YES

We show the termination of the TRS R:

  a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
  a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
  a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
  a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
  a__nats(N) -> cons(mark(N),nats(s(N)))
  a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
  mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
  mark(sieve(X)) -> a__sieve(mark(X))
  mark(nats(X)) -> a__nats(mark(X))
  mark(zprimes()) -> a__zprimes()
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(|0|()) -> |0|()
  mark(s(X)) -> s(mark(X))
  a__filter(X1,X2,X3) -> filter(X1,X2,X3)
  a__sieve(X) -> sieve(X)
  a__nats(X) -> nats(X)
  a__zprimes() -> zprimes()

-- SCC decomposition.

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

p1: a__filter#(cons(X,Y),s(N),M) -> mark#(X)
p2: a__sieve#(cons(s(N),Y)) -> mark#(N)
p3: a__nats#(N) -> mark#(N)
p4: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))
p5: a__zprimes#() -> a__nats#(s(s(|0|())))
p6: mark#(filter(X1,X2,X3)) -> a__filter#(mark(X1),mark(X2),mark(X3))
p7: mark#(filter(X1,X2,X3)) -> mark#(X1)
p8: mark#(filter(X1,X2,X3)) -> mark#(X2)
p9: mark#(filter(X1,X2,X3)) -> mark#(X3)
p10: mark#(sieve(X)) -> a__sieve#(mark(X))
p11: mark#(sieve(X)) -> mark#(X)
p12: mark#(nats(X)) -> a__nats#(mark(X))
p13: mark#(nats(X)) -> mark#(X)
p14: mark#(zprimes()) -> a__zprimes#()
p15: mark#(cons(X1,X2)) -> mark#(X1)
p16: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16}


-- Reduction pair.

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

p1: a__filter#(cons(X,Y),s(N),M) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(zprimes()) -> a__zprimes#()
p5: a__zprimes#() -> a__nats#(s(s(|0|())))
p6: a__nats#(N) -> mark#(N)
p7: mark#(nats(X)) -> mark#(X)
p8: mark#(nats(X)) -> a__nats#(mark(X))
p9: mark#(sieve(X)) -> mark#(X)
p10: mark#(sieve(X)) -> a__sieve#(mark(X))
p11: a__sieve#(cons(s(N),Y)) -> mark#(N)
p12: mark#(filter(X1,X2,X3)) -> mark#(X3)
p13: mark#(filter(X1,X2,X3)) -> mark#(X2)
p14: mark#(filter(X1,X2,X3)) -> mark#(X1)
p15: mark#(filter(X1,X2,X3)) -> a__filter#(mark(X1),mark(X2),mark(X3))
p16: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__filter#_A(x1,x2,x3) = max{21, x1 + 18, x2 + 17, x3 - 1}
        cons_A(x1,x2) = max{23, x1 + 13}
        s_A(x1) = x1 + 6
        mark#_A(x1) = max{21, x1}
        zprimes_A = 64
        a__zprimes#_A = 45
        a__nats#_A(x1) = x1 + 22
        |0|_A = 10
        nats_A(x1) = x1 + 41
        mark_A(x1) = x1
        sieve_A(x1) = max{22, x1}
        a__sieve#_A(x1) = max{21, x1 - 18}
        filter_A(x1,x2,x3) = max{x1 + 18, x2 + 20, x3 + 21}
        a__nats_A(x1) = x1 + 41
        a__filter_A(x1,x2,x3) = max{x1 + 18, x2 + 20, x3 + 21}
        a__sieve_A(x1) = max{22, x1}
        a__zprimes_A = 64
  
    precedence:
    
      a__zprimes# = a__nats# > nats = a__sieve# = a__nats > cons = mark > zprimes = |0| = a__zprimes > s > sieve = a__sieve > a__filter > filter > mark# > a__filter#
    
    partial status:
  
      pi(a__filter#) = []
      pi(cons) = []
      pi(s) = []
      pi(mark#) = [1]
      pi(zprimes) = []
      pi(a__zprimes#) = []
      pi(a__nats#) = []
      pi(|0|) = []
      pi(nats) = []
      pi(mark) = [1]
      pi(sieve) = [1]
      pi(a__sieve#) = []
      pi(filter) = [3]
      pi(a__nats) = []
      pi(a__filter) = [1, 2]
      pi(a__sieve) = [1]
      pi(a__zprimes) = []

