YES

We show the termination of the TRS R:

  terms(N) -> cons(recip(sqr(N)),n__terms(s(N)))
  sqr(|0|()) -> |0|()
  sqr(s(X)) -> s(n__add(sqr(activate(X)),dbl(activate(X))))
  dbl(|0|()) -> |0|()
  dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
  add(|0|(),X) -> X
  add(s(X),Y) -> s(n__add(activate(X),Y))
  first(|0|(),X) -> nil()
  first(s(X),cons(Y,Z)) -> cons(Y,n__first(activate(X),activate(Z)))
  terms(X) -> n__terms(X)
  add(X1,X2) -> n__add(X1,X2)
  s(X) -> n__s(X)
  dbl(X) -> n__dbl(X)
  first(X1,X2) -> n__first(X1,X2)
  activate(n__terms(X)) -> terms(X)
  activate(n__add(X1,X2)) -> add(X1,X2)
  activate(n__s(X)) -> s(X)
  activate(n__dbl(X)) -> dbl(X)
  activate(n__first(X1,X2)) -> first(X1,X2)
  activate(X) -> X

-- SCC decomposition.

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

p1: terms#(N) -> sqr#(N)
p2: terms#(N) -> s#(N)
p3: sqr#(s(X)) -> s#(n__add(sqr(activate(X)),dbl(activate(X))))
p4: sqr#(s(X)) -> sqr#(activate(X))
p5: sqr#(s(X)) -> activate#(X)
p6: sqr#(s(X)) -> dbl#(activate(X))
p7: dbl#(s(X)) -> s#(n__s(n__dbl(activate(X))))
p8: dbl#(s(X)) -> activate#(X)
p9: add#(s(X),Y) -> s#(n__add(activate(X),Y))
p10: add#(s(X),Y) -> activate#(X)
p11: first#(s(X),cons(Y,Z)) -> activate#(X)
p12: first#(s(X),cons(Y,Z)) -> activate#(Z)
p13: activate#(n__terms(X)) -> terms#(X)
p14: activate#(n__add(X1,X2)) -> add#(X1,X2)
p15: activate#(n__s(X)) -> s#(X)
p16: activate#(n__dbl(X)) -> dbl#(X)
p17: activate#(n__first(X1,X2)) -> first#(X1,X2)

and R consists of:

r1: terms(N) -> cons(recip(sqr(N)),n__terms(s(N)))
r2: sqr(|0|()) -> |0|()
r3: sqr(s(X)) -> s(n__add(sqr(activate(X)),dbl(activate(X))))
r4: dbl(|0|()) -> |0|()
r5: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r6: add(|0|(),X) -> X
r7: add(s(X),Y) -> s(n__add(activate(X),Y))
r8: first(|0|(),X) -> nil()
r9: first(s(X),cons(Y,Z)) -> cons(Y,n__first(activate(X),activate(Z)))
r10: terms(X) -> n__terms(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: s(X) -> n__s(X)
r13: dbl(X) -> n__dbl(X)
r14: first(X1,X2) -> n__first(X1,X2)
r15: activate(n__terms(X)) -> terms(X)
r16: activate(n__add(X1,X2)) -> add(X1,X2)
r17: activate(n__s(X)) -> s(X)
r18: activate(n__dbl(X)) -> dbl(X)
r19: activate(n__first(X1,X2)) -> first(X1,X2)
r20: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p4, p5, p6, p8, p10, p11, p12, p13, p14, p16, p17}


-- Reduction pair.

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

p1: terms#(N) -> sqr#(N)
p2: sqr#(s(X)) -> dbl#(activate(X))
p3: dbl#(s(X)) -> activate#(X)
p4: activate#(n__first(X1,X2)) -> first#(X1,X2)
p5: first#(s(X),cons(Y,Z)) -> activate#(Z)
p6: activate#(n__dbl(X)) -> dbl#(X)
p7: activate#(n__add(X1,X2)) -> add#(X1,X2)
p8: add#(s(X),Y) -> activate#(X)
p9: activate#(n__terms(X)) -> terms#(X)
p10: first#(s(X),cons(Y,Z)) -> activate#(X)
p11: sqr#(s(X)) -> activate#(X)
p12: sqr#(s(X)) -> sqr#(activate(X))

and R consists of:

