YES

We show the termination of the TRS R:

  active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
  active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
  active(from(X)) -> mark(cons(X,from(s(X))))
  mark(|2nd|(X)) -> active(|2nd|(mark(X)))
  mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
  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)
  cons1(mark(X1),X2) -> cons1(X1,X2)
  cons1(X1,mark(X2)) -> cons1(X1,X2)
  cons1(active(X1),X2) -> cons1(X1,X2)
  cons1(X1,active(X2)) -> cons1(X1,X2)
  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|(cons1(X,cons(Y,Z)))) -> mark#(Y)
p2: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p3: active#(|2nd|(cons(X,X1))) -> |2nd|#(cons1(X,X1))
p4: active#(|2nd|(cons(X,X1))) -> cons1#(X,X1)
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: active#(from(X)) -> cons#(X,from(s(X)))
p7: active#(from(X)) -> from#(s(X))
p8: active#(from(X)) -> s#(X)
p9: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p10: mark#(|2nd|(X)) -> |2nd|#(mark(X))
p11: mark#(|2nd|(X)) -> mark#(X)
p12: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p13: mark#(cons1(X1,X2)) -> cons1#(mark(X1),mark(X2))
p14: mark#(cons1(X1,X2)) -> mark#(X1)
p15: mark#(cons1(X1,X2)) -> mark#(X2)
p16: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p17: mark#(cons(X1,X2)) -> cons#(mark(X1),X2)
p18: mark#(cons(X1,X2)) -> mark#(X1)
p19: mark#(from(X)) -> active#(from(mark(X)))
p20: mark#(from(X)) -> from#(mark(X))
p21: mark#(from(X)) -> mark#(X)
p22: mark#(s(X)) -> active#(s(mark(X)))
p23: mark#(s(X)) -> s#(mark(X))
p24: mark#(s(X)) -> mark#(X)
p25: |2nd|#(mark(X)) -> |2nd|#(X)
p26: |2nd|#(active(X)) -> |2nd|#(X)
p27: cons1#(mark(X1),X2) -> cons1#(X1,X2)
p28: cons1#(X1,mark(X2)) -> cons1#(X1,X2)
p29: cons1#(active(X1),X2) -> cons1#(X1,X2)
p30: cons1#(X1,active(X2)) -> cons1#(X1,X2)
p31: cons#(mark(X1),X2) -> cons#(X1,X2)
p32: cons#(X1,mark(X2)) -> cons#(X1,X2)
p33: cons#(active(X1),X2) -> cons#(X1,X2)
p34: cons#(X1,active(X2)) -> cons#(X1,X2)
p35: from#(mark(X)) -> from#(X)
p36: from#(active(X)) -> from#(X)
p37: s#(mark(X)) -> s#(X)
p38: s#(active(X)) -> s#(X)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: s(active(X)) -> s(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p5, p9, p11, p12, p14, p15, p16, p18, p19, p21, p22, p24}
  {p25, p26}
  {p27, p28, p29, p30}
  {p31, p32, p33, p34}
  {p35, p36}
  {p37, p38}


-- Reduction pair.

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

p1: active#(|2nd|(cons1(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: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p10: mark#(cons1(X1,X2)) -> mark#(X2)
p11: mark#(cons1(X1,X2)) -> mark#(X1)
p12: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p13: mark#(|2nd|(X)) -> mark#(X)
p14: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = x1 + 11
          |2nd|_A(x1) = max{5, x1 + 3}
          cons1_A(x1,x2) = max{6, x1, x2}
          cons_A(x1,x2) = max{7, x1, x2}
          mark#_A(x1) = x1 + 11
          s_A(x1) = max{7, x1}
          mark_A(x1) = max{1, x1}
          from_A(x1) = max{7, x1}
          active_A(x1) = x1
    
      precedence:
      
        active# = |2nd| = cons1 = cons = mark# = s = mark = from = active
      
      partial status:
    
