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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = ((0,0),(1,0)) x1 + (4,1)
            |2nd|_A(x1) = (2,2)
            cons_A(x1,x2) = (1,0)
            mark#_A(x1) = (4,3)
            s_A(x1) = (1,4)
            mark_A(x1) = (3,1)
            from_A(x1) = (2,4)
            active_A(x1) = (3,1)
    
      precedence:
      
        cons > active# = mark# = s > |2nd| > from > mark = 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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = (2,2)
            |2nd|_A(x1) = (0,1)
            cons_A(x1,x2) = (3,3)
            mark#_A(x1) = (2,2)
            s_A(x1) = (1,1)
            mark_A(x1) = (4,0)
            from_A(x1) = (5,0)
            active_A(x1) = (4,0)
    
      precedence:
      
        |2nd| > from > active# = mark# > cons = mark = active > s
      
      partial status:
    
        pi(active#) = []
        pi(|2nd|) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(mark) = []
        pi(from) = []
        pi(active) = []
    

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

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            |2nd|_A(x1) = (3,5)
            cons_A(x1,x2) = (2,5)
            mark#_A(x1) = (4,1)
            mark_A(x1) = (1,4)
            from_A(x1) = (3,3)
            s_A(x1) = (2,5)
            active_A(x1) = ((0,0),(1,0)) x1 + (1,1)
    
      precedence:
      
        from > cons > active# = |2nd| = mark# = mark = s = active
      
      partial status:
    
        pi(active#) = []
        pi(|2nd|) = []
        pi(cons) = []
        pi(mark#) = []
        pi(mark) = []
        pi(from) = []
        pi(s) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = (0,2)
            |2nd|_A(x1) = (1,1)
            cons_A(x1,x2) = (1,1)
            mark#_A(x1) = (0,2)
            mark_A(x1) = (2,3)
            from_A(x1) = (0,0)
            s_A(x1) = (2,0)
            active_A(x1) = (2,3)
    
      precedence:
      
        |2nd| > mark = active > from > s > active# = cons = mark#
      
      partial status:
    
        pi(active#) = []
        pi(|2nd|) = []
        pi(cons) = []
        pi(mark#) = []
        pi(mark) = []
        pi(from) = []
        pi(s) = []
        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: 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)) -> mark#(X1)
p6: mark#(from(X)) -> active#(from(mark(X)))
p7: mark#(from(X)) -> 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#(from(X)) -> mark#(X)
p4: mark#(from(X)) -> active#(from(mark(X)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(cons(X1,X2)) -> mark#(X1)
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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            active#_A(x1) = ((0,1),(0,0)) x1 + (1,2)
            |2nd|_A(x1) = ((1,0),(1,1)) x1 + (2,0)
            cons_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(0,0)) x2 + (4,0)
            mark#_A(x1) = ((0,1),(0,0)) x1 + (3,2)
            s_A(x1) = ((0,0),(0,1)) x1 + (5,3)
            from_A(x1) = ((0,1),(1,1)) x1 + (16,3)
            mark_A(x1) = x1
            active_A(x1) = x1
    
      precedence:
      
        from > cons > active# = |2nd| = mark > mark# = s > active
      
      partial status:
    
        pi(active#) = []
        pi(|2nd|) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            active#_A(x1) = (0,0)
            |2nd|_A(x1) = (0,1)
            cons_A(x1,x2) = (2,1)
            mark#_A(x1) = (0,0)
            s_A(x1) = (0,1)
            from_A(x1) = (0,0)
            mark_A(x1) = (1,2)
            active_A(x1) = (1,2)
    
      precedence:
      
        cons > |2nd| > active# = mark# = s = from = mark = active
      
      partial status:
    
        pi(active#) = []
        pi(|2nd|) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
        pi(mark) = []
        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: active#(|2nd|(cons(X,cons(Y,Z)))) -> mark#(Y)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(from(X)) -> mark#(X)
p4: mark#(from(X)) -> active#(from(mark(X)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(cons(X1,X2)) -> mark#(X1)
p7: 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, 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)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(from(X)) -> active#(from(mark(X)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(from(X)) -> 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = x1 + (1,3)
            |2nd|_A(x1) = (2,5)
            cons_A(x1,x2) = (1,3)
            mark#_A(x1) = (3,4)
            from_A(x1) = (2,1)
            mark_A(x1) = (2,2)
            s_A(x1) = (1,3)
            active_A(x1) = ((1,0),(0,0)) x1 + (0,2)
    
      precedence:
      
        active# = |2nd| = cons = mark# = mark = active > from = s
      
      partial status:
    
        pi(active#) = []
        pi(|2nd|) = []
        pi(cons) = []
        pi(mark#) = []
        pi(from) = []
        pi(mark) = []
        pi(s) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = (4,4)
            |2nd|_A(x1) = (1,0)
            cons_A(x1,x2) = (2,3)
            mark#_A(x1) = (4,4)
            from_A(x1) = (5,2)
            mark_A(x1) = (2,3)
            s_A(x1) = (3,1)
            active_A(x1) = (2,3)
    
      precedence:
      
        active# = cons = mark# = from = mark = s = active > |2nd|
      
      partial status:
    
        pi(active#) = []
        pi(|2nd|) = []
        pi(cons) = []
        pi(mark#) = []
        pi(from) = []
        pi(mark) = []
        pi(s) = []
        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#(|2nd|(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(from(X)) -> active#(from(mark(X)))
p4: active#(from(X)) -> mark#(cons(X,from(s(X))))
p5: mark#(from(X)) -> 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)) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(from(X)) -> mark#(X)
p4: mark#(from(X)) -> active#(from(mark(X)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(cons(X1,X2)) -> mark#(X1)