r1: terms(N) -> cons(recip(sqr(N)),n__terms(s(N)))
r2: sqr(|0|()) -> |0|()
r3: sqr(s(X)) -> s(n__add(sqr(activate(X)),dbl(activate(X))))
r4: dbl(|0|()) -> |0|()
r5: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r6: add(|0|(),X) -> X
r7: add(s(X),Y) -> s(n__add(activate(X),Y))
r8: first(|0|(),X) -> nil()
r9: first(s(X),cons(Y,Z)) -> cons(Y,n__first(activate(X),activate(Z)))
r10: terms(X) -> n__terms(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: s(X) -> n__s(X)
r13: dbl(X) -> n__dbl(X)
r14: first(X1,X2) -> n__first(X1,X2)
r15: activate(n__terms(X)) -> terms(X)
r16: activate(n__add(X1,X2)) -> add(X1,X2)
r17: activate(n__s(X)) -> s(X)
r18: activate(n__dbl(X)) -> dbl(X)
r19: activate(n__first(X1,X2)) -> first(X1,X2)
r20: activate(X) -> X

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          terms#_A(x1) = x1 + 18
          sqr#_A(x1) = x1 + 18
          s_A(x1) = max{19, x1}
          dbl#_A(x1) = x1 + 6
          activate_A(x1) = max{2, x1}
          activate#_A(x1) = max{5, x1}
          n__first_A(x1,x2) = max{47, x1 + 46, x2 + 43}
          first#_A(x1,x2) = max{42, x1 + 6, x2 + 6}
          cons_A(x1,x2) = max{42, x1 + 6, x2}
          n__dbl_A(x1) = x1 + 7
          n__add_A(x1,x2) = max{x1, x2 + 7}
          add#_A(x1,x2) = max{x1, x2 + 6}
          n__terms_A(x1) = max{42, x1 + 23}
          sqr_A(x1) = x1 + 16
          |0|_A = 45
          dbl_A(x1) = x1 + 7
          terms_A(x1) = max{42, x1 + 23}
          recip_A(x1) = max{22, x1}
          n__s_A(x1) = max{19, x1}
          add_A(x1,x2) = max{x1, x2 + 7}
          first_A(x1,x2) = max{47, x1 + 46, x2 + 43}
          nil_A = 44
    
      precedence:
      
        n__dbl = dbl > sqr > n__add = add > first# > s = activate# = add# = n__s > activate = terms > first > cons > n__first > n__terms > nil > |0| > terms# > sqr# > dbl# > recip
      
      partial status:
    
        pi(terms#) = [1]
        pi(sqr#) = [1]
        pi(s) = [1]
        pi(dbl#) = [1]
        pi(activate) = [1]
        pi(activate#) = [1]
        pi(n__first) = [1, 2]
        pi(first#) = [1, 2]
        pi(cons) = [1, 2]
        pi(n__dbl) = [1]
        pi(n__add) = [1, 2]
        pi(add#) = [1, 2]
        pi(n__terms) = [1]
        pi(sqr) = [1]
        pi(|0|) = []
        pi(dbl) = [1]
        pi(terms) = [1]
        pi(recip) = [1]
        pi(n__s) = [1]
        pi(add) = [1, 2]
        pi(first) = []
        pi(nil) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          terms#_A(x1) = x1 + 86
          sqr#_A(x1) = x1 + 86
          s_A(x1) = x1 + 110
          dbl#_A(x1) = max{16, x1}
          activate_A(x1) = x1 + 24
          activate#_A(x1) = x1 + 69
          n__first_A(x1,x2) = max{537, x1 + 355, x2 + 182}
          first#_A(x1,x2) = max{x1 + 424, x2 + 250}
          cons_A(x1,x2) = max{16, x1 - 17, x2 - 181}
          n__dbl_A(x1) = max{134, x1 + 8}
          n__add_A(x1,x2) = max{120, x1 - 24, x2 - 50}
          add#_A(x1,x2) = max{18, x1 - 41}
          n__terms_A(x1) = x1 + 86
          sqr_A(x1) = max{123, x1 + 120}
          |0|_A = 160
          dbl_A(x1) = max{159, x1 + 109}
          terms_A(x1) = max{16, x1 + 15}
          recip_A(x1) = 17
          n__s_A(x1) = 0
          add_A(x1,x2) = max{230, x1 + 120, x2 + 101}
          first_A(x1,x2) = 355
          nil_A = 161
    