        pi(active#) = []
        pi(|2nd|) = []
        pi(cons1) = []
        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) = 27
          |2nd|_A(x1) = 0
          cons1_A(x1,x2) = 0
          cons_A(x1,x2) = 0
          mark#_A(x1) = 27
          s_A(x1) = 0
          mark_A(x1) = 26
          from_A(x1) = 0
          active_A(x1) = x1 + 26
    
      precedence:
      
        active# = mark# > from > |2nd| = cons1 = cons = s = mark = active
      
      partial status:
    
        pi(active#) = []
        pi(|2nd|) = []
        pi(cons1) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(mark) = []
        pi(from) = []
        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#(s(X)) -> active#(s(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: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p7: mark#(cons(X1,X2)) -> mark#(X1)
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p9: mark#(cons1(X1,X2)) -> mark#(X2)
p10: mark#(cons1(X1,X2)) -> mark#(X1)
p11: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p12: mark#(|2nd|(X)) -> mark#(X)
p13: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: s(active(X)) -> s(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13}


-- 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#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p4: mark#(|2nd|(X)) -> mark#(X)
p5: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p6: active#(from(X)) -> mark#(cons(X,from(s(X))))
p7: mark#(cons1(X1,X2)) -> mark#(X1)
p8: mark#(cons1(X1,X2)) -> mark#(X2)
p9: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p10: mark#(cons(X1,X2)) -> mark#(X1)
p11: mark#(from(X)) -> active#(from(mark(X)))
p12: mark#(from(X)) -> mark#(X)
p13: mark#(s(X)) -> active#(s(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 4
          s_A(x1) = max{6, x1}
          |2nd|_A(x1) = max{12, x1 + 7}
          active#_A(x1) = x1 + 4
          mark_A(x1) = max{4, x1}
          cons_A(x1,x2) = max{7, x1, x2}
          cons1_A(x1,x2) = max{6, x1, x2}
          from_A(x1) = max{7, x1}
          active_A(x1) = max{6, x1}
    
      precedence:
      
        mark# = s = |2nd| = active# = mark = cons = cons1 = from = active
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(|2nd|) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        pi(cons1) = []
        pi(from) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 111
          s_A(x1) = 109
          |2nd|_A(x1) = 124
          active#_A(x1) = 111
          mark_A(x1) = 96
          cons_A(x1,x2) = 81
          cons1_A(x1,x2) = 86
          from_A(x1) = 111
          active_A(x1) = 96
    
      precedence:
      
        mark# = s = |2nd| = active# = mark = from = active > cons1 > cons
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(|2nd|) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        pi(cons1) = []
        pi(from) = []
        pi(active) = []
    

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#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p3: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p4: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(cons1(X1,X2)) -> mark#(X1)
p7: mark#(cons1(X1,X2)) -> mark#(X2)
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(from(X)) -> active#(from(mark(X)))
p11: mark#(from(X)) -> mark#(X)
p12: mark#(s(X)) -> active#(s(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: s(active(X)) -> s(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

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#(from(X)) -> mark#(X)
p5: mark#(from(X)) -> active#(from(mark(X)))
p6: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p7: mark#(cons(X1,X2)) -> mark#(X1)
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p9: mark#(cons1(X1,X2)) -> mark#(X2)
p10: mark#(cons1(X1,X2)) -> mark#(X1)
p11: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p12: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 4
          s_A(x1) = max{1, x1}
          active#_A(x1) = x1 + 4
          mark_A(x1) = x1
          from_A(x1) = max{17, x1 + 15}
          cons_A(x1,x2) = max{x1 + 15, x2}
          |2nd|_A(x1) = max{0, x1 - 5}
          cons1_A(x1,x2) = max{8, x1, x2}
          active_A(x1) = x1
    
      precedence:
      
        s = cons1 > mark# = active# = from > mark = cons = |2nd| = active
      
      partial status:
    
