1,X2)
p2: cons#(active(X1),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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons#_A(x1,x2) = x2 + 1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
  
    precedence:
    
      cons# = mark = active
    
    partial status:
  
      pi(cons#) = []
      pi(mark) = [1]
      pi(active) = [1]

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: cons#(active(X1),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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: cons#(active(X1),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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        cons#_A(x1,x2) = max{1, x1, x2}
        active_A(x1) = x1 + 1
  
    precedence:
    
      cons# = active
    
    partial status:
  
      pi(cons#) = [1, 2]
      pi(active) = []

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: 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:

  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

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        from#_A(x1) = x1 + 1
        active_A(x1) = x1
  
    precedence:
    
      from# = active
    
    partial status:
  
      pi(from#) = [1]
      pi(active) = [1]

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: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        s#_A(x1) = max{3, x1 + 2}
        mark_A(x1) = x1
        active_A(x1) = x1 + 1
  
    precedence:
    
      s# = mark = active
    
    partial status:
  
      pi(s#) = [1]
      pi(mark) = [1]
      pi(active) = [1]

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: 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}


-- Reduction pair.

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

p1: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        s#_A(x1) = x1 + 1
        active_A(x1) = x1
  
    precedence:
    
      s# = active
    
    partial status:
  
      pi(s#) = [1]
      pi(active) = [1]

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: |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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        |2nd|#_A(x1) = max{3, x1 + 2}
        mark_A(x1) = x1 + 3
        active_A(x1) = max{1, x1}
  
    precedence:
    
      |2nd|# = mark = active
    
    partial status:
  
      pi(|2nd|#) = [1]
      pi(mark) = [1]
      pi(active) = [1]

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: |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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: |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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        |2nd|#_A(x1) = x1 + 2
        active_A(x1) = x1 + 2
  
    precedence:
    
      |2nd|# = active
    
    partial status:
  
      pi(|2nd|#) = [1]
      pi(active) = [1]

The next rules are strictly ordered:

  p1

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