      precedence:
      
        terms > activate > dbl = add = first = nil > n__s > n__first > |0| > s = n__dbl = n__add = n__terms = sqr > add# > activate# = first# > terms# > sqr# > dbl# = recip > cons
      
      partial status:
    
        pi(terms#) = [1]
        pi(sqr#) = [1]
        pi(s) = []
        pi(dbl#) = [1]
        pi(activate) = [1]
        pi(activate#) = [1]
        pi(n__first) = []
        pi(first#) = [1, 2]
        pi(cons) = []
        pi(n__dbl) = [1]
        pi(n__add) = []
        pi(add#) = []
        pi(n__terms) = []
        pi(sqr) = [1]
        pi(|0|) = []
        pi(dbl) = []
        pi(terms) = [1]
        pi(recip) = []
        pi(n__s) = []
        pi(add) = [1]
        pi(first) = []
        pi(nil) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p9, p10, p11, p12

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__add(X1,X2)) -> add#(X1,X2)
p2: add#(s(X),Y) -> activate#(X)

and R consists of:

r1: terms(N) -> cons(recip(sqr(N)),n__terms(s(N)))
r2: sqr(|0|()) -> |0|()
r3: sqr(s(X)) -> s(n__add(sqr(activate(X)),dbl(activate(X))))
r4: dbl(|0|()) -> |0|()
r5: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r6: add(|0|(),X) -> X
r7: add(s(X),Y) -> s(n__add(activate(X),Y))
r8: first(|0|(),X) -> nil()
r9: first(s(X),cons(Y,Z)) -> cons(Y,n__first(activate(X),activate(Z)))
r10: terms(X) -> n__terms(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: s(X) -> n__s(X)
r13: dbl(X) -> n__dbl(X)
r14: first(X1,X2) -> n__first(X1,X2)
r15: activate(n__terms(X)) -> terms(X)
r16: activate(n__add(X1,X2)) -> add(X1,X2)
r17: activate(n__s(X)) -> s(X)
r18: activate(n__dbl(X)) -> dbl(X)
r19: activate(n__first(X1,X2)) -> first(X1,X2)
r20: activate(X) -> 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: activate#(n__add(X1,X2)) -> add#(X1,X2)
p2: add#(s(X),Y) -> activate#(X)

and R consists of:

r1: terms(N) -> cons(recip(sqr(N)),n__terms(s(N)))
r2: sqr(|0|()) -> |0|()
r3: sqr(s(X)) -> s(n__add(sqr(activate(X)),dbl(activate(X))))
r4: dbl(|0|()) -> |0|()
r5: dbl(s(X)) -> s(n__s(n__dbl(activate(X))))
r6: add(|0|(),X) -> X
r7: add(s(X),Y) -> s(n__add(activate(X),Y))
r8: first(|0|(),X) -> nil()
r9: first(s(X),cons(Y,Z)) -> cons(Y,n__first(activate(X),activate(Z)))
r10: terms(X) -> n__terms(X)
r11: add(X1,X2) -> n__add(X1,X2)
r12: s(X) -> n__s(X)
r13: dbl(X) -> n__dbl(X)
r14: first(X1,X2) -> n__first(X1,X2)
r15: activate(n__terms(X)) -> terms(X)
r16: activate(n__add(X1,X2)) -> add(X1,X2)
r17: activate(n__s(X)) -> s(X)
r18: activate(n__dbl(X)) -> dbl(X)
r19: activate(n__first(X1,X2)) -> first(X1,X2)
r20: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{3, x1 + 2}
          n__add_A(x1,x2) = max{x1 + 5, x2 + 5}
          add#_A(x1,x2) = max{x1 + 5, x2 + 4}
          s_A(x1) = x1 + 4
    
      precedence:
      
        activate# = n__add = add# = s
      
      partial status:
    
        pi(activate#) = [1]
        pi(n__add) = [1, 2]
        pi(add#) = [1, 2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{5, x1 + 4}
          n__add_A(x1,x2) = max{x1 + 1, x2 + 3}
          add#_A(x1,x2) = max{x1 + 4, x2 + 6}
          s_A(x1) = x1
    
      precedence:
      
        activate# = n__add = add# = s
      
      partial status:
    
        pi(activate#) = []
        pi(n__add) = [2]
        pi(add#) = [2]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1, p2

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