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: mark#(s(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(zprimes()) -> a__zprimes#()
p4: a__zprimes#() -> a__nats#(s(s(|0|())))
p5: a__nats#(N) -> mark#(N)
p6: mark#(nats(X)) -> mark#(X)
p7: mark#(nats(X)) -> a__nats#(mark(X))
p8: mark#(sieve(X)) -> mark#(X)
p9: mark#(sieve(X)) -> a__sieve#(mark(X))
p10: a__sieve#(cons(s(N),Y)) -> mark#(N)
p11: mark#(filter(X1,X2,X3)) -> mark#(X3)
p12: mark#(filter(X1,X2,X3)) -> mark#(X2)
p13: mark#(filter(X1,X2,X3)) -> mark#(X1)
p14: mark#(filter(X1,X2,X3)) -> a__filter#(mark(X1),mark(X2),mark(X3))
p15: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p15}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(filter(X1,X2,X3)) -> mark#(X1)
p3: mark#(filter(X1,X2,X3)) -> mark#(X2)
p4: mark#(filter(X1,X2,X3)) -> mark#(X3)
p5: mark#(sieve(X)) -> a__sieve#(mark(X))
p6: a__sieve#(cons(s(N),Y)) -> mark#(N)
p7: mark#(sieve(X)) -> mark#(X)
p8: mark#(nats(X)) -> a__nats#(mark(X))
p9: a__nats#(N) -> mark#(N)
p10: mark#(nats(X)) -> mark#(X)
p11: mark#(zprimes()) -> a__zprimes#()
p12: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))
p13: a__zprimes#() -> a__nats#(s(s(|0|())))
p14: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 26
        s_A(x1) = max{18, x1 + 15}
        filter_A(x1,x2,x3) = max{58, x1 + 27, x2 + 26, x3 + 41}
        sieve_A(x1) = x1
        a__sieve#_A(x1) = max{26, x1 + 23}
        mark_A(x1) = max{2, x1}
        cons_A(x1,x2) = max{40, x1 + 11}
        nats_A(x1) = max{43, x1 + 41}
        a__nats#_A(x1) = x1 + 38
        zprimes_A = 111
        a__zprimes#_A = 110
        a__nats_A(x1) = max{43, x1 + 41}
        |0|_A = 15
        a__filter_A(x1,x2,x3) = max{58, x1 + 27, x2 + 26, x3 + 41}
        a__sieve_A(x1) = x1
        a__zprimes_A = 111
  
    precedence:
    
      s = mark = cons = zprimes = a__nats = |0| = a__filter = a__zprimes > a__sieve > filter = sieve > mark# > a__sieve# > a__zprimes# > nats = a__nats#
    
    partial status:
  
      pi(mark#) = [1]
      pi(s) = []
      pi(filter) = []
      pi(sieve) = [1]
      pi(a__sieve#) = [1]
      pi(mark) = [1]
      pi(cons) = []
      pi(nats) = [1]
      pi(a__nats#) = []
      pi(zprimes) = []
      pi(a__zprimes#) = []
      pi(a__nats) = [1]
      pi(|0|) = []
      pi(a__filter) = [1]
      pi(a__sieve) = [1]
      pi(a__zprimes) = []

The next rules are strictly ordered:

  p13

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(filter(X1,X2,X3)) -> mark#(X1)
p3: mark#(filter(X1,X2,X3)) -> mark#(X2)
p4: mark#(filter(X1,X2,X3)) -> mark#(X3)
p5: mark#(sieve(X)) -> a__sieve#(mark(X))
p6: a__sieve#(cons(s(N),Y)) -> mark#(N)
p7: mark#(sieve(X)) -> mark#(X)
p8: mark#(nats(X)) -> a__nats#(mark(X))
p9: a__nats#(N) -> mark#(N)
p10: mark#(nats(X)) -> mark#(X)
p11: mark#(zprimes()) -> a__zprimes#()
p12: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))
p13: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(zprimes()) -> a__zprimes#()
p4: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))
p5: a__sieve#(cons(s(N),Y)) -> mark#(N)
p6: mark#(nats(X)) -> mark#(X)
p7: mark#(nats(X)) -> a__nats#(mark(X))
p8: a__nats#(N) -> mark#(N)
p9: mark#(sieve(X)) -> mark#(X)
p10: mark#(sieve(X)) -> a__sieve#(mark(X))
p11: mark#(filter(X1,X2,X3)) -> mark#(X3)
p12: mark#(filter(X1,X2,X3)) -> mark#(X2)
p13: mark#(filter(X1,X2,X3)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 28
        s_A(x1) = x1 + 23
        cons_A(x1,x2) = x1 + 11
        zprimes_A = 118
        a__zprimes#_A = 118
        a__sieve#_A(x1) = max{50, x1 + 27}
        a__nats_A(x1) = max{51, x1 + 33}
        |0|_A = 12
        nats_A(x1) = x1 + 33
        a__nats#_A(x1) = max{61, x1 + 29}
        mark_A(x1) = x1 + 22
        sieve_A(x1) = x1 + 22
        filter_A(x1,x2,x3) = max{25, x1 + 22, x2 + 1, x3}
        a__filter_A(x1,x2,x3) = max{25, x1 + 22, x2 + 1, x3}
        a__sieve_A(x1) = x1 + 22
        a__zprimes_A = 119
  
