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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a__first#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
        cons_A(x1,x2) = x1 + 1
        mark#_A(x1) = x1
        from_A(x1) = x1 + 2
        a__from#_A(x1) = x1 + 1
        mark_A(x1) = x1
        first_A(x1,x2) = x1 + x2 + 1
        a__first_A(x1,x2) = x1 + x2 + 1
        |0|_A() = 1
        nil_A() = 1
        a__from_A(x1) = x1 + 2
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a__first#_A(x1,x2) = 1
        s_A(x1) = 1
        cons_A(x1,x2) = 3
        mark#_A(x1) = 0
        from_A(x1) = 1
        a__from#_A(x1) = x1 + 1
        mark_A(x1) = 4
        first_A(x1,x2) = 1
        a__first_A(x1,x2) = 2
        |0|_A() = 0
        nil_A() = 1
        a__from_A(x1) = 2
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9

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