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. lexicographic path order with precedence:
    
      precedence:
      
        filter# > nats > cons > |0| > activate# > s > n__sieve > n__nats > activate > sieve# > sieve > filter > n__filter
      
      argument filter:
    
        pi(filter#) = 1
        pi(cons) = 2
        pi(|0|) = []
        pi(activate#) = 1
        pi(n__sieve) = 1
        pi(sieve#) = 1
        pi(s) = 1
        pi(n__filter) = 1
        pi(activate) = 1
        pi(filter) = 1
        pi(sieve) = 1
        pi(nats) = 1
        pi(n__nats) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        filter#_A(x1,x2,x3) = x1
        cons_A(x1,x2) = x2
        |0|_A() = 1
        activate#_A(x1) = x1
        n__sieve_A(x1) = x1 + 2
        sieve#_A(x1) = x1 + 1
        s_A(x1) = x1 + 1
        n__filter_A(x1,x2,x3) = x1
        activate_A(x1) = x1
        filter_A(x1,x2,x3) = x1
        sieve_A(x1) = x1 + 2
        nats_A(x1) = 1
        n__nats_A(x1) = 1
    

The next rules are strictly ordered:

  p2, p3, p6, p7

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__filter(X1,X2,X3)) -> filter#(X1,X2,X3)
p3: 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(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}


-- 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#(X1,X2,X3)
p3: 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(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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        activate# > s > |0| > filter# > n__filter > cons
      
      argument filter:
    
        pi(filter#) = 1
        pi(cons) = [2]
        pi(|0|) = []
        pi(activate#) = 1
        pi(n__filter) = 1
        pi(s) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        filter#_A(x1,x2,x3) = 0
        cons_A(x1,x2) = x2 + 1
        |0|_A() = 1
        activate#_A(x1) = x1 + 1
        n__filter_A(x1,x2,x3) = x1 + 1
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1, p2, p3

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