    precedence:
    
      mark# = |0| = a__nats# > cons = a__nats = nats = mark = filter = a__filter = a__sieve > zprimes = a__zprimes# = a__zprimes > s = a__sieve# > sieve
    
    partial status:
  
      pi(mark#) = []
      pi(s) = []
      pi(cons) = []
      pi(zprimes) = []
      pi(a__zprimes#) = []
      pi(a__sieve#) = [1]
      pi(a__nats) = []
      pi(|0|) = []
      pi(nats) = []
      pi(a__nats#) = []
      pi(mark) = []
      pi(sieve) = []
      pi(filter) = []
      pi(a__filter) = []
      pi(a__sieve) = []
      pi(a__zprimes) = []

The next rules are strictly ordered:

  p10

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(zprimes()) -> a__zprimes#()
p4: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))
p5: a__sieve#(cons(s(N),Y)) -> mark#(N)
p6: mark#(nats(X)) -> mark#(X)
p7: mark#(nats(X)) -> a__nats#(mark(X))
p8: a__nats#(N) -> mark#(N)
p9: mark#(sieve(X)) -> mark#(X)
p10: mark#(filter(X1,X2,X3)) -> mark#(X3)
p11: mark#(filter(X1,X2,X3)) -> mark#(X2)
p12: mark#(filter(X1,X2,X3)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(filter(X1,X2,X3)) -> mark#(X1)
p3: mark#(filter(X1,X2,X3)) -> mark#(X2)
p4: mark#(filter(X1,X2,X3)) -> mark#(X3)
p5: mark#(sieve(X)) -> mark#(X)
p6: mark#(nats(X)) -> a__nats#(mark(X))
p7: a__nats#(N) -> mark#(N)
p8: mark#(nats(X)) -> mark#(X)
p9: mark#(zprimes()) -> a__zprimes#()
p10: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))
p11: a__sieve#(cons(s(N),Y)) -> mark#(N)
p12: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{41, x1 - 95}
        s_A(x1) = x1 + 41
        filter_A(x1,x2,x3) = max{161, x1 + 45, x2 + 137, x3 + 21}
        sieve_A(x1) = x1 + 51
        nats_A(x1) = x1 + 159
        a__nats#_A(x1) = max{43, x1 - 79}
        mark_A(x1) = x1 + 40
        zprimes_A = 276
        a__zprimes#_A = 41
        a__sieve#_A(x1) = max{41, x1 - 222}
        a__nats_A(x1) = x1 + 159
        |0|_A = 22
        cons_A(x1,x2) = max{159, x1 + 87}
        a__filter_A(x1,x2,x3) = max{176, x1 + 45, x2 + 137, x3 + 21}
        a__sieve_A(x1) = max{52, x1 + 51}
        a__zprimes_A = 315
  
    precedence:
    
      mark = a__zprimes > a__zprimes# > zprimes = a__sieve# > mark# = s = filter = a__nats# = |0| = a__filter > a__nats > cons > sieve = nats = a__sieve
    
    partial status:
  
      pi(mark#) = []
      pi(s) = [1]
      pi(filter) = []
      pi(sieve) = []
      pi(nats) = []
      pi(a__nats#) = []
      pi(mark) = []
      pi(zprimes) = []
      pi(a__zprimes#) = []
      pi(a__sieve#) = []
      pi(a__nats) = []
      pi(|0|) = []
      pi(cons) = [1]
      pi(a__filter) = []
      pi(a__sieve) = [1]
      pi(a__zprimes) = []