        pi(mark#) = [1]
        pi(s) = []
        pi(active#) = [1]
        pi(mark) = []
        pi(from) = []
        pi(cons) = []
        pi(|2nd|) = []
        pi(cons1) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 46
          s_A(x1) = 82
          active#_A(x1) = 46
          mark_A(x1) = 49
          from_A(x1) = 237
          cons_A(x1,x2) = 43
          |2nd|_A(x1) = 45
          cons1_A(x1,x2) = 44
          active_A(x1) = 49
    
      precedence:
      
        mark# = active# > s = mark = from = active > cons1 > cons = |2nd|
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(active#) = []
        pi(mark) = []
        pi(from) = []
        pi(cons) = []
        pi(|2nd|) = []
        pi(cons1) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p4, p7

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#(from(X)) -> active#(from(mark(X)))
p5: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p6: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p7: mark#(cons1(X1,X2)) -> mark#(X2)
p8: mark#(cons1(X1,X2)) -> mark#(X1)
p9: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p10: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: s(active(X)) -> s(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p3: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p4: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(cons1(X1,X2)) -> mark#(X1)
p7: mark#(cons1(X1,X2)) -> mark#(X2)
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p9: mark#(from(X)) -> active#(from(mark(X)))
p10: mark#(s(X)) -> active#(s(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 17
          s_A(x1) = max{3, x1}
          |2nd|_A(x1) = max{0, x1 - 4}
          active#_A(x1) = x1 + 17
          mark_A(x1) = max{3, x1}
          cons_A(x1,x2) = max{8, x1 + 5, x2}
          cons1_A(x1,x2) = max{8, x1 + 3, x2}
          from_A(x1) = max{11, x1 + 7}
          active_A(x1) = x1
    
      precedence:
      
        s = cons1 > mark# = active# > from > cons > |2nd| = mark = active
      
      partial status:
    
        pi(mark#) = [1]
        pi(s) = []
        pi(|2nd|) = []
        pi(active#) = [1]
        pi(mark) = []
        pi(cons) = []
        pi(cons1) = []
        pi(from) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 145
          s_A(x1) = 144
          |2nd|_A(x1) = 140
          active#_A(x1) = 145
          mark_A(x1) = 144
          cons_A(x1,x2) = 140
          cons1_A(x1,x2) = 144
          from_A(x1) = 144
          active_A(x1) = 144
    
      precedence:
      
        cons1 > mark = active > from > mark# = s = active# > cons > |2nd|
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(|2nd|) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        pi(cons1) = []
        pi(from) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p6

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#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p3: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p4: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(cons1(X1,X2)) -> mark#(X2)
p7: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p8: mark#(from(X)) -> active#(from(mark(X)))
p9: mark#(s(X)) -> active#(s(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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: 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#(from(X)) -> active#(from(mark(X)))
p5: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p6: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p7: mark#(cons1(X1,X2)) -> mark#(X2)
p8: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p9: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 16
          s_A(x1) = x1
          active#_A(x1) = x1 + 16
          mark_A(x1) = x1
          from_A(x1) = max{24, x1 + 7}
          cons_A(x1,x2) = max{24, x1 + 7, x2}
          |2nd|_A(x1) = max{10, x1 - 5}
          cons1_A(x1,x2) = max{9, x1 + 6, x2}
          active_A(x1) = x1
    
      precedence:
      
        mark# = s = active# = mark = from = cons = |2nd| = cons1 = active
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(active#) = []
        pi(mark) = []
        pi(from) = []
        pi(cons) = []
        pi(|2nd|) = []
        pi(cons1) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 51
          s_A(x1) = 50
          active#_A(x1) = max{21, x1}
          mark_A(x1) = 67
          from_A(x1) = 51
          cons_A(x1,x2) = 51
          |2nd|_A(x1) = 51
          cons1_A(x1,x2) = 4
          active_A(x1) = 67
    
      precedence:
      
        |2nd| > mark# = active# = mark = active > from > s > cons > cons1
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(active#) = []
        pi(mark) = []
        pi(from) = []
        pi(cons) = []
        pi(|2nd|) = []
        pi(cons1) = []
        pi(active) = []
    

