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(filter(activate(Y),N,N)))
  nats(N) -> cons(N,n__nats(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)
  activate(n__filter(X1,X2,X3)) -> filter(X1,X2,X3)
  activate(n__sieve(X)) -> sieve(X)
  activate(n__nats(X)) -> nats(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)) -> filter#(activate(Y),N,N)
p5: sieve#(cons(s(N),Y)) -> activate#(Y)
p6: zprimes#() -> sieve#(nats(s(s(|0|()))))
p7: zprimes#() -> nats#(s(s(|0|())))
p8: activate#(n__filter(X1,X2,X3)) -> filter#(X1,X2,X3)
p9: activate#(n__sieve(X)) -> sieve#(X)
p10: activate#(n__nats(X)) -> nats#(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(filter(activate(Y),N,N)))
r5: nats(N) -> cons(N,n__nats(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: activate(n__filter(X1,X2,X3)) -> filter(X1,X2,X3)
r11: activate(n__sieve(X)) -> sieve(X)
r12: activate(n__nats(X)) -> nats(X)
r13: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p8, p9}


-- 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__sieve(X)) -> sieve#(X)
p3: sieve#(cons(s(N),Y)) -> activate#(Y)
p4: activate#(n__filter(X1,X2,X3)) -> filter#(X1,X2,X3)
p5: filter#(cons(X,Y),s(N),M) -> activate#(Y)
p6: sieve#(cons(s(N),Y)) -> filter#(activate(Y),N,N)
p7: 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(filter(activate(Y),N,N)))
r5: nats(N) -> cons(N,n__nats(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: activate(n__filter(X1,X2,X3)) -> filter(X1,X2,X3)
r11: activate(n__sieve(X)) -> sieve(X)
r12: activate(n__nats(X)) -> nats(X)
r13: activate(X) -> X

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          filter#_A(x1,x2,x3) = x1 + 8
          cons_A(x1,x2) = max{x1 + 7, x2}
          |0|_A = 0
          activate#_A(x1) = max{15, x1 + 8}
          n__sieve_A(x1) = max{17, x1 + 9}
          sieve#_A(x1) = x1 + 8
          s_A(x1) = 0
          n__filter_A(x1,x2,x3) = max{1, x1}
          activate_A(x1) = max{7, x1}
          filter_A(x1,x2,x3) = max{4, x1}
          sieve_A(x1) = max{17, x1 + 9}
          nats_A(x1) = x1 + 7
          n__nats_A(x1) = x1 + 7
    
      precedence:
      
        sieve# = activate = filter > nats > n__nats > sieve > n__filter > filter# = activate# > n__sieve > |0| = s > cons
      
      partial status:
    
        pi(filter#) = [1]
        pi(cons) = [2]
        pi(|0|) = []
        pi(activate#) = [1]
        pi(n__sieve) = [1]
        pi(sieve#) = [1]
        pi(s) = []
        pi(n__filter) = [1]
        pi(activate) = [1]
        pi(filter) = [1]
        pi(sieve) = [1]
        pi(nats) = [1]
        pi(n__nats) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          filter#_A(x1,x2,x3) = max{3, x1 - 7}
          cons_A(x1,x2) = x2 + 24
          |0|_A = 3
          activate#_A(x1) = x1 + 8
          n__sieve_A(x1) = max{25, x1 - 6}
          sieve#_A(x1) = x1 + 2
          s_A(x1) = 51
          n__filter_A(x1,x2,x3) = max{8, x1 - 15}
          activate_A(x1) = x1 + 31
          filter_A(x1,x2,x3) = max{31, x1 - 15}
          sieve_A(x1) = max{32, x1 + 25}
          nats_A(x1) = max{76, x1 + 1}
          n__nats_A(x1) = max{45, x1}
    
      precedence:
      
        activate = sieve > filter > cons > |0| = n__filter > filter# = activate# = s = nats > n__sieve = sieve# = n__nats
      
      partial status:
    
        pi(filter#) = []
        pi(cons) = [2]
        pi(|0|) = []
        pi(activate#) = [1]
        pi(n__sieve) = []
        pi(sieve#) = [1]
        pi(s) = []
        pi(n__filter) = []
        pi(activate) = [1]
        pi(filter) = []
        pi(sieve) = []
        pi(nats) = [1]
        pi(n__nats) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7

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