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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        filter#_A(x1,x2,x3) = max{3, x1 - 2, x3}
        cons_A(x1,x2) = max{40, x1 + 20, x2}
        |0|_A = 10
        activate#_A(x1) = max{12, x1 - 2}
        n__s_A(x1) = max{8, x1}
        n__nats_A(x1) = max{41, x1 + 20}
        n__sieve_A(x1) = x1 + 11
        sieve#_A(x1) = max{12, x1 + 8}
        activate_A(x1) = x1
        s_A(x1) = max{8, x1}
        n__filter_A(x1,x2,x3) = max{x1, x2 + 6, x3 + 20}
        filter_A(x1,x2,x3) = max{x1, x2 + 6, x3 + 20}
        sieve_A(x1) = x1 + 11
        nats_A(x1) = max{41, x1 + 20}
  
    precedence:
    
      filter# = activate# = activate = n__filter = filter = nats > |0| = s > n__s = sieve > sieve# > cons = n__nats = n__sieve
    
    partial status:
  
      pi(filter#) = []
      pi(cons) = []
      pi(|0|) = []
      pi(activate#) = []
      pi(n__s) = []
      pi(n__nats) = [1]
      pi(n__sieve) = [1]
      pi(sieve#) = [1]
      pi(activate) = []
      pi(s) = []
      pi(n__filter) = []
      pi(filter) = []
      pi(sieve) = []
      pi(nats) = []

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: filter#(cons(X,Y),|0|(),M) -> activate#(Y)
p2: activate#(n__s(X)) -> activate#(X)
p3: activate#(n__sieve(X)) -> activate#(X)
p4: activate#(n__sieve(X)) -> sieve#(activate(X))
p5: sieve#(cons(s(N),Y)) -> activate#(Y)
p6: activate#(n__filter(X1,X2,X3)) -> activate#(X3)
p7: activate#(n__filter(X1,X2,X3)) -> activate#(X2)
p8: activate#(n__filter(X1,X2,X3)) -> activate#(X1)
p9: activate#(n__filter(X1,X2,X3)) -> filter#(activate(X1),activate(X2),activate(X3))
p10: filter#(cons(X,Y),s(N),M) -> activate#(Y)
p11: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11}


-- 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__filter(X1,X2,X3)) -> activate#(X2)
p6: activate#(n__filter(X1,X2,X3)) -> activate#(X3)
p7: activate#(n__sieve(X)) -> sieve#(activate(X))
p8: sieve#(cons(|0|(),Y)) -> activate#(Y)
p9: activate#(n__sieve(X)) -> activate#(X)
p10: activate#(n__s(X)) -> activate#(X)
p11: sieve#(cons(s(N),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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        filter#_A(x1,x2,x3) = max{x1 + 6, x2 + 5, x3 + 5}
        cons_A(x1,x2) = max{x1 + 4, x2}
        |0|_A = 1
        activate#_A(x1) = x1 + 6
        n__filter_A(x1,x2,x3) = max{x1, x2 + 4, x3 + 4}
        activate_A(x1) = x1
        s_A(x1) = x1
        n__sieve_A(x1) = max{3, x1 + 2}
        sieve#_A(x1) = x1 + 6
        n__s_A(x1) = x1
        filter_A(x1,x2,x3) = max{x1, x2 + 4, x3 + 4}
        sieve_A(x1) = max{3, x1 + 2}
        nats_A(x1) = x1 + 4
        n__nats_A(x1) = x1 + 4
  
    precedence:
    
      n__filter = sieve# = filter > cons = s = n__sieve = n__s = sieve = nats = n__nats > filter# = activate# > |0| = activate
    
    partial status:
  
      pi(filter#) = []
      pi(cons) = []
      pi(|0|) = []
      pi(activate#) = []
      pi(n__filter) = []
      pi(activate) = [1]
      pi(s) = []
      pi(n__sieve) = []
      pi(sieve#) = [1]
      pi(n__s) = []
      pi(filter) = []
      pi(sieve) = []
      pi(nats) = []
      pi(n__nats) = []

The next rules are strictly ordered:

  p11

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#(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__filter(X1,X2,X3)) -> activate#(X2)
p6: activate#(n__filter(X1,X2,X3)) -> activate#(X3)
p7: activate#(n__sieve(X)) -> sieve#(activate(X))
p8: sieve#(cons(|0|(),Y)) -> activate#(Y)
p9: activate#(n__sieve(X)) -> activate#(X)
p10: 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, p5, p6, p7, p8, p9, p10}