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#(s(X)) -> mark#(X)
p2: active#(from(X)) -> mark#(cons(X,from(s(X))))
p3: mark#(from(X)) -> active#(from(mark(X)))
p4: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p5: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: mark#(cons1(X1,X2)) -> mark#(X2)
p7: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p8: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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: mark#(s(X)) -> mark#(X)
p2: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p3: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p4: mark#(cons1(X1,X2)) -> active#(cons1(mark(X1),mark(X2)))
p5: active#(from(X)) -> mark#(cons(X,from(s(X))))
p6: mark#(cons1(X1,X2)) -> mark#(X2)
p7: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p8: mark#(from(X)) -> active#(from(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 3
          s_A(x1) = x1
          |2nd|_A(x1) = max{0, x1 - 5}
          active#_A(x1) = x1 + 3
          mark_A(x1) = x1
          cons_A(x1,x2) = max{x1 + 9, x2}
          cons1_A(x1,x2) = max{x1 + 6, x2}
          from_A(x1) = x1 + 10
          active_A(x1) = x1
    
      precedence:
      
        s = cons1 = from > cons > |2nd| > mark# = active# > mark = active
      
      partial status:
    
        pi(mark#) = [1]
        pi(s) = []
        pi(|2nd|) = []
        pi(active#) = [1]
        pi(mark) = []
        pi(cons) = []
        pi(cons1) = []
        pi(from) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 19
          s_A(x1) = 19
          |2nd|_A(x1) = 19
          active#_A(x1) = max{11, x1}
          mark_A(x1) = 19
          cons_A(x1,x2) = 19
          cons1_A(x1,x2) = 10
          from_A(x1) = 19
          active_A(x1) = 19
    
      precedence:
      
        from > mark# = s = |2nd| = active# = mark = cons = active > cons1
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(|2nd|) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        pi(cons1) = []
        pi(from) = []
        pi(active) = []
    

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#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p3: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p4: active#(from(X)) -> mark#(cons(X,from(s(X))))
p5: mark#(cons1(X1,X2)) -> mark#(X2)
p6: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p7: mark#(from(X)) -> active#(from(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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: mark#(s(X)) -> mark#(X)
p2: mark#(from(X)) -> active#(from(mark(X)))
p3: active#(from(X)) -> mark#(cons(X,from(s(X))))
p4: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p5: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p6: mark#(cons1(X1,X2)) -> mark#(X2)
p7: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 4
          s_A(x1) = x1
          from_A(x1) = x1
          active#_A(x1) = x1 + 4
          mark_A(x1) = x1
          cons_A(x1,x2) = max{x1 - 2, x2}
          |2nd|_A(x1) = x1 + 7
          cons1_A(x1,x2) = x2
          active_A(x1) = x1
    
      precedence:
      
        s = cons1 > from > mark# = active# > mark = cons = |2nd| = active
      
      partial status:
    
        pi(mark#) = [1]
        pi(s) = []
        pi(from) = []
        pi(active#) = [1]
        pi(mark) = []
        pi(cons) = []
        pi(|2nd|) = []
        pi(cons1) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 12
          s_A(x1) = 9
          from_A(x1) = 39
          active#_A(x1) = max{10, x1 - 27}
          mark_A(x1) = 0
          cons_A(x1,x2) = 25
          |2nd|_A(x1) = 39
          cons1_A(x1,x2) = 0
          active_A(x1) = 0
    
      precedence:
      
        cons > mark# = active# = |2nd| > from = mark = cons1 = active > s
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        pi(|2nd|) = []
        pi(cons1) = []
        pi(active) = []
    

