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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        a__terms#_A(x1) = x1 + (1,2,4,0)
        mark#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,0,1)) x1 + (0,1,3,1)
        recip_A(x1) = x1 + (0,7,6,1)
        cons_A(x1,x2) = x1 + ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x2 + (0,2,1,0)
        first_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,0,0,0)) x2 + (2,0,1,5)
        a__first#_A(x1,x2) = x1 + ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x2 + (1,0,0,0)
        mark_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(0,0,0,0)) x1 + (0,5,6,3)
        s_A(x1) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x1 + (0,0,3,0)
        dbl_A(x1) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (5,0,0,1)
        add_A(x1,x2) = x1 + x2 + (2,0,1,1)
        a__add#_A(x1,x2) = x2 + (1,2,2,3)
        |0|_A() = (1,4,9,1)
        sqr_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,0,0,0)) x1 + (1,1,1,1)
        terms_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(1,1,0,0)) x1 + (1,17,8,1)
        a__terms_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(0,0,0,0)) x1 + (1,17,9,2)
        a__sqr_A(x1) = x1 + (1,6,2,2)
        a__dbl_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,0,0,0)) x1 + (5,1,7,1)
        a__add_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,0,0)) x2 + (2,1,4,4)
        a__first_A(x1,x2) = x1 + x2 + (2,2,2,4)
        nil_A() = (1,1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        a__first# > mark# > a__add > a__terms# > mark > a__dbl > a__sqr > a__first > sqr > nil > a__terms > terms > s > cons > first > add > dbl > |0| > recip > a__add#
      
      argument filter:
    
        pi(a__terms#) = []
        pi(mark#) = []
        pi(recip) = []
        pi(cons) = []
        pi(first) = []
        pi(a__first#) = []
        pi(mark) = []
        pi(s) = []
        pi(dbl) = []
        pi(add) = []
        pi(a__add#) = []
        pi(|0|) = []
        pi(sqr) = []
        pi(terms) = []
        pi(a__terms) = []
        pi(a__sqr) = []
        pi(a__dbl) = []
        pi(a__add) = []
        pi(a__first) = []
        pi(nil) = []
    

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.