The next rules are strictly ordered:

  p9

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(filter(X1,X2,X3)) -> mark#(X1)
p3: mark#(filter(X1,X2,X3)) -> mark#(X2)
p4: mark#(filter(X1,X2,X3)) -> mark#(X3)
p5: mark#(sieve(X)) -> mark#(X)
p6: mark#(nats(X)) -> a__nats#(mark(X))
p7: a__nats#(N) -> mark#(N)
p8: mark#(nats(X)) -> mark#(X)
p9: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))
p10: a__sieve#(cons(s(N),Y)) -> mark#(N)
p11: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(nats(X)) -> mark#(X)
p4: mark#(nats(X)) -> a__nats#(mark(X))
p5: a__nats#(N) -> mark#(N)
p6: mark#(sieve(X)) -> mark#(X)
p7: mark#(filter(X1,X2,X3)) -> mark#(X3)
p8: mark#(filter(X1,X2,X3)) -> mark#(X2)
p9: mark#(filter(X1,X2,X3)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 1
        s_A(x1) = max{4, x1 + 2}
        cons_A(x1,x2) = max{88, x1 + 16}
        nats_A(x1) = x1 + 112
        a__nats#_A(x1) = x1 + 95
        mark_A(x1) = x1 + 17
        sieve_A(x1) = max{111, x1 + 22}
        filter_A(x1,x2,x3) = max{44, x1 + 42, x2 + 32, x3 + 39}
        a__filter_A(x1,x2,x3) = max{45, x1 + 42, x2 + 32, x3 + 39}
        |0|_A = 1
        a__sieve_A(x1) = max{111, x1 + 22}
        a__nats_A(x1) = x1 + 112
        a__zprimes_A = 140
        zprimes_A = 124
  
    precedence:
    
      cons = mark > nats = a__nats > mark# = a__nats# = sieve = |0| = a__sieve = a__zprimes = zprimes > s = filter = a__filter
    
    partial status:
  
      pi(mark#) = [1]
      pi(s) = [1]
      pi(cons) = []
      pi(nats) = []
      pi(a__nats#) = [1]
      pi(mark) = [1]
      pi(sieve) = []
      pi(filter) = []
      pi(a__filter) = []
      pi(|0|) = []
      pi(a__sieve) = []
      pi(a__nats) = []
      pi(a__zprimes) = []
      pi(zprimes) = []

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: mark#(s(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(nats(X)) -> mark#(X)
p4: a__nats#(N) -> mark#(N)
p5: mark#(sieve(X)) -> mark#(X)
p6: mark#(filter(X1,X2,X3)) -> mark#(X3)
p7: mark#(filter(X1,X2,X3)) -> mark#(X2)
p8: mark#(filter(X1,X2,X3)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p5, p6, p7, p8}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(filter(X1,X2,X3)) -> mark#(X1)
p3: mark#(filter(X1,X2,X3)) -> mark#(X2)
p4: mark#(filter(X1,X2,X3)) -> mark#(X3)
p5: mark#(sieve(X)) -> mark#(X)
p6: mark#(nats(X)) -> mark#(X)
p7: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 3
        s_A(x1) = x1
        filter_A(x1,x2,x3) = max{x1 + 2, x2 + 4, x3 + 2}
        sieve_A(x1) = x1
        nats_A(x1) = x1
        cons_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      mark# = s = filter = sieve = nats = cons
    
    partial status:
  
      pi(mark#) = [1]
      pi(s) = [1]
      pi(filter) = [2, 3]
      pi(sieve) = [1]
      pi(nats) = [1]
      pi(cons) = [1, 2]

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: mark#(s(X)) -> mark#(X)
p2: mark#(filter(X1,X2,X3)) -> mark#(X2)
p3: mark#(filter(X1,X2,X3)) -> mark#(X3)
p4: mark#(sieve(X)) -> mark#(X)
p5: mark#(nats(X)) -> mark#(X)
p6: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

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: mark#(s(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(nats(X)) -> mark#(X)
p4: mark#(sieve(X)) -> mark#(X)
p5: mark#(filter(X1,X2,X3)) -> mark#(X3)
p6: mark#(filter(X1,X2,X3)) -> mark#(X2)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 4
        s_A(x1) = x1
        cons_A(x1,x2) = max{x1 + 1, x2}
        nats_A(x1) = x1 + 3
        sieve_A(x1) = x1
        filter_A(x1,x2,x3) = max{x1 + 1, x2 + 2, x3}
  
    precedence:
    
      mark# = s = cons = nats = sieve = filter
    
    partial status:
  
      pi(mark#) = []
      pi(s) = [1]
      pi(cons) = [2]
      pi(nats) = [1]
      pi(sieve) = [1]
      pi(filter) = [3]

