YES

We show the termination of the TRS R:

  filter(cons(X,Y),|0|(),M) -> cons(|0|(),n__filter(activate(Y),M,M))
  filter(cons(X,Y),s(N),M) -> cons(X,n__filter(activate(Y),N,M))
  sieve(cons(|0|(),Y)) -> cons(|0|(),n__sieve(activate(Y)))
  sieve(cons(s(N),Y)) -> cons(s(N),n__sieve(n__filter(activate(Y),N,N)))
  nats(N) -> cons(N,n__nats(n__s(N)))
  zprimes() -> sieve(nats(s(s(|0|()))))
  filter(X1,X2,X3) -> n__filter(X1,X2,X3)
  sieve(X) -> n__sieve(X)
  nats(X) -> n__nats(X)
  s(X) -> n__s(X)
  activate(n__filter(X1,X2,X3)) -> filter(activate(X1),activate(X2),activate(X3))
  activate(n__sieve(X)) -> sieve(activate(X))
  activate(n__nats(X)) -> nats(activate(X))
  activate(n__s(X)) -> s(activate(X))
  activate(X) -> X

-- SCC decomposition.

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

p1: filter#(cons(X,Y),|0|(),M) -> activate#(Y)
p2: filter#(cons(X,Y),s(N),M) -> activate#(Y)
p3: sieve#(cons(|0|(),Y)) -> activate#(Y)
p4: sieve#(cons(s(N),Y)) -> activate#(Y)
p5: zprimes#() -> sieve#(nats(s(s(|0|()))))
p6: zprimes#() -> nats#(s(s(|0|())))
p7: zprimes#() -> s#(s(|0|()))
p8: zprimes#() -> s#(|0|())
p9: activate#(n__filter(X1,X2,X3)) -> filter#(activate(X1),activate(X2),activate(X3))
p10: activate#(n__filter(X1,X2,X3)) -> activate#(X1)
p11: activate#(n__filter(X1,X2,X3)) -> activate#(X2)
p12: activate#(n__filter(X1,X2,X3)) -> activate#(X3)
p13: activate#(n__sieve(X)) -> sieve#(activate(X))
p14: activate#(n__sieve(X)) -> activate#(X)
p15: activate#(n__nats(X)) -> nats#(activate(X))
p16: activate#(n__nats(X)) -> activate#(X)
p17: activate#(n__s(X)) -> s#(activate(X))
p18: activate#(n__s(X)) -> activate#(X)

and R consists of:

r1: filter(cons(X,Y),|0|(),M) -> cons(|0|(),n__filter(activate(Y),M,M))
r2: filter(cons(X,Y),s(N),M) -> cons(X,n__filter(activate(Y),N,M))
r3: sieve(cons(|0|(),Y)) -> cons(|0|(),n__sieve(activate(Y)))
r4: sieve(cons(s(N),Y)) -> cons(s(N),n__sieve(n__filter(activate(Y),N,N)))
r5: nats(N) -> cons(N,n__nats(n__s(N)))
r6: zprimes() -> sieve(nats(s(s(|0|()))))
r7: filter(X1,X2,X3) -> n__filter(X1,X2,X3)
r8: sieve(X) -> n__sieve(X)
r9: nats(X) -> n__nats(X)
r10: s(X) -> n__s(X)
r11: activate(n__filter(X1,X2,X3)) -> filter(activate(X1),activate(X2),activate(X3))
r12: activate(n__sieve(X)) -> sieve(activate(X))
r13: activate(n__nats(X)) -> nats(activate(X))
r14: activate(n__s(X)) -> s(activate(X))
r15: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p9, p10, p11, p12, p13, p14, p16, p18}


-- Reduction pair.

