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:
    
        max/plus interpretations on natural numbers:
        
          a__first#_A(x1,x2) = max{14, x1 + 2, x2 + 5}
          s_A(x1) = max{1, x1}
          cons_A(x1,x2) = max{20, x1 + 11, x2}
          mark#_A(x1) = max{1, x1}
          from_A(x1) = max{23, x1 + 21}
          a__from#_A(x1) = x1 + 1
          mark_A(x1) = max{12, x1 + 10}
          first_A(x1,x2) = max{24, x1 + 18, x2 + 15}
          a__first_A(x1,x2) = max{24, x1 + 18, x2 + 15}
          |0|_A = 13
          nil_A = 0
          a__from_A(x1) = max{23, x1 + 21}
    
      precedence:
      
        mark = a__first > a__from > s = cons = from > |0| > first > mark# = a__from# > a__first# = nil
      
      partial status:
    
        pi(a__first#) = [1, 2]
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(mark#) = [1]
        pi(from) = [1]
        pi(a__from#) = [1]
        pi(mark) = [1]
        pi(first) = [1, 2]
        pi(a__first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__first#_A(x1,x2) = max{2, x1 - 2, x2}
          s_A(x1) = max{12, x1 + 4}
          cons_A(x1,x2) = max{x1 + 2, x2 - 6}
          mark#_A(x1) = max{2, x1 + 1}
          from_A(x1) = max{62, x1 + 32}
          a__from#_A(x1) = x1 + 3
          mark_A(x1) = x1 + 28
          first_A(x1,x2) = max{x1 + 26, x2 + 28}
          a__first_A(x1,x2) = 19
          |0|_A = 20
          nil_A = 31
          a__from_A(x1) = max{61, x1 + 30}
    
      precedence:
      
        first > s = mark = a__from > cons > a__first# = mark# = a__from# > a__first = |0| = nil > from
      
      partial status:
    
        pi(a__first#) = [2]
        pi(s) = []
        pi(cons) = [1]
        pi(mark#) = [1]
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = [1]
        pi(first) = [2]
        pi(a__first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = []
    

The next rules are strictly ordered:

  p2, p3, p4, p5, p6, p7, p8

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))

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}


-- 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))

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:
    
        max/plus interpretations on natural numbers:
        
          a__first#_A(x1,x2) = max{32, x1 + 30, x2 + 26}
          s_A(x1) = max{3, x1}
          cons_A(x1,x2) = max{25, x1 + 9, x2}
          mark#_A(x1) = x1 + 10
          first_A(x1,x2) = max{x1 + 29, x2 + 26}
          mark_A(x1) = x1 + 8
          a__first_A(x1,x2) = max{30, x1 + 29, x2 + 26}
          |0|_A = 26
          nil_A = 25
          a__from_A(x1) = max{25, x1 + 17}
          from_A(x1) = x1 + 17
    
      precedence:
      
        a__first# = mark# = mark > s = a__first > first = |0| = nil = a__from > cons = from
      
      partial status:
    
        pi(a__first#) = [1, 2]
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(mark#) = [1]
        pi(first) = [1, 2]
        pi(mark) = [1]
        pi(a__first) = [1, 2]
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = [1]
        pi(from) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__first#_A(x1,x2) = max{x1 + 10, x2 + 10}
          s_A(x1) = x1 + 6
          cons_A(x1,x2) = max{x1 + 8, x2 + 8}
          mark#_A(x1) = x1 + 10
          first_A(x1,x2) = max{x1 + 4, x2 + 1}
          mark_A(x1) = x1
          a__first_A(x1,x2) = max{x1 + 5, x2 + 7}
          |0|_A = 2
          nil_A = 6
          a__from_A(x1) = max{4, x1 + 2}
          from_A(x1) = 1
    
      precedence:
      
        a__first# = s = cons = mark# = first = mark = a__first = |0| = nil = a__from = from
      
      partial status:
    
        pi(a__first#) = []
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(mark#) = [1]
        pi(first) = [2]
        pi(mark) = [1]
        pi(a__first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(a__from) = []
        pi(from) = []
    

The next rules are strictly ordered:

  p1, p2

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