YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(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: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(from(X)) -> mark#(X)
p5: mark#(from(X)) -> a__from#(mark(X))
p6: a__from#(X) -> mark#(X)
p7: mark#(first(X1,X2)) -> mark#(X2)
p8: mark#(first(X1,X2)) -> mark#(X1)
p9: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__first#_A(x1,x2) = ((0,0),(1,0)) x2 + (5,10)
            s_A(x1) = ((1,0),(1,0)) x1 + (6,24)
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (0,10)
            mark#_A(x1) = ((0,0),(1,0)) x1 + (5,6)
            from_A(x1) = ((1,0),(1,0)) x1 + (6,8)
            a__from#_A(x1) = ((0,0),(1,0)) x1 + (5,7)
            mark_A(x1) = ((1,0),(1,1)) x1 + (0,9)
            first_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (4,0)
            a__first_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (4,1)
            |0|_A() = (2,2)
            nil_A() = (1,1)
            a__from_A(x1) = ((1,0),(1,0)) x1 + (6,23)
    
      precedence:
      
        a__first# = mark# = a__from# > nil > s = mark = a__first = a__from > from > cons = first = |0|
      
      partial status:
    
        pi(a__first#) = []
        pi(s) = []
        pi(cons) = []
        pi(mark#) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = [1]
        pi(first) = []
        pi(a__first) = [1]
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__first#_A(x1,x2) = (1,1)
            s_A(x1) = (3,7)
            cons_A(x1,x2) = (8,12)
            mark#_A(x1) = (1,1)
            from_A(x1) = (2,2)
            a__from#_A(x1) = (1,1)
            mark_A(x1) = ((0,0),(1,0)) x1 + (10,3)
            first_A(x1,x2) = (7,8)
            a__first_A(x1,x2) = (10,9)
            |0|_A() = (9,11)
            nil_A() = (9,10)
            a__from_A(x1) = (9,6)
    
      precedence:
      
        mark > a__first = a__from > from > a__first# = s = cons = mark# = a__from# = first > |0| = nil
      
      partial status:
    
        pi(a__first#) = []
        pi(s) = []
        pi(cons) = []
        pi(mark#) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = []
        pi(first) = []
        pi(a__first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = []
    

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

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(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: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p3: mark#(first(X1,X2)) -> mark#(X1)
p4: mark#(first(X1,X2)) -> mark#(X2)
p5: mark#(from(X)) -> a__from#(mark(X))
p6: a__from#(X) -> mark#(X)
p7: mark#(from(X)) -> mark#(X)
p8: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__first#_A(x1,x2) = ((1,0),(0,0)) x2 + (8,4)
            s_A(x1) = (0,0)
            cons_A(x1,x2) = x1 + (0,1)
            mark#_A(x1) = x1 + (3,4)
            first_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (5,0)
            mark_A(x1) = ((1,0),(1,1)) x1 + (0,2)
            from_A(x1) = ((1,0),(1,1)) x1 + (5,0)
            a__from#_A(x1) = x1 + (4,1)
            a__first_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (5,4)
            |0|_A() = (0,1)
            nil_A() = (1,0)
            a__from_A(x1) = ((1,0),(1,1)) x1 + (5,4)
    
      precedence:
      
        s = first = mark = a__first > from = |0| = a__from > cons > a__from# = nil > a__first# = mark#
      
      partial status:
    
        pi(a__first#) = []
        pi(s) = []
        pi(cons) = []
        pi(mark#) = [1]
        pi(first) = []
        pi(mark) = []
        pi(from) = []
        pi(a__from#) = []
        pi(a__first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__first#_A(x1,x2) = (0,0)
            s_A(x1) = (1,1)
            cons_A(x1,x2) = (0,3)
            mark#_A(x1) = (0,0)
            first_A(x1,x2) = (0,4)
            mark_A(x1) = (3,5)
            from_A(x1) = (0,1)
            a__from#_A(x1) = (0,0)
            a__first_A(x1,x2) = (0,4)
            |0|_A() = (1,2)
            nil_A() = (0,1)
            a__from_A(x1) = (2,2)
    
      precedence:
      
        mark > a__from > |0| = nil > a__first# = cons = mark# > a__first > first > s > from = a__from#
      
      partial status:
    
        pi(a__first#) = []
        pi(s) = []
        pi(cons) = []
        pi(mark#) = []
        pi(first) = []
        pi(mark) = []
        pi(from) = []
        pi(a__from#) = []
        pi(a__first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = []
    