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: mark#(s(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(sieve(X)) -> mark#(X)
p4: mark#(filter(X1,X2,X3)) -> mark#(X3)
p5: mark#(filter(X1,X2,X3)) -> mark#(X2)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

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: mark#(s(X)) -> mark#(X)
p2: mark#(filter(X1,X2,X3)) -> mark#(X2)
p3: mark#(filter(X1,X2,X3)) -> mark#(X3)
p4: mark#(sieve(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{3, x1 + 2}
        s_A(x1) = x1
        filter_A(x1,x2,x3) = max{x2, x3}
        sieve_A(x1) = max{3, x1 + 2}
        cons_A(x1,x2) = max{x1 + 1, x2 + 1}
  
    precedence:
    
      mark# = s = filter = sieve = cons
    
    partial status:
  
      pi(mark#) = []
      pi(s) = [1]
      pi(filter) = [3]
      pi(sieve) = [1]
      pi(cons) = [2]

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: mark#(s(X)) -> mark#(X)
p2: mark#(filter(X1,X2,X3)) -> mark#(X2)
p3: mark#(filter(X1,X2,X3)) -> mark#(X3)
p4: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(filter(X1,X2,X3)) -> mark#(X3)
p4: mark#(filter(X1,X2,X3)) -> mark#(X2)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 3
        s_A(x1) = x1 + 1
        cons_A(x1,x2) = max{x1, x2}
        filter_A(x1,x2,x3) = max{2, x2 + 1, x3}
  
    precedence:
    
      mark# = s = cons = filter
    
    partial status:
  
      pi(mark#) = []
      pi(s) = [1]
      pi(cons) = [2]
      pi(filter) = [3]

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: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(filter(X1,X2,X3)) -> mark#(X3)
p3: mark#(filter(X1,X2,X3)) -> mark#(X2)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

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: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(filter(X1,X2,X3)) -> mark#(X2)
p3: mark#(filter(X1,X2,X3)) -> mark#(X3)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 4
        cons_A(x1,x2) = max{x1 + 1, x2}
        filter_A(x1,x2,x3) = max{x1 - 1, x2 + 1, x3}
  
    precedence:
    
      mark# = cons = filter
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]
      pi(filter) = [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: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(filter(X1,X2,X3)) -> mark#(X3)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(filter(X1,X2,X3)) -> mark#(X3)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{4, x1 + 3}
        cons_A(x1,x2) = max{x1, x2 + 1}
        filter_A(x1,x2,x3) = max{3, x1 + 2, x2 + 2, x3 + 2}
  
    precedence:
    
      mark# = cons = filter
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]
      pi(filter) = [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: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__filter(cons(X,Y),|0|(),M) -> cons(|0|(),filter(Y,M,M))
r2: a__filter(cons(X,Y),s(N),M) -> cons(mark(X),filter(Y,N,M))
r3: a__sieve(cons(|0|(),Y)) -> cons(|0|(),sieve(Y))
r4: a__sieve(cons(s(N),Y)) -> cons(s(mark(N)),sieve(filter(Y,N,N)))
r5: a__nats(N) -> cons(mark(N),nats(s(N)))
r6: a__zprimes() -> a__sieve(a__nats(s(s(|0|()))))
r7: mark(filter(X1,X2,X3)) -> a__filter(mark(X1),mark(X2),mark(X3))
r8: mark(sieve(X)) -> a__sieve(mark(X))
r9: mark(nats(X)) -> a__nats(mark(X))
r10: mark(zprimes()) -> a__zprimes()
r11: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r12: mark(|0|()) -> |0|()
r13: mark(s(X)) -> s(mark(X))
r14: a__filter(X1,X2,X3) -> filter(X1,X2,X3)
r15: a__sieve(X) -> sieve(X)
r16: a__nats(X) -> nats(X)
r17: a__zprimes() -> zprimes()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 2
        cons_A(x1,x2) = max{x1 + 1, x2 + 1}
  
    precedence:
    
      mark# = cons
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]

The next rules are strictly ordered:

  p1

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