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#(from(X)) -> active#(from(mark(X)))
p3: active#(from(X)) -> mark#(cons(X,from(s(X))))
p4: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p5: mark#(cons1(X1,X2)) -> mark#(X2)
p6: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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#(s(X)) -> mark#(X)
p2: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p3: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p4: mark#(cons1(X1,X2)) -> mark#(X2)
p5: mark#(from(X)) -> active#(from(mark(X)))
p6: active#(from(X)) -> mark#(cons(X,from(s(X))))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 6
          s_A(x1) = x1
          |2nd|_A(x1) = x1 + 11
          active#_A(x1) = max{9, x1 + 6}
          mark_A(x1) = x1
          cons_A(x1,x2) = max{x1 + 3, x2}
          cons1_A(x1,x2) = max{x1 + 2, x2}
          from_A(x1) = x1 + 3
          active_A(x1) = x1
    
      precedence:
      
        s = cons1 > from > mark# = |2nd| = active# = mark = cons = active
      
      partial status:
    
        pi(mark#) = [1]
        pi(s) = []
        pi(|2nd|) = []
        pi(active#) = [1]
        pi(mark) = []
        pi(cons) = []
        pi(cons1) = []
        pi(from) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 54
          s_A(x1) = 85
          |2nd|_A(x1) = 167
          active#_A(x1) = 54
          mark_A(x1) = 70
          cons_A(x1,x2) = 70
          cons1_A(x1,x2) = 53
          from_A(x1) = 81
          active_A(x1) = 70
    
      precedence:
      
        s = cons > |2nd| = mark = cons1 = from = active > mark# = active#
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(|2nd|) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        pi(cons1) = []
        pi(from) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p6

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#(|2nd|(X)) -> active#(|2nd|(mark(X)))
p3: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p4: mark#(cons1(X1,X2)) -> mark#(X2)
p5: mark#(from(X)) -> active#(from(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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#(from(X)) -> active#(from(mark(X)))
p3: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p4: mark#(cons1(X1,X2)) -> mark#(X2)
p5: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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, r17, r18, r19, r20, r21, r22

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 5
          s_A(x1) = x1
          from_A(x1) = x1 + 3
          active#_A(x1) = x1 + 5
          mark_A(x1) = x1
          |2nd|_A(x1) = x1
          cons_A(x1,x2) = max{x1 + 3, x2}
          cons1_A(x1,x2) = max{x1 + 1, x2}
          active_A(x1) = x1
    
      precedence:
      
        mark# = s = from = active# = mark = |2nd| = cons = cons1 = active
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
        pi(active#) = []
        pi(mark) = []
        pi(|2nd|) = []
        pi(cons) = []
        pi(cons1) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 63
          s_A(x1) = 0
          from_A(x1) = 46
          active#_A(x1) = max{52, x1 - 4}
          mark_A(x1) = max{70, x1 + 3}
          |2nd|_A(x1) = 67
          cons_A(x1,x2) = 62
          cons1_A(x1,x2) = 67
          active_A(x1) = 70
    
      precedence:
      
        from = |2nd| > mark# = active# = cons > s = mark = cons1 = active
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
        pi(active#) = []
        pi(mark) = []
        pi(|2nd|) = []
        pi(cons) = []
        pi(cons1) = []
        pi(active) = []
    

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#(s(X)) -> mark#(X)
p2: active#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p3: mark#(cons1(X1,X2)) -> mark#(X2)
p4: mark#(|2nd|(X)) -> active#(|2nd|(mark(X)))

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: s(active(X)) -> s(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- 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#(|2nd|(cons(X,X1))) -> mark#(|2nd|(cons1(X,X1)))
p4: mark#(cons1(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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, r17, r18, r19, r20, r21, r22

