YES

We show the termination of the TRS R:

  a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
  a__sqr(|0|()) -> |0|()
  a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
  a__dbl(|0|()) -> |0|()
  a__dbl(s(X)) -> s(s(dbl(X)))
  a__add(|0|(),X) -> mark(X)
  a__add(s(X),Y) -> s(add(X,Y))
  a__first(|0|(),X) -> nil()
  a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
  mark(terms(X)) -> a__terms(mark(X))
  mark(sqr(X)) -> a__sqr(mark(X))
  mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
  mark(dbl(X)) -> a__dbl(mark(X))
  mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(recip(X)) -> recip(mark(X))
  mark(s(X)) -> s(X)
  mark(|0|()) -> |0|()
  mark(nil()) -> nil()
  a__terms(X) -> terms(X)
  a__sqr(X) -> sqr(X)
  a__add(X1,X2) -> add(X1,X2)
  a__dbl(X) -> dbl(X)
  a__first(X1,X2) -> first(X1,X2)

-- SCC decomposition.

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

p1: a__terms#(N) -> a__sqr#(mark(N))
p2: a__terms#(N) -> mark#(N)
p3: a__add#(|0|(),X) -> mark#(X)
p4: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p5: mark#(terms(X)) -> a__terms#(mark(X))
p6: mark#(terms(X)) -> mark#(X)
p7: mark#(sqr(X)) -> a__sqr#(mark(X))
p8: mark#(sqr(X)) -> mark#(X)
p9: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p10: mark#(add(X1,X2)) -> mark#(X1)
p11: mark#(add(X1,X2)) -> mark#(X2)
p12: mark#(dbl(X)) -> a__dbl#(mark(X))
p13: mark#(dbl(X)) -> mark#(X)
p14: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p15: mark#(first(X1,X2)) -> mark#(X1)
p16: mark#(first(X1,X2)) -> mark#(X2)
p17: mark#(cons(X1,X2)) -> mark#(X1)
p18: mark#(recip(X)) -> mark#(X)

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5, p6, p8, p9, p10, p11, p13, p14, p15, p16, p17, p18}


-- Reduction pair.

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

p1: a__terms#(N) -> mark#(N)
p2: mark#(recip(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
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))
p7: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p8: mark#(dbl(X)) -> mark#(X)
p9: mark#(add(X1,X2)) -> mark#(X2)
p10: mark#(add(X1,X2)) -> mark#(X1)
p11: mark#(add(X1,X2)) -> a__add#(mark(X1),mark(X2))
p12: a__add#(|0|(),X) -> mark#(X)
p13: mark#(sqr(X)) -> mark#(X)
p14: mark#(terms(X)) -> mark#(X)
p15: mark#(terms(X)) -> a__terms#(mark(X))

and R consists of:

r1: a__terms(N) -> cons(recip(a__sqr(mark(N))),terms(s(N)))
r2: a__sqr(|0|()) -> |0|()
r3: a__sqr(s(X)) -> s(add(sqr(X),dbl(X)))
r4: a__dbl(|0|()) -> |0|()
r5: a__dbl(s(X)) -> s(s(dbl(X)))
r6: a__add(|0|(),X) -> mark(X)
r7: a__add(s(X),Y) -> s(add(X,Y))
r8: a__first(|0|(),X) -> nil()
r9: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r10: mark(terms(X)) -> a__terms(mark(X))
r11: mark(sqr(X)) -> a__sqr(mark(X))
r12: mark(add(X1,X2)) -> a__add(mark(X1),mark(X2))
r13: mark(dbl(X)) -> a__dbl(mark(X))
r14: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r15: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r16: mark(recip(X)) -> recip(mark(X))
r17: mark(s(X)) -> s(X)
r18: mark(|0|()) -> |0|()
r19: mark(nil()) -> nil()
r20: a__terms(X) -> terms(X)
r21: a__sqr(X) -> sqr(X)
r22: a__add(X1,X2) -> add(X1,X2)
r23: a__dbl(X) -> dbl(X)
r24: a__first(X1,X2) -> first(X1,X2)

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, r23, r24

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      a__terms#_A(x1) = x1 + (1,4,0)
      mark#_A(x1) = ((1,0,0),(0,1,0),(1,0,0)) x1 + (0,3,3)
      recip_A(x1) = x1 + (1,0,0)
      cons_A(x1,x2) = ((1,0,0),(0,1,0),(1,0,1)) x1 + (0,3,2)
      first_A(x1,x2) = ((1,0,0),(0,0,0),(0,1,0)) x1 + x2 + (2,1,0)
      a__first#_A(x1,x2) = ((0,0,0),(1,0,0),(1,1,0)) x1 + x2 + (1,0,0)
      mark_A(x1) = ((1,0,0),(0,1,0),(1,0,1)) x1 + (0,3,1)
      s_A(x1) = ((0,0,0),(0,0,0),(1,0,0)) x1 + (1,1,0)
      dbl_A(x1) = x1 + (3,1,0)
      add_A(x1,x2) = x1 + x2 + (2,2,0)
      a__add#_A(x1,x2) = x2 + (2,4,4)
      |0|_A() = (0,0,1)
      sqr_A(x1) = ((1,0,0),(0,1,0),(1,0,1)) x1 + (2,1,0)
      terms_A(x1) = ((1,0,0),(1,1,0),(0,0,1)) x1 + (4,5,0)
      a__terms_A(x1) = ((1,0,0),(1,1,0),(0,0,1)) x1 + (4,5,0)
      a__sqr_A(x1) = ((1,0,0),(0,1,0),(1,0,1)) x1 + (2,1,1)
      a__dbl_A(x1) = x1 + (3,1,2)
      a__add_A(x1,x2) = x1 + ((1,0,0),(0,0,0),(1,0,0)) x2 + (2,4,4)
      a__first_A(x1,x2) = ((1,0,0),(0,0,0),(0,1,0)) x1 + x2 + (2,2,0)
      nil_A() = (0,1,1)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15

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