YES

Problem:
 terms(N) -> cons(recip(sqr(N)))
 sqr(0()) -> 0()
 sqr(s(X)) -> s(add(sqr(X),dbl(X)))
 dbl(0()) -> 0()
 dbl(s(X)) -> s(s(dbl(X)))
 add(0(),X) -> X
 add(s(X),Y) -> s(add(X,Y))
 first(0(),X) -> nil()
 first(s(X),cons(Y)) -> cons(Y)

Proof:
 DP Processor:
  DPs:
   terms#(N) -> sqr#(N)
   sqr#(s(X)) -> dbl#(X)
   sqr#(s(X)) -> sqr#(X)
   sqr#(s(X)) -> add#(sqr(X),dbl(X))
   dbl#(s(X)) -> dbl#(X)
   add#(s(X),Y) -> add#(X,Y)
  TRS:
   terms(N) -> cons(recip(sqr(N)))
   sqr(0()) -> 0()
   sqr(s(X)) -> s(add(sqr(X),dbl(X)))
   dbl(0()) -> 0()
   dbl(s(X)) -> s(s(dbl(X)))
   add(0(),X) -> X
   add(s(X),Y) -> s(add(X,Y))
   first(0(),X) -> nil()
   first(s(X),cons(Y)) -> cons(Y)
  TDG Processor:
   DPs:
    terms#(N) -> sqr#(N)
    sqr#(s(X)) -> dbl#(X)
    sqr#(s(X)) -> sqr#(X)
    sqr#(s(X)) -> add#(sqr(X),dbl(X))
    dbl#(s(X)) -> dbl#(X)
    add#(s(X),Y) -> add#(X,Y)
   TRS:
    terms(N) -> cons(recip(sqr(N)))
    sqr(0()) -> 0()
    sqr(s(X)) -> s(add(sqr(X),dbl(X)))
    dbl(0()) -> 0()
    dbl(s(X)) -> s(s(dbl(X)))
    add(0(),X) -> X
    add(s(X),Y) -> s(add(X,Y))
    first(0(),X) -> nil()
    first(s(X),cons(Y)) -> cons(Y)
   graph:
    add#(s(X),Y) -> add#(X,Y) -> add#(s(X),Y) -> add#(X,Y)
    dbl#(s(X)) -> dbl#(X) -> dbl#(s(X)) -> dbl#(X)
    sqr#(s(X)) -> add#(sqr(X),dbl(X)) -> add#(s(X),Y) -> add#(X,Y)
    sqr#(s(X)) -> dbl#(X) -> dbl#(s(X)) -> dbl#(X)
    sqr#(s(X)) -> sqr#(X) -> sqr#(s(X)) -> add#(sqr(X),dbl(X))
    sqr#(s(X)) -> sqr#(X) -> sqr#(s(X)) -> sqr#(X)
    sqr#(s(X)) -> sqr#(X) -> sqr#(s(X)) -> dbl#(X)
    terms#(N) -> sqr#(N) -> sqr#(s(X)) -> add#(sqr(X),dbl(X))
    terms#(N) -> sqr#(N) -> sqr#(s(X)) -> sqr#(X)
    terms#(N) -> sqr#(N) -> sqr#(s(X)) -> dbl#(X)
   SCC Processor:
    #sccs: 3
    #rules: 3
    #arcs: 10/36
    DPs:
     sqr#(s(X)) -> sqr#(X)
    TRS:
     terms(N) -> cons(recip(sqr(N)))
     sqr(0()) -> 0()
     sqr(s(X)) -> s(add(sqr(X),dbl(X)))
     dbl(0()) -> 0()
     dbl(s(X)) -> s(s(dbl(X)))
     add(0(),X) -> X
     add(s(X),Y) -> s(add(X,Y))
     first(0(),X) -> nil()
     first(s(X),cons(Y)) -> cons(Y)
    LPO Processor:
     argument filtering:
      pi(terms) = [0]
      pi(sqr) = [0]
      pi(recip) = 0
      pi(cons) = 0
      pi(0) = []
      pi(s) = [0]
      pi(dbl) = [0]
      pi(add) = [0,1]
      pi(first) = [1]
      pi(nil) = []
      pi(sqr#) = 0
     precedence:
      terms > sqr > first ~ add ~ dbl > sqr# ~ nil ~ s ~ 0 ~ cons ~ recip
     problem:
      DPs:
       