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:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            mark#_A(x1) = ((1,1),(0,0)) x1 + (1,0)
            |2nd|_A(x1) = ((1,1),(0,0)) x1 + (2,1)
            s_A(x1) = ((0,0),(1,1)) x1 + (2,1)
            from_A(x1) = ((1,1),(0,0)) x1
            active#_A(x1) = x1 + (1,0)
            mark_A(x1) = ((0,0),(1,1)) x1
            cons_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(0,0)) x2
            active_A(x1) = ((0,0),(1,1)) x1
    
      precedence:
      
        mark# = |2nd| = s = from = active# = mark = cons = active
      
      partial status:
    
        pi(mark#) = []
        pi(|2nd|) = []
        pi(s) = []
        pi(from) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            mark#_A(x1) = (0,0)
            |2nd|_A(x1) = (0,0)
            s_A(x1) = (1,3)
            from_A(x1) = (3,1)
            active#_A(x1) = (0,0)
            mark_A(x1) = (2,2)
            cons_A(x1,x2) = (1,0)
            active_A(x1) = (2,2)
    
      precedence:
      
        s > |2nd| > mark = cons = active > mark# = from = active#
      
      partial status:
    
        pi(mark#) = []
        pi(|2nd|) = []
        pi(s) = []
        pi(from) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        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#(s(X)) -> mark#(X)
p2: mark#(from(X)) -> mark#(X)
p3: mark#(from(X)) -> active#(from(mark(X)))
p4: active#(from(X)) -> mark#(cons(X,from(s(X))))
p5: mark#(cons(X1,X2)) -> mark#(X1)

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#(cons(X1,X2)) -> mark#(X1)
p3: mark#(from(X)) -> active#(from(mark(X)))
p4: active#(from(X)) -> mark#(cons(X,from(s(X))))
p5: mark#(from(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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            mark#_A(x1) = ((1,1),(1,1)) x1 + (5,11)
            s_A(x1) = ((1,1),(0,0)) x1 + (2,18)
            cons_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (4,1)
            from_A(x1) = ((0,0),(1,1)) x1 + (1,10)
            active#_A(x1) = ((0,1),(0,1)) x1 + (2,7)
            mark_A(x1) = ((1,1),(0,0)) x1 + (3,0)
            active_A(x1) = ((1,1),(0,0)) x1
            |2nd|_A(x1) = ((1,1),(0,0)) x1
    
      precedence:
      
        mark# = active# > cons > s > from = mark > active > |2nd|
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(cons) = []
        pi(from) = []
        pi(active#) = []
        pi(mark) = []
        pi(active) = []
        pi(|2nd|) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            mark#_A(x1) = (4,2)
            s_A(x1) = (2,0)
            cons_A(x1,x2) = (1,0)
            from_A(x1) = (5,4)
            active#_A(x1) = (3,1)
            mark_A(x1) = (2,3)
            active_A(x1) = (2,3)
            |2nd|_A(x1) = (2,3)
    
      precedence:
      
        mark# = from = active# > mark > s = active > |2nd| > cons
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(cons) = []
        pi(from) = []
        pi(active#) = []
        pi(mark) = []
        pi(active) = []
        pi(|2nd|) = []
    

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#(s(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(from(X)) -> active#(from(mark(X)))
p4: mark#(from(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, p4}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(from(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)

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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
            s_A(x1) = x1 + (1,3)
            from_A(x1) = ((1,0),(0,0)) x1 + (3,1)
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (3,1)
    
      precedence:
      
        mark# = s = from = cons
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
        pi(cons) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = (0,0)
            s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            from_A(x1) = (1,1)
            cons_A(x1,x2) = x2 + (1,1)
    
      precedence:
      
        mark# = from > s = cons
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
        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#(s(X)) -> mark#(X)
p2: mark#(from(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}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(from(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 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 + (1,2)
            s_A(x1) = x1 + (2,1)
            from_A(x1) = ((1,0),(1,1)) x1 + (2,1)
    
      precedence:
      
        from > mark# = s
      
      partial status:
    
        pi(mark#) = [1]
        pi(s) = [1]
        pi(from) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,0)) x1 + (0,2)
            s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            from_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        mark# = s = from
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
    

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(|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:
    
        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: 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 reduction pair:

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

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

The next rules are strictly ordered:

  p1, p2

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