YES

We show the termination of the TRS R:

  active(first(|0|(),X)) -> mark(nil())
  active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
  active(from(X)) -> mark(cons(X,from(s(X))))
  mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
  mark(|0|()) -> active(|0|())
  mark(nil()) -> active(nil())
  mark(s(X)) -> active(s(mark(X)))
  mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
  mark(from(X)) -> active(from(mark(X)))
  first(mark(X1),X2) -> first(X1,X2)
  first(X1,mark(X2)) -> first(X1,X2)
  first(active(X1),X2) -> first(X1,X2)
  first(X1,active(X2)) -> first(X1,X2)
  s(mark(X)) -> s(X)
  s(active(X)) -> s(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)

-- SCC decomposition.

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

p1: active#(first(|0|(),X)) -> mark#(nil())
p2: active#(first(s(X),cons(Y,Z))) -> mark#(cons(Y,first(X,Z)))
p3: active#(first(s(X),cons(Y,Z))) -> cons#(Y,first(X,Z))
p4: active#(first(s(X),cons(Y,Z))) -> first#(X,Z)
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#(first(X1,X2)) -> active#(first(mark(X1),mark(X2)))
p10: mark#(first(X1,X2)) -> first#(mark(X1),mark(X2))
p11: mark#(first(X1,X2)) -> mark#(X1)
p12: mark#(first(X1,X2)) -> mark#(X2)
p13: mark#(|0|()) -> active#(|0|())
p14: mark#(nil()) -> active#(nil())
p15: mark#(s(X)) -> active#(s(mark(X)))
p16: mark#(s(X)) -> s#(mark(X))
p17: mark#(s(X)) -> mark#(X)
p18: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p19: mark#(cons(X1,X2)) -> cons#(mark(X1),X2)
p20: mark#(cons(X1,X2)) -> mark#(X1)
p21: mark#(from(X)) -> active#(from(mark(X)))
p22: mark#(from(X)) -> from#(mark(X))
p23: mark#(from(X)) -> mark#(X)
p24: first#(mark(X1),X2) -> first#(X1,X2)
p25: first#(X1,mark(X2)) -> first#(X1,X2)
p26: first#(active(X1),X2) -> first#(X1,X2)
p27: first#(X1,active(X2)) -> first#(X1,X2)
p28: s#(mark(X)) -> s#(X)
p29: s#(active(X)) -> s#(X)
p30: cons#(mark(X1),X2) -> cons#(X1,X2)
p31: cons#(X1,mark(X2)) -> cons#(X1,X2)
p32: cons#(active(X1),X2) -> cons#(X1,X2)
p33: cons#(X1,active(X2)) -> cons#(X1,X2)
p34: from#(mark(X)) -> from#(X)
p35: from#(active(X)) -> from#(X)

and R consists of:

r1: active(first(|0|(),X)) -> mark(nil())
r2: active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
r5: mark(|0|()) -> active(|0|())
r6: mark(nil()) -> active(nil())
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(from(X)) -> active(from(mark(X)))
r10: first(mark(X1),X2) -> first(X1,X2)
r11: first(X1,mark(X2)) -> first(X1,X2)
r12: first(active(X1),X2) -> first(X1,X2)
r13: first(X1,active(X2)) -> first(X1,X2)
r14: s(mark(X)) -> s(X)
r15: s(active(X)) -> s(X)
r16: cons(mark(X1),X2) -> cons(X1,X2)
r17: cons(X1,mark(X2)) -> cons(X1,X2)
r18: cons(active(X1),X2) -> cons(X1,X2)
r19: cons(X1,active(X2)) -> cons(X1,X2)
r20: from(mark(X)) -> from(X)
r21: from(active(X)) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p2, p5, p9, p11, p12, p15, p17, p18, p20, p21, p23}
  {p30, p31, p32, p33}
  {p24, p25, p26, p27}
  {p34, p35}
  {p28, p29}


-- Reduction pair.

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

p1: mark#(from(X)) -> active#(from(mark(X)))
p2: active#(from(X)) -> mark#(cons(X,from(s(X))))
p3: mark#(from(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p6: active#(first(s(X),cons(Y,Z))) -> mark#(cons(Y,first(X,Z)))
p7: mark#(s(X)) -> mark#(X)
p8: mark#(s(X)) -> active#(s(mark(X)))
p9: mark#(first(X1,X2)) -> mark#(X2)
p10: mark#(first(X1,X2)) -> mark#(X1)
p11: mark#(first(X1,X2)) -> active#(first(mark(X1),mark(X2)))

and R consists of:

r1: active(first(|0|(),X)) -> mark(nil())
r2: active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
r5: mark(|0|()) -> active(|0|())
r6: mark(nil()) -> active(nil())
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(from(X)) -> active(from(mark(X)))
r10: first(mark(X1),X2) -> first(X1,X2)
r11: first(X1,mark(X2)) -> first(X1,X2)
r12: first(active(X1),X2) -> first(X1,X2)
r13: first(X1,active(X2)) -> first(X1,X2)
r14: s(mark(X)) -> s(X)
r15: s(active(X)) -> s(X)
r16: cons(mark(X1),X2) -> cons(X1,X2)
r17: cons(X1,mark(X2)) -> cons(X1,X2)
r18: cons(active(X1),X2) -> cons(X1,X2)
r19: cons(X1,active(X2)) -> cons(X1,X2)
r20: from(mark(X)) -> from(X)
r21: from(active(X)) -> from(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

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 + 14
          from_A(x1) = max{20, x1 + 19}
          active#_A(x1) = max{21, x1 + 14}
          mark_A(x1) = x1
          cons_A(x1,x2) = max{x1 + 13, x2 - 17}
          s_A(x1) = max{18, x1 + 14}
          first_A(x1,x2) = max{x1 + 7, x2 + 7}
          active_A(x1) = max{4, x1}
          |0|_A = 5
          nil_A = 5
    
      precedence:
      
        from = mark = active = |0| = nil > cons = first > mark# = active# = s
      
      partial status:
    
        pi(mark#) = []
        pi(from) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        pi(s) = [1]
        pi(first) = [1]
        pi(active) = []
        pi(|0|) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 147
          from_A(x1) = 235
          active#_A(x1) = 147
          mark_A(x1) = 140
          cons_A(x1,x2) = 89
          s_A(x1) = 141
          first_A(x1,x2) = 143
          active_A(x1) = 140
          |0|_A = 121
          nil_A = 144
    
      precedence:
      
        from = mark = cons = s = first = active = |0| = nil > mark# = active#
      
      partial status:
    
        pi(mark#) = []
        pi(from) = []
        pi(active#) = []
        pi(mark) = []
        pi(cons) = []
        pi(s) = []
        pi(first) = []
        pi(active) = []
        pi(|0|) = []
        pi(nil) = []
    

The next rules are strictly ordered:

  p10

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: active(first(|0|(),X)) -> mark(nil())
r2: active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
r5: mark(|0|()) -> active(|0|())
r6: mark(nil()) -> active(nil())
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(from(X)) -> active(from(mark(X)))
r10: first(mark(X1),X2) -> first(X1,X2)
r11: first(X1,mark(X2)) -> first(X1,X2)
r12: first(active(X1),X2) -> first(X1,X2)
r13: first(X1,active(X2)) -> first(X1,X2)
r14: s(mark(X)) -> s(X)
r15: s(active(X)) -> s(X)
r16: cons(mark(X1),X2) -> cons(X1,X2)
r17: cons(X1,mark(X2)) -> cons(X1,X2)
r18: cons(active(X1),X2) -> cons(X1,X2)
r19: cons(X1,active(X2)) -> cons(X1,X2)
r20: from(mark(X)) -> from(X)
r21: from(active(X)) -> from(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#(from(X)) -> active#(from(mark(X)))
p2: active#(first(s(X),cons(Y,Z))) -> mark#(cons(Y,first(X,Z)))
p3: mark#(first(X1,X2)) -> active#(first(mark(X1),mark(X2)))
p4: active#(from(X)) -> mark#(cons(X,from(s(X))))
p5: mark#(first(X1,X2)) -> mark#(X2)
p6: mark#(s(X)) -> active#(s(mark(X)))
p7: mark#(s(X)) -> mark#(X)
p8: mark#(cons(X1,X2)) -> active#(cons(mark(X1),X2))
p9: mark#(cons(X1,X2)) -> mark#(X1)
p10: mark#(from(X)) -> mark#(X)

and R consists of:

r1: active(first(|0|(),X)) -> mark(nil())
r2: active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
r5: mark(|0|()) -> active(|0|())
r6: mark(nil()) -> active(nil())
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(from(X)) -> active(from(mark(X)))
r10: first(mark(X1),X2) -> first(X1,X2)
r11: first(X1,mark(X2)) -> first(X1,X2)
r12: first(active(X1),X2) -> first(X1,X2)
r13: first(X1,active(X2)) -> first(X1,X2)
r14: s(mark(X)) -> s(X)
r15: s(active(X)) -> s(X)
r16: cons(mark(X1),X2) -> cons(X1,X2)
r17: cons(X1,mark(X2)) -> cons(X1,X2)
r18: cons(active(X1),X2) -> cons(X1,X2)
r19: cons(X1,active(X2)) -> cons(X1,X2)
r20: from(mark(X)) -> from(X)
r21: from(active(X)) -> from(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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{2, x1}
          from_A(x1) = max{9, x1 + 5}
          active#_A(x1) = x1
          mark_A(x1) = max{1, x1}
          first_A(x1,x2) = max{10, x1 + 3, x2 + 5}
          s_A(x1) = max{4, x1}
          cons_A(x1,x2) = max{4, x1 + 3, x2}
          active_A(x1) = max{4, x1}
          |0|_A = 5
          nil_A = 4
    
      precedence:
      
        |0| > from = first > mark# = mark > s = active = nil > active# = cons
      
      partial status:
    
        pi(mark#) = [1]
        pi(from) = []
        pi(active#) = [1]
        pi(mark) = [1]
        pi(first) = []
        pi(s) = [1]
        pi(cons) = []
        pi(active) = [1]
        pi(|0|) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 159
          from_A(x1) = 179
          active#_A(x1) = max{151, x1 + 6}
          mark_A(x1) = max{160, x1 + 8}
          first_A(x1,x2) = 179
          s_A(x1) = 152
          cons_A(x1,x2) = 78
          active_A(x1) = max{160, x1 - 19}
          |0|_A = 0
          nil_A = 180
    
      precedence:
      
        from > mark = s = active = nil > mark# = active# = |0| > first = cons
      
      partial status:
    
        pi(mark#) = []
        pi(from) = []
        pi(active#) = [1]
        pi(mark) = []
        pi(first) = []
        pi(s) = []
        pi(cons) = []
        pi(active) = []
        pi(|0|) = []
        pi(nil) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: active(first(|0|(),X)) -> mark(nil())
r2: active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
r5: mark(|0|()) -> active(|0|())
r6: mark(nil()) -> active(nil())
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(from(X)) -> active(from(mark(X)))
r10: first(mark(X1),X2) -> first(X1,X2)
r11: first(X1,mark(X2)) -> first(X1,X2)
r12: first(active(X1),X2) -> first(X1,X2)
r13: first(X1,active(X2)) -> first(X1,X2)
r14: s(mark(X)) -> s(X)
r15: s(active(X)) -> s(X)
r16: cons(mark(X1),X2) -> cons(X1,X2)
r17: cons(X1,mark(X2)) -> cons(X1,X2)
r18: cons(active(X1),X2) -> cons(X1,X2)
r19: cons(X1,active(X2)) -> cons(X1,X2)
r20: from(mark(X)) -> from(X)
r21: from(active(X)) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p3, p5, p6}


-- Reduction pair.

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

p1: mark#(first(X1,X2)) -> mark#(X2)
p2: mark#(from(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(s(X)) -> mark#(X)

and R consists of:

r1: active(first(|0|(),X)) -> mark(nil())
r2: active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
r5: mark(|0|()) -> active(|0|())
r6: mark(nil()) -> active(nil())
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(from(X)) -> active(from(mark(X)))
r10: first(mark(X1),X2) -> first(X1,X2)
r11: first(X1,mark(X2)) -> first(X1,X2)
r12: first(active(X1),X2) -> first(X1,X2)
r13: first(X1,active(X2)) -> first(X1,X2)
r14: s(mark(X)) -> s(X)
r15: s(active(X)) -> s(X)
r16: cons(mark(X1),X2) -> cons(X1,X2)
r17: cons(X1,mark(X2)) -> cons(X1,X2)
r18: cons(active(X1),X2) -> cons(X1,X2)
r19: cons(X1,active(X2)) -> cons(X1,X2)
r20: from(mark(X)) -> from(X)
r21: from(active(X)) -> from(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) = max{3, x1 + 2}
          first_A(x1,x2) = max{1, x2}
          from_A(x1) = max{1, x1}
          cons_A(x1,x2) = max{x1, x2}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        mark# = first = from = cons = s
      
      partial status:
    
        pi(mark#) = [1]
        pi(first) = [2]
        pi(from) = [1]
        pi(cons) = [1, 2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 2
          first_A(x1,x2) = x2
          from_A(x1) = x1
          cons_A(x1,x2) = max{x1, x2}
          s_A(x1) = x1
    
      precedence:
      
        mark# = first = from = cons = s
      
      partial status:
    