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

p1: filter#(cons(X,Y),|0|(),M) -> activate#(Y)
p2: activate#(n__s(X)) -> activate#(X)
p3: activate#(n__nats(X)) -> activate#(X)
p4: activate#(n__sieve(X)) -> activate#(X)
p5: activate#(n__sieve(X)) -> sieve#(activate(X))
p6: sieve#(cons(s(N),Y)) -> activate#(Y)
p7: activate#(n__filter(X1,X2,X3)) -> activate#(X3)
p8: activate#(n__filter(X1,X2,X3)) -> activate#(X2)
p9: activate#(n__filter(X1,X2,X3)) -> activate#(X1)
p10: activate#(n__filter(X1,X2,X3)) -> filter#(activate(X1),activate(X2),activate(X3))
p11: filter#(cons(X,Y),s(N),M) -> activate#(Y)
p12: sieve#(cons(|0|(),Y)) -> activate#(Y)

and R consists of:

r1: filter(cons(X,Y),|0|(),M) -> cons(|0|(),n__filter(activate(Y),M,M))
r2: filter(cons(X,Y),s(N),M) -> cons(X,n__filter(activate(Y),N,M))
r3: sieve(cons(|0|(),Y)) -> cons(|0|(),n__sieve(activate(Y)))
r4: sieve(cons(s(N),Y)) -> cons(s(N),n__sieve(n__filter(activate(Y),N,N)))
r5: nats(N) -> cons(N,n__nats(n__s(N)))
r6: zprimes() -> sieve(nats(s(s(|0|()))))
r7: filter(X1,X2,X3) -> n__filter(X1,X2,X3)
r8: sieve(X) -> n__sieve(X)
r9: nats(X) -> n__nats(X)
r10: s(X) -> n__s(X)
r11: activate(n__filter(X1,X2,X3)) -> filter(activate(X1),activate(X2),activate(X3))
r12: activate(n__sieve(X)) -> sieve(activate(X))
r13: activate(n__nats(X)) -> nats(activate(X))
r14: activate(n__s(X)) -> s(activate(X))
r15: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      filter#_A(x1,x2,x3) = max{x1 + 2, x2 + 3, x3 + 1}
      cons_A(x1,x2) = max{x1 + 9, x2}
      |0|_A = 0
      activate#_A(x1) = x1 + 2
      n__s_A(x1) = x1
      n__nats_A(x1) = max{10, x1 + 9}
      n__sieve_A(x1) = max{5, x1 + 4}
      sieve#_A(x1) = x1 + 2
      activate_A(x1) = x1
      s_A(x1) = x1
      n__filter_A(x1,x2,x3) = max{x1, x2 + 8, x3 + 8}
      filter_A(x1,x2,x3) = max{x1, x2 + 8, x3 + 8}
      sieve_A(x1) = max{5, x1 + 4}
      nats_A(x1) = max{10, x1 + 9}
    
    2. max/plus interpretations on natural numbers:
    
      filter#_A(x1,x2,x3) = 0
      cons_A(x1,x2) = 6
      |0|_A = 0
      activate#_A(x1) = 0
      n__s_A(x1) = 0
      n__nats_A(x1) = 0
      n__sieve_A(x1) = 10
      sieve#_A(x1) = max{2, x1}
      activate_A(x1) = max{7, x1 + 2}
      s_A(x1) = 3
      n__filter_A(x1,x2,x3) = 3
      filter_A(x1,x2,x3) = 6
      sieve_A(x1) = 11
      nats_A(x1) = 6
    

The next rules are strictly ordered:

  p3, p4, p5, p6, p7, p8, p12

We remove them from the problem.

-- SCC decomposition.

