YES

We show the termination of the TRS R:

  terms(N) -> cons(recip(sqr(N)),n__terms(n__s(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,Z)) -> cons(Y,n__first(X,activate(Z)))
  terms(X) -> n__terms(X)
  s(X) -> n__s(X)
  first(X1,X2) -> n__first(X1,X2)
  activate(n__terms(X)) -> terms(activate(X))
  activate(n__s(X)) -> s(activate(X))
  activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
  activate(X) -> X

-- SCC decomposition.

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

p1: terms#(N) -> sqr#(N)
p2: sqr#(s(X)) -> s#(add(sqr(X),dbl(X)))
p3: sqr#(s(X)) -> add#(sqr(X),dbl(X))
p4: sqr#(s(X)) -> sqr#(X)
p5: sqr#(s(X)) -> dbl#(X)
p6: dbl#(s(X)) -> s#(s(dbl(X)))
p7: dbl#(s(X)) -> s#(dbl(X))
p8: dbl#(s(X)) -> dbl#(X)
p9: add#(s(X),Y) -> s#(add(X,Y))
p10: add#(s(X),Y) -> add#(X,Y)
p11: first#(s(X),cons(Y,Z)) -> activate#(Z)
p12: activate#(n__terms(X)) -> terms#(activate(X))
p13: activate#(n__terms(X)) -> activate#(X)
p14: activate#(n__s(X)) -> s#(activate(X))
p15: activate#(n__s(X)) -> activate#(X)
p16: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p17: activate#(n__first(X1,X2)) -> activate#(X1)
p18: activate#(n__first(X1,X2)) -> activate#(X2)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p11, p13, p15, p16, p17, p18}
  {p4}
  {p10}
  {p8}


-- Reduction pair.

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

p1: activate#(n__first(X1,X2)) -> activate#(X2)
p2: activate#(n__first(X1,X2)) -> activate#(X1)
p3: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p4: first#(s(X),cons(Y,Z)) -> activate#(Z)
p5: activate#(n__s(X)) -> activate#(X)
p6: activate#(n__terms(X)) -> activate#(X)

and R consists of:

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          activate#_A(x1) = ((1,0),(0,0)) x1 + (0,3)
          n__first_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,1)
          first#_A(x1,x2) = ((1,0),(0,0)) x2 + (1,2)
          activate_A(x1) = ((1,0),(1,1)) x1
          s_A(x1) = ((1,0),(0,0)) x1
          cons_A(x1,x2) = ((1,0),(0,0)) x2 + (0,2)
          n__s_A(x1) = ((1,0),(0,0)) x1
          n__terms_A(x1) = ((1,0),(1,0)) x1 + (4,4)
          dbl_A(x1) = (2,2)
          |0|_A() = (1,1)
          add_A(x1,x2) = x2
          sqr_A(x1) = (3,3)
          terms_A(x1) = ((1,0),(1,0)) x1 + (4,5)
          recip_A(x1) = (1,0)
          first_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (2,2)
          nil_A() = (0,0)
  
    precedence:
    
      activate = cons = dbl = |0| = add = terms = first > n__first > activate# = first# = s = n__s = sqr > n__terms = recip = nil
    
    partial status:
  
      pi(activate#) = []
      pi(n__first) = [2]
      pi(first#) = []
      pi(activate) = []
      pi(s) = []
      pi(cons) = []
      pi(n__s) = []
      pi(n__terms) = []
      pi(dbl) = []
      pi(|0|) = []
      pi(add) = []
      pi(sqr) = []
      pi(terms) = []
      pi(recip) = []
      pi(first) = []
      pi(nil) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__first(X1,X2)) -> activate#(X1)
p2: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p3: first#(s(X),cons(Y,Z)) -> activate#(Z)
p4: activate#(n__s(X)) -> activate#(X)
p5: activate#(n__terms(X)) -> activate#(X)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: activate#(n__first(X1,X2)) -> activate#(X1)
p2: activate#(n__terms(X)) -> activate#(X)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p5: first#(s(X),cons(Y,Z)) -> activate#(Z)

and R consists of:

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          activate#_A(x1) = ((1,0),(1,1)) x1 + (1,7)
          n__first_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (4,2)
          n__terms_A(x1) = ((1,0),(1,1)) x1 + (2,6)
          n__s_A(x1) = x1
          first#_A(x1,x2) = ((1,0),(0,0)) x2 + (2,8)
          activate_A(x1) = ((1,0),(1,1)) x1 + (0,3)
          s_A(x1) = x1
          cons_A(x1,x2) = ((1,0),(0,0)) x2 + (0,4)
          dbl_A(x1) = (2,1)
          |0|_A() = (2,0)
          add_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (0,1)
          sqr_A(x1) = (3,6)
          terms_A(x1) = ((1,0),(1,1)) x1 + (2,7)
          recip_A(x1) = ((0,0),(1,0)) x1 + (1,1)
          first_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (4,3)
          nil_A() = (1,5)
  
    precedence:
    
      n__first = n__terms = n__s = activate = s = cons = dbl = |0| = add = sqr = terms = recip = first > nil > activate# = first#
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__first) = []
      pi(n__terms) = []
      pi(n__s) = []
      pi(first#) = []
      pi(activate) = []
      pi(s) = []
      pi(cons) = []
      pi(dbl) = []
      pi(|0|) = []
      pi(add) = []
      pi(sqr) = []
      pi(terms) = []
      pi(recip) = []
      pi(first) = []
      pi(nil) = []

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__first(X1,X2)) -> activate#(X1)
p2: activate#(n__terms(X)) -> activate#(X)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: activate#(n__first(X1,X2)) -> activate#(X1)
p2: activate#(n__s(X)) -> activate#(X)
p3: activate#(n__terms(X)) -> activate#(X)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          activate#_A(x1) = x1 + (1,1)
          n__first_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,2)
          n__s_A(x1) = x1 + (2,1)
          n__terms_A(x1) = x1 + (2,2)
  
    precedence:
    
      n__first = n__terms > activate# = n__s
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__first) = []
      pi(n__s) = [1]
      pi(n__terms) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__first(X1,X2)) -> activate#(X1)
p2: activate#(n__s(X)) -> activate#(X)

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          activate#_A(x1) = x1 + (1,1)
          n__first_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (2,2)
          n__s_A(x1) = ((1,0),(1,1)) x1 + (2,2)
  
    precedence:
    
      activate# = n__first = n__s
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__first) = [1, 2]
      pi(n__s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__s(X)) -> activate#(X)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: activate#(n__s(X)) -> activate#(X)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          activate#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      activate# = n__s
    
    partial status:
  
      pi(activate#) = []
      pi(n__s) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: sqr#(s(X)) -> sqr#(X)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          sqr#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      sqr# = s
    
    partial status:
  
      pi(sqr#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: add#(s(X),Y) -> add#(X,Y)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          add#_A(x1,x2) = x1
          s_A(x1) = x1 + (1,1)
  
    precedence:
    
      add# = s
    
    partial status:
  
      pi(add#) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: dbl#(s(X)) -> dbl#(X)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          dbl#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      dbl# = s
    
    partial status:
  
      pi(dbl#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

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