      TRS:
       terms(N) -> cons(recip(sqr(N)))
       sqr(0()) -> 0()
       sqr(s(X)) -> s(add(sqr(X),dbl(X)))
       dbl(0()) -> 0()
       dbl(s(X)) -> s(s(dbl(X)))
       add(0(),X) -> X
       add(s(X),Y) -> s(add(X,Y))
       first(0(),X) -> nil()
       first(s(X),cons(Y)) -> cons(Y)
     Qed
    
    DPs:
     dbl#(s(X)) -> dbl#(X)
    TRS:
     terms(N) -> cons(recip(sqr(N)))
     sqr(0()) -> 0()
     sqr(s(X)) -> s(add(sqr(X),dbl(X)))
     dbl(0()) -> 0()
     dbl(s(X)) -> s(s(dbl(X)))
     add(0(),X) -> X
     add(s(X),Y) -> s(add(X,Y))
     first(0(),X) -> nil()
     first(s(X),cons(Y)) -> cons(Y)
    LPO Processor:
     argument filtering:
      pi(terms) = [0]
      pi(sqr) = [0]
      pi(recip) = 0
      pi(cons) = 0
      pi(0) = []
      pi(s) = [0]
      pi(dbl) = [0]
      pi(add) = [0,1]
      pi(first) = [1]
      pi(nil) = []
      pi(dbl#) = 0
     precedence:
      terms > sqr > first ~ add ~ dbl > dbl# ~ nil ~ s ~ 0 ~ cons ~ recip
     problem:
      DPs:
       
      TRS:
       terms(N) -> cons(recip(sqr(N)))
       sqr(0()) -> 0()
       sqr(s(X)) -> s(add(sqr(X),dbl(X)))
       dbl(0()) -> 0()
       dbl(s(X)) -> s(s(dbl(X)))
       add(0(),X) -> X
       add(s(X),Y) -> s(add(X,Y))
       first(0(),X) -> nil()
       first(s(X),cons(Y)) -> cons(Y)
     Qed
    
    DPs:
     add#(s(X),Y) -> add#(X,Y)
    TRS:
     terms(N) -> cons(recip(sqr(N)))
     sqr(0()) -> 0()
     sqr(s(X)) -> s(add(sqr(X),dbl(X)))
     dbl(0()) -> 0()
     dbl(s(X)) -> s(s(dbl(X)))
     add(0(),X) -> X
     add(s(X),Y) -> s(add(X,Y))
     first(0(),X) -> nil()
     first(s(X),cons(Y)) -> cons(Y)
    LPO Processor:
     argument filtering:
      pi(terms) = [0]
      pi(sqr) = [0]
      pi(recip) = 0
      pi(cons) = 0
      pi(0) = []
      pi(s) = [0]
      pi(dbl) = [0]
      pi(add) = [0,1]
      pi(first) = [1]
      pi(nil) = []
      pi(add#) = 0
     precedence:
      terms > sqr > first ~ add ~ dbl > add# ~ nil ~ s ~ 0 ~ cons ~ recip
     problem:
      DPs:
       
      TRS:
       terms(N) -> cons(recip(sqr(N)))
       sqr(0()) -> 0()
       sqr(s(X)) -> s(add(sqr(X),dbl(X)))
       dbl(0()) -> 0()
       dbl(s(X)) -> s(s(dbl(X)))
       add(0(),X) -> X
       add(s(X),Y) -> s(add(X,Y))
       first(0(),X) -> nil()
       first(s(X),cons(Y)) -> cons(Y)
     Qed