        pi(mark#) = [1]
        pi(first) = [2]
        pi(from) = [1]
        pi(cons) = [1, 2]
        pi(s) = [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(first(|0|(),X)) -> mark(nil())
r2: active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
r5: mark(|0|()) -> active(|0|())
r6: mark(nil()) -> active(nil())
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(from(X)) -> active(from(mark(X)))
r10: first(mark(X1),X2) -> first(X1,X2)
r11: first(X1,mark(X2)) -> first(X1,X2)
r12: first(active(X1),X2) -> first(X1,X2)
r13: first(X1,active(X2)) -> first(X1,X2)
r14: s(mark(X)) -> s(X)
r15: s(active(X)) -> s(X)
r16: cons(mark(X1),X2) -> cons(X1,X2)
r17: cons(X1,mark(X2)) -> cons(X1,X2)
r18: cons(active(X1),X2) -> cons(X1,X2)
r19: cons(X1,active(X2)) -> cons(X1,X2)
r20: from(mark(X)) -> from(X)
r21: from(active(X)) -> from(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: first#(mark(X1),X2) -> first#(X1,X2)
p2: first#(X1,active(X2)) -> first#(X1,X2)
p3: first#(active(X1),X2) -> first#(X1,X2)
p4: first#(X1,mark(X2)) -> first#(X1,X2)

and R consists of:

r1: active(first(|0|(),X)) -> mark(nil())
r2: active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
r5: mark(|0|()) -> active(|0|())
r6: mark(nil()) -> active(nil())
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(from(X)) -> active(from(mark(X)))
r10: first(mark(X1),X2) -> first(X1,X2)
r11: first(X1,mark(X2)) -> first(X1,X2)
r12: first(active(X1),X2) -> first(X1,X2)
r13: first(X1,active(X2)) -> first(X1,X2)
r14: s(mark(X)) -> s(X)
r15: s(active(X)) -> s(X)
r16: cons(mark(X1),X2) -> cons(X1,X2)
r17: cons(X1,mark(X2)) -> cons(X1,X2)
r18: cons(active(X1),X2) -> cons(X1,X2)
r19: cons(X1,active(X2)) -> cons(X1,X2)
r20: from(mark(X)) -> from(X)
r21: from(active(X)) -> from(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:
        
          first#_A(x1,x2) = max{2, x1 + 1, x2 + 1}
          mark_A(x1) = max{1, x1}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        first# = mark = active
      
      partial status:
    
        pi(first#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          first#_A(x1,x2) = max{x1 + 1, x2 - 1}
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        first# = mark = active
      
      partial status:
    
        pi(first#) = [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(first(|0|(),X)) -> mark(nil())
r2: active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
r5: mark(|0|()) -> active(|0|())
r6: mark(nil()) -> active(nil())
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(from(X)) -> active(from(mark(X)))
r10: first(mark(X1),X2) -> first(X1,X2)
r11: first(X1,mark(X2)) -> first(X1,X2)
r12: first(active(X1),X2) -> first(X1,X2)
r13: first(X1,active(X2)) -> first(X1,X2)
r14: s(mark(X)) -> s(X)
r15: s(active(X)) -> s(X)
r16: cons(mark(X1),X2) -> cons(X1,X2)
r17: cons(X1,mark(X2)) -> cons(X1,X2)
r18: cons(active(X1),X2) -> cons(X1,X2)
r19: cons(X1,active(X2)) -> cons(X1,X2)
r20: from(mark(X)) -> from(X)
r21: from(active(X)) -> from(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(first(|0|(),X)) -> mark(nil())
r2: active(first(s(X),cons(Y,Z))) -> mark(cons(Y,first(X,Z)))
r3: active(from(X)) -> mark(cons(X,from(s(X))))
r4: mark(first(X1,X2)) -> active(first(mark(X1),mark(X2)))
r5: mark(|0|()) -> active(|0|())
r6: mark(nil()) -> active(nil())
r7: mark(s(X)) -> active(s(mark(X)))
r8: mark(cons(X1,X2)) -> active(cons(mark(X1),X2))
r9: mark(from(X)) -> active(from(mark(X)))
r10: first(mark(X1),X2) -> first(X1,X2)
r11: first(X1,mark(X2)) -> first(X1,X2)
r12: first(active(X1),X2) -> first(X1,X2)
r13: first(X1,active(X2)) -> first(X1,X2)
r14: s(mark(X)) -> s(X)
r15: s(active(X)) -> s(X)
r16: cons(mark(X1),X2) -> cons(X1,X2)
r17: cons(X1,mark(X2)) -> cons(X1,X2)
r18: cons(active(X1),X2) -> cons(X1,X2)
r19: cons(X1,active(X2)) -> cons(X1,X2)
r20: from(mark(X)) -> from(X)
r21: from(active(X)) -> from(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.