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

p1: filter#(cons(X,Y),|0|(),M) -> activate#(Y)
p2: activate#(n__s(X)) -> activate#(X)
p3: activate#(n__filter(X1,X2,X3)) -> activate#(X1)
p4: activate#(n__filter(X1,X2,X3)) -> filter#(activate(X1),activate(X2),activate(X3))
p5: filter#(cons(X,Y),s(N),M) -> activate#(Y)

and R consists of:

r1: filter(cons(X,Y),|0|(),M) -> cons(|0|(),n__filter(activate(Y),M,M))
r2: filter(cons(X,Y),s(N),M) -> cons(X,n__filter(activate(Y),N,M))
r3: sieve(cons(|0|(),Y)) -> cons(|0|(),n__sieve(activate(Y)))
r4: sieve(cons(s(N),Y)) -> cons(s(N),n__sieve(n__filter(activate(Y),N,N)))
r5: nats(N) -> cons(N,n__nats(n__s(N)))
r6: zprimes() -> sieve(nats(s(s(|0|()))))
r7: filter(X1,X2,X3) -> n__filter(X1,X2,X3)
r8: sieve(X) -> n__sieve(X)
r9: nats(X) -> n__nats(X)
r10: s(X) -> n__s(X)
r11: activate(n__filter(X1,X2,X3)) -> filter(activate(X1),activate(X2),activate(X3))
r12: activate(n__sieve(X)) -> sieve(activate(X))
r13: activate(n__nats(X)) -> nats(activate(X))
r14: activate(n__s(X)) -> s(activate(X))
r15: 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: filter#(cons(X,Y),|0|(),M) -> activate#(Y)
p2: activate#(n__filter(X1,X2,X3)) -> filter#(activate(X1),activate(X2),activate(X3))
p3: filter#(cons(X,Y),s(N),M) -> activate#(Y)
p4: activate#(n__filter(X1,X2,X3)) -> activate#(X1)
p5: activate#(n__s(X)) -> activate#(X)

and R consists of:

r1: filter(cons(X,Y),|0|(),M) -> cons(|0|(),n__filter(activate(Y),M,M))
r2: filter(cons(X,Y),s(N),M) -> cons(X,n__filter(activate(Y),N,M))
r3: sieve(cons(|0|(),Y)) -> cons(|0|(),n__sieve(activate(Y)))
r4: sieve(cons(s(N),Y)) -> cons(s(N),n__sieve(n__filter(activate(Y),N,N)))
r5: nats(N) -> cons(N,n__nats(n__s(N)))
r6: zprimes() -> sieve(nats(s(s(|0|()))))
r7: filter(X1,X2,X3) -> n__filter(X1,X2,X3)
r8: sieve(X) -> n__sieve(X)
r9: nats(X) -> n__nats(X)
r10: s(X) -> n__s(X)
r11: activate(n__filter(X1,X2,X3)) -> filter(activate(X1),activate(X2),activate(X3))
r12: activate(n__sieve(X)) -> sieve(activate(X))
r13: activate(n__nats(X)) -> nats(activate(X))
r14: activate(n__s(X)) -> s(activate(X))
r15: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      filter#_A(x1,x2,x3) = max{x1 + 7, x2 - 1, x3 + 6}
      cons_A(x1,x2) = max{6, x2 - 5}
      |0|_A = 2
      activate#_A(x1) = x1
      n__filter_A(x1,x2,x3) = max{x1 + 7, x2 + 7, x3 + 7}
      activate_A(x1) = x1
      s_A(x1) = max{5, x1 + 4}
      n__s_A(x1) = max{5, x1 + 4}
      filter_A(x1,x2,x3) = max{x1 + 7, x2 + 7, x3 + 7}
      sieve_A(x1) = 18
      n__sieve_A(x1) = 18
      nats_A(x1) = x1 + 6
      n__nats_A(x1) = x1 + 6
    
    2. max/plus interpretations on natural numbers:
    
      filter#_A(x1,x2,x3) = 0
      cons_A(x1,x2) = 6
      |0|_A = 0
      activate#_A(x1) = x1 + 1
      n__filter_A(x1,x2,x3) = 0
      activate_A(x1) = max{8, x1 + 6}
      s_A(x1) = 6
      n__s_A(x1) = 5
      filter_A(x1,x2,x3) = 7
      sieve_A(x1) = 8
      n__sieve_A(x1) = 0
      nats_A(x1) = 7
      n__nats_A(x1) = 0
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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