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: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p3: mark#(first(X1,X2)) -> mark#(X1)
p4: mark#(first(X1,X2)) -> mark#(X2)
p5: mark#(from(X)) -> a__from#(mark(X))
p6: mark#(from(X)) -> mark#(X)
p7: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p6, p7}


-- Reduction pair.

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

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

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__first#_A(x1,x2) = ((1,0),(1,1)) x2 + (1,1)
            s_A(x1) = (5,0)
            cons_A(x1,x2) = ((1,0),(1,0)) x1 + (3,0)
            mark#_A(x1) = ((1,0),(0,0)) x1 + (2,5)
            from_A(x1) = ((1,0),(0,0)) x1 + (4,6)
            first_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (0,3)
            mark_A(x1) = ((1,0),(1,1)) x1 + (0,8)
            a__first_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (0,4)
            |0|_A() = (2,2)
            nil_A() = (1,1)
            a__from_A(x1) = ((1,0),(0,0)) x1 + (4,7)
    
      precedence:
      
        cons = mark = a__first = |0| = nil > first > mark# > a__first# = s = from = a__from
      
      partial status:
    
        pi(a__first#) = [2]
        pi(s) = []
        pi(cons) = []
        pi(mark#) = []
        pi(from) = []
        pi(first) = []
        pi(mark) = [1]
        pi(a__first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__first#_A(x1,x2) = (6,1)
            s_A(x1) = (4,4)
            cons_A(x1,x2) = (1,0)
            mark#_A(x1) = (5,5)
            from_A(x1) = (0,0)
            first_A(x1,x2) = (0,3)
            mark_A(x1) = ((1,0),(1,0)) x1 + (3,2)
            a__first_A(x1,x2) = (3,2)
            |0|_A() = (0,2)
            nil_A() = (1,1)
            a__from_A(x1) = (2,5)
    
      precedence:
      
        nil > |0| > a__first# > mark# = mark > a__first > s = first = a__from > cons = from
      
      partial status:
    
        pi(a__first#) = []
        pi(s) = []
        pi(cons) = []
        pi(mark#) = []
        pi(from) = []
        pi(first) = []
        pi(mark) = []
        pi(a__first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = []
    

The next rules are strictly ordered:

  p1, p6

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(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#(cons(X1,X2)) -> mark#(X1)
p2: mark#(first(X1,X2)) -> mark#(X1)
p3: mark#(first(X1,X2)) -> mark#(X2)
p4: mark#(from(X)) -> mark#(X)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = x1 + (1,2)
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,3)
            first_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (2,1)
            from_A(x1) = ((1,0),(0,0)) x1 + (2,1)
    
      precedence:
      
        cons = first = from > mark#
      
      partial status:
    
        pi(mark#) = [1]
        pi(cons) = [1]
        pi(first) = [2]
        pi(from) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = (0,0)
            cons_A(x1,x2) = (1,1)
            first_A(x1,x2) = ((1,0),(1,1)) x2 + (1,1)
            from_A(x1) = (1,1)
    
      precedence:
      
        mark# = first > from > cons
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = []
        pi(first) = []
        pi(from) = []
    

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

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

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

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,1)) x1 + (0,1)
            cons_A(x1,x2) = x1 + x2 + (1,2)
            from_A(x1) = x1 + (1,2)
            first_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,2)
    
      precedence:
      
        from > mark# = cons = first
      
      partial status:
    
        pi(mark#) = [1]
        pi(cons) = [1, 2]
        pi(from) = [1]
        pi(first) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(0,0)) x1 + (2,2)
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,1)
            from_A(x1) = ((1,0),(0,0)) x1 + (3,1)
            first_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
    
      precedence:
      
        mark# = from > first > cons
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = []
        pi(from) = []
        pi(first) = []
    

The next rules are strictly ordered:

  p1, p2, p3

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