-- 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__sieve(X)) -> activate#(X)
p4: activate#(n__sieve(X)) -> sieve#(activate(X))
p5: sieve#(cons(|0|(),Y)) -> activate#(Y)
p6: activate#(n__filter(X1,X2,X3)) -> activate#(X3)
p7: activate#(n__filter(X1,X2,X3)) -> activate#(X2)
p8: activate#(n__filter(X1,X2,X3)) -> activate#(X1)
p9: activate#(n__filter(X1,X2,X3)) -> filter#(activate(X1),activate(X2),activate(X3))
p10: 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 set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        filter#_A(x1,x2,x3) = max{x1 + 2, x2 - 1, x3 + 9}
        cons_A(x1,x2) = max{x1 + 9, x2}
        |0|_A = 0
        activate#_A(x1) = max{7, x1 + 2}
        n__s_A(x1) = x1
        n__sieve_A(x1) = x1 + 11
        sieve#_A(x1) = max{6, x1 + 2}
        activate_A(x1) = x1
        n__filter_A(x1,x2,x3) = max{x1, x2, x3 + 8}
        s_A(x1) = x1
        filter_A(x1,x2,x3) = max{x1, x2, x3 + 8}
        sieve_A(x1) = x1 + 11
        nats_A(x1) = max{11, x1 + 10}
        n__nats_A(x1) = max{11, x1 + 10}
  
    precedence:
    
      |0| = sieve# > filter# = activate# = n__sieve = sieve > nats = n__nats > activate = filter > cons > n__s = n__filter = s
    
    partial status:
  
      pi(filter#) = []
      pi(cons) = []
      pi(|0|) = []
      pi(activate#) = []
      pi(n__s) = [1]
      pi(n__sieve) = []
      pi(sieve#) = []
      pi(activate) = [1]
      pi(n__filter) = [3]
      pi(s) = [1]
      pi(filter) = []
      pi(sieve) = []
      pi(nats) = []
      pi(n__nats) = []

The next rules are strictly ordered:

  p4

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__sieve(X)) -> activate#(X)
p4: sieve#(cons(|0|(),Y)) -> activate#(Y)
p5: activate#(n__filter(X1,X2,X3)) -> activate#(X3)
p6: activate#(n__filter(X1,X2,X3)) -> activate#(X2)
p7: activate#(n__filter(X1,X2,X3)) -> activate#(X1)
p8: activate#(n__filter(X1,X2,X3)) -> filter#(activate(X1),activate(X2),activate(X3))
p9: 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, p5, p6, p7, 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__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__filter(X1,X2,X3)) -> activate#(X2)
p6: activate#(n__filter(X1,X2,X3)) -> activate#(X3)
p7: activate#(n__sieve(X)) -> activate#(X)
p8: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        filter#_A(x1,x2,x3) = max{x1 - 2, x2 + 2, x3 - 3}
        cons_A(x1,x2) = max{x1 + 6, x2}
        |0|_A = 2
        activate#_A(x1) = max{4, x1 - 2}
        n__filter_A(x1,x2,x3) = max{15, x1, x2 + 5, x3 + 6}
        activate_A(x1) = x1
        s_A(x1) = max{10, x1}
        n__sieve_A(x1) = x1 + 3
        n__s_A(x1) = max{10, x1}
        filter_A(x1,x2,x3) = max{15, x1, x2 + 5, x3 + 6}
        sieve_A(x1) = x1 + 3
        nats_A(x1) = max{21, x1 + 11}
        n__nats_A(x1) = max{21, x1 + 11}
  
    precedence:
    
      cons = n__filter = n__sieve = filter = sieve = nats = n__nats > activate > s > filter# = activate# = n__s > |0|
    
    partial status:
  
      pi(filter#) = []
      pi(cons) = []
      pi(|0|) = []
      pi(activate#) = []
      pi(n__filter) = []
      pi(activate) = [1]
      pi(s) = [1]
      pi(n__sieve) = []
      pi(n__s) = [1]
      pi(filter) = []
      pi(sieve) = []
      pi(nats) = []
      pi(n__nats) = []

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: 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__filter(X1,X2,X3)) -> activate#(X3)
p6: activate#(n__sieve(X)) -> activate#(X)
p7: 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, p5, p6, p7}


-- 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__sieve(X)) -> activate#(X)
p4: activate#(n__filter(X1,X2,X3)) -> activate#(X3)
p5: activate#(n__filter(X1,X2,X3)) -> activate#(X1)
p6: activate#(n__filter(X1,X2,X3)) -> filter#(activate(X1),activate(X2),activate(X3))
p7: 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 set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        filter#_A(x1,x2,x3) = max{x1, x2 + 1, x3 - 1}
        cons_A(x1,x2) = max{x1 + 6, x2}
        |0|_A = 4
        activate#_A(x1) = max{5, x1}
        n__s_A(x1) = x1
        n__sieve_A(x1) = x1 + 12
        n__filter_A(x1,x2,x3) = max{x1, x2 + 6, x3 + 6}
        activate_A(x1) = x1
        s_A(x1) = x1
        filter_A(x1,x2,x3) = max{x1, x2 + 6, x3 + 6}
        sieve_A(x1) = x1 + 12
        nats_A(x1) = x1 + 7
        n__nats_A(x1) = x1 + 7
  