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) = ((0,1),(0,1)) x1 + (2,0)
            s_A(x1) = ((0,0),(0,1)) x1 + (1,1)
            |2nd|_A(x1) = ((1,0),(1,0)) x1 + (4,4)
            active#_A(x1) = x1 + (1,0)
            mark_A(x1) = x1
            cons_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(0,0)) x2 + (6,1)
            cons1_A(x1,x2) = ((0,1),(0,1)) x2 + (3,11)
            active_A(x1) = x1
            from_A(x1) = ((0,1),(1,1)) x1 + (10,1)
    
      precedence:
      
        mark > mark# = cons1 > s = |2nd| = active# = cons = active = from
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(|2nd|) = []
        pi(active#) = [1]
        pi(mark) = []
        pi(cons) = []
        pi(cons1) = []
        pi(active) = []
        pi(from) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            mark#_A(x1) = (3,2)
            s_A(x1) = (4,1)
            |2nd|_A(x1) = (2,0)
            active#_A(x1) = (1,1)
            mark_A(x1) = (1,0)
            cons_A(x1,x2) = (0,0)
            cons1_A(x1,x2) = (0,0)
            active_A(x1) = (1,0)
            from_A(x1) = (5,1)
    
      precedence:
      
        mark# = |2nd| = mark = active > from > cons1 > cons > active# > s
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(|2nd|) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        pi(cons1) = []
        pi(active) = []
        pi(from) = []
    

The next rules are strictly ordered:

  p2, 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#(cons1(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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#(cons1(X1,X2)) -> mark#(X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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:
        
          mark#_A(x1) = x1 + 4
          s_A(x1) = max{1, x1}
          cons1_A(x1,x2) = max{x1 - 1, x2}
    
      precedence:
      
        mark# = s = cons1
      
      partial status:
    
        pi(mark#) = [1]
        pi(s) = [1]
        pi(cons1) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{1, x1}
          s_A(x1) = max{1, x1}
          cons1_A(x1,x2) = x2
    
      precedence:
      
        mark# = s = cons1
      
      partial status:
    
        pi(mark#) = [1]
        pi(s) = [1]
        pi(cons1) = [2]
    

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

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

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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.

-- Reduction pair.

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

p1: cons1#(mark(X1),X2) -> cons1#(X1,X2)
p2: cons1#(X1,active(X2)) -> cons1#(X1,X2)
p3: cons1#(active(X1),X2) -> cons1#(X1,X2)
p4: cons1#(X1,mark(X2)) -> cons1#(X1,X2)

and R consists of:

r1: active(|2nd|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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:
        
          cons1#_A(x1,x2) = max{2, x1 + 1, x2 + 1}
          mark_A(x1) = max{1, x1}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        cons1# = mark = active
      
      partial status:
    
        pi(cons1#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          cons1#_A(x1,x2) = max{x1 + 1, x2 - 1}
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        cons1# = mark = active
      
      partial status:
    
        pi(cons1#) = [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: 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|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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|(cons1(X,cons(Y,Z)))) -> mark(Y)
r2: active(|2nd|(cons(X,X1))) -> mark(|2nd|(cons1(X,X1)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(|2nd|(X)) -> active(|2nd|(mark(X)))
r5: mark(cons1(X1,X2)) -> active(cons1(mark(X1),mark(X2)))
r6: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r7: mark(from(X)) -> active(from(mark(X)))
r8: mark(s(X)) -> active(s(mark(X)))
r9: |2nd|(mark(X)) -> |2nd|(X)
r10: |2nd|(active(X)) -> |2nd|(X)
r11: cons1(mark(X1),X2) -> cons1(X1,X2)
r12: cons1(X1,mark(X2)) -> cons1(X1,X2)
r13: cons1(active(X1),X2) -> cons1(X1,X2)
r14: cons1(X1,active(X2)) -> cons1(X1,X2)
r15: cons(mark(X1),X2) -> cons(X1,X2)
r16: cons(X1,mark(X2)) -> cons(X1,X2)
r17: cons(active(X1),X2) -> cons(X1,X2)
r18: cons(X1,active(X2)) -> cons(X1,X2)
r19: from(mark(X)) -> from(X)
r20: from(active(X)) -> from(X)
r21: s(mark(X)) -> s(X)
r22: 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.