    precedence:
    
      |0| = n__filter = filter > cons = n__sieve = sieve = nats = n__nats > filter# = activate# > activate > s > n__s
    
    partial status:
  
      pi(filter#) = []
      pi(cons) = []
      pi(|0|) = []
      pi(activate#) = []
      pi(n__s) = [1]
      pi(n__sieve) = []
      pi(n__filter) = []
      pi(activate) = [1]
      pi(s) = [1]
      pi(filter) = []
      pi(sieve) = []
      pi(nats) = []
      pi(n__nats) = []

The next rules are strictly ordered:

  p4

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__sieve(X)) -> activate#(X)
p4: activate#(n__filter(X1,X2,X3)) -> activate#(X1)
p5: activate#(n__filter(X1,X2,X3)) -> filter#(activate(X1),activate(X2),activate(X3))
p6: 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, p6}


-- 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__sieve(X)) -> activate#(X)
p6: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        filter#_A(x1,x2,x3) = max{x1 + 15, x2 + 14, x3 + 14}
        cons_A(x1,x2) = max{11, x1 + 4, x2}
        |0|_A = 2
        activate#_A(x1) = x1 + 15
        n__filter_A(x1,x2,x3) = max{x1, x2 + 2, x3 + 3}
        activate_A(x1) = x1
        s_A(x1) = x1
        n__sieve_A(x1) = max{12, x1 + 7}
        n__s_A(x1) = x1
        filter_A(x1,x2,x3) = max{x1, x2 + 2, x3 + 3}
        sieve_A(x1) = max{12, x1 + 7}
        nats_A(x1) = x1 + 11
        n__nats_A(x1) = x1 + 11
  
    precedence:
    
      |0| = activate > sieve > nats = n__nats > filter# = cons = activate# = n__filter = s = n__sieve = n__s = filter
    
    partial status:
  
      pi(filter#) = []
      pi(cons) = []
      pi(|0|) = []
      pi(activate#) = []
      pi(n__filter) = [1, 3]
      pi(activate) = [1]
      pi(s) = []
      pi(n__sieve) = [1]
      pi(n__s) = []
      pi(filter) = [1, 3]
      pi(sieve) = [1]
      pi(nats) = [1]
      pi(n__nats) = [1]

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: 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 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__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 set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        filter#_A(x1,x2,x3) = max{x1, x2, x3 + 2}
        cons_A(x1,x2) = max{2, x2}
        |0|_A = 1
        activate#_A(x1) = max{2, x1}
        n__s_A(x1) = max{2, x1}
        n__filter_A(x1,x2,x3) = max{17, x1 + 7, x2 + 12, x3 + 12}
        activate_A(x1) = max{4, x1}
        s_A(x1) = max{2, x1}
        filter_A(x1,x2,x3) = max{17, x1 + 7, x2 + 12, x3 + 12}
        sieve_A(x1) = 4
        n__sieve_A(x1) = 3
        nats_A(x1) = max{4, x1 - 3}
        n__nats_A(x1) = max{0, x1 - 3}
  
    precedence:
    
      activate = nats > |0| = s = filter > n__s = n__filter = sieve > cons > filter# = activate# = n__sieve > n__nats
    
    partial status:
  
      pi(filter#) = [1, 2, 3]
      pi(cons) = [2]
      pi(|0|) = []
      pi(activate#) = [1]
      pi(n__s) = [1]
      pi(n__filter) = []
      pi(activate) = [1]
      pi(s) = [1]
      pi(filter) = []
      pi(sieve) = []
      pi(n__sieve) = []
      pi(nats) = []
      pi(n__nats) = []

The next rules are strictly ordered:

  p4

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: 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:

  {p2, p3}


-- Reduction pair.

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

p1: activate#(n__s(X)) -> activate#(X)
p2: activate#(n__filter(X1,X2,X3)) -> activate#(X1)

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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        activate#_A(x1) = max{0, x1 - 2}
        n__s_A(x1) = x1
        n__filter_A(x1,x2,x3) = max{x1 + 1, x2 + 3, x3 + 3}
  
    precedence:
    
      n__filter > activate# = n__s
    
    partial status:
  
      pi(activate#) = []
      pi(n__s) = [1]
      pi(n__filter) = [1, 3]

The next rules are strictly ordered:

  p2

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: 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}


-- 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: 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

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        activate#_A(x1) = x1 + 2
        n__s_A(x1) = x1 + 2
  
    precedence:
    
      activate# = n__s
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__s) = [1]

The next rules are strictly ordered:

  p1

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