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:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          a__filter#_A(x1,x2,x3) = x1 + x2
          cons_A(x1,x2) = x1 + 2
          s_A(x1) = x1
          mark#_A(x1) = x1 + 1
          zprimes_A() = 5
          a__zprimes#_A() = 4
          a__nats#_A(x1) = x1 + 3
          |0|_A() = 0
          nats_A(x1) = x1 + 3
          mark_A(x1) = x1
          sieve_A(x1) = x1
          a__sieve#_A(x1) = x1
          filter_A(x1,x2,x3) = x1 + x2 + x3
          a__nats_A(x1) = x1 + 3
          a__filter_A(x1,x2,x3) = x1 + x2 + x3
          a__sieve_A(x1) = x1
          a__zprimes_A() = 5
  
    precedence:
    
      |0| > zprimes = nats = mark = sieve = filter = a__nats = a__filter = a__sieve = a__zprimes > cons > a__sieve# > a__filter# = mark# = a__zprimes# > s = a__nats#
    
    partial status:
  
      pi(a__filter#) = [1]
      pi(cons) = []
      pi(s) = [1]
      pi(mark#) = [1]
      pi(zprimes) = []
      pi(a__zprimes#) = []
      pi(a__nats#) = []
      pi(|0|) = []
      pi(nats) = [1]
      pi(mark) = [1]
      pi(sieve) = [1]
      pi(a__sieve#) = []
      pi(filter) = [1, 2, 3]
      pi(a__nats) = [1]
      pi(a__filter) = [1, 2, 3]
      pi(a__sieve) = [1]
      pi(a__zprimes) = []

The next rules are strictly ordered:

  p15

We remove them from the problem.

-- 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: 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: 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:

  {p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, 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:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          s_A(x1) = x1
          filter_A(x1,x2,x3) = x1 + x2 + x3 + 1
          sieve_A(x1) = x1 + 6
          a__sieve#_A(x1) = x1
          mark_A(x1) = x1
          cons_A(x1,x2) = x1 + 2
          nats_A(x1) = x1 + 3
          a__nats#_A(x1) = x1 + 2
          zprimes_A() = 13
          a__zprimes#_A() = 7
          a__nats_A(x1) = x1 + 3
          |0|_A() = 3
          a__filter_A(x1,x2,x3) = x1 + x2 + x3 + 1
          a__sieve_A(x1) = x1 + 6
          a__zprimes_A() = 13
  
    precedence:
    
      zprimes = a__zprimes > mark# = s = mark = nats = a__zprimes# = a__nats = |0| = a__filter > filter > sieve = a__sieve# = cons = a__nats# = a__sieve
    
    partial status:
  
      pi(mark#) = [1]
      pi(s) = [1]
      pi(filter) = [1, 2, 3]
      pi(sieve) = []
      pi(a__sieve#) = [1]
      pi(mark) = [1]
      pi(cons) = [1]
      pi(nats) = [1]
      pi(a__nats#) = [1]
      pi(zprimes) = []
      pi(a__zprimes#) = []
      pi(a__nats) = [1]
      pi(|0|) = []
      pi(a__filter) = [2]
      pi(a__sieve) = []
      pi(a__zprimes) = []

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: 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: a__zprimes#() -> a__sieve#(a__nats(s(s(|0|()))))
p12: a__zprimes#() -> 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, 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#(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#(sieve(X)) -> a__sieve#(mark(X))
p8: a__sieve#(cons(s(N),Y)) -> mark#(N)
p9: mark#(filter(X1,X2,X3)) -> mark#(X3)
p10: mark#(filter(X1,X2,X3)) -> mark#(X2)
p11: 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:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1 + 10
          s_A(x1) = x1 + 3
          cons_A(x1,x2) = x1 + 6
          nats_A(x1) = x1 + 9
          a__nats#_A(x1) = x1 + 11
          mark_A(x1) = x1
          sieve_A(x1) = x1 + 3
          a__sieve#_A(x1) = x1 + 2
          filter_A(x1,x2,x3) = x1 + x2 + x3 + 5
          a__filter_A(x1,x2,x3) = x1 + x2 + x3 + 5
          |0|_A() = 4
          a__sieve_A(x1) = x1 + 3
          a__nats_A(x1) = x1 + 9
          a__zprimes_A() = 23
          zprimes_A() = 23
  
    precedence:
    
      a__nats# > mark# = a__sieve# > mark = sieve = |0| = a__sieve > a__filter > cons = filter = a__nats > s > nats = a__zprimes = zprimes
    
    partial status:
  
      pi(mark#) = []
      pi(s) = [1]
      pi(cons) = []
      pi(nats) = [1]
      pi(a__nats#) = [1]
      pi(mark) = []
      pi(sieve) = []
      pi(a__sieve#) = []
      pi(filter) = [2]
      pi(a__filter) = [2]
      pi(|0|) = []
      pi(a__sieve) = []
      pi(a__nats) = []
      pi(a__zprimes) = []
      pi(zprimes) = []

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


-- 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)

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:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1 + 3
          s_A(x1) = x1 + 1
          filter_A(x1,x2,x3) = x1 + x2 + x3 + 4
          sieve_A(x1) = x1 + 2
          a__sieve#_A(x1) = x1 + 1
          mark_A(x1) = x1
          cons_A(x1,x2) = x1 + 6
          nats_A(x1) = x1 + 7
          a__nats#_A(x1) = x1 + 4
          a__filter_A(x1,x2,x3) = x1 + x2 + x3 + 4
          |0|_A() = 5
          a__sieve_A(x1) = x1 + 2
          a__nats_A(x1) = x1 + 7
          a__zprimes_A() = 17
          zprimes_A() = 17
  
    precedence:
    
      a__zprimes = zprimes > cons = nats = a__nats > a__nats# > mark# = s = filter = a__sieve# = mark = a__filter > sieve = |0| = a__sieve
    
    partial status:
  
      pi(mark#) = [1]
      pi(s) = [1]
      pi(filter) = []
      pi(sieve) = []
      pi(a__sieve#) = [1]
      pi(mark) = [1]
      pi(cons) = []
      pi(nats) = []
      pi(a__nats#) = []
      pi(a__filter) = [2]
      pi(|0|) = []
      pi(a__sieve) = [1]
      pi(a__nats) = []
      pi(a__zprimes) = []
      pi(zprimes) = []

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: 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: a__sieve#(cons(s(N),Y)) -> mark#(N)
p6: mark#(sieve(X)) -> mark#(X)
p7: mark#(nats(X)) -> a__nats#(mark(X))
p8: a__nats#(N) -> mark#(N)
p9: mark#(nats(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, p6, p7, p8, p9}


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(nats(X)) -> mark#(X)
p3: mark#(nats(X)) -> a__nats#(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 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:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1 + 1
          s_A(x1) = x1 + 6
          nats_A(x1) = x1 + 7
          a__nats#_A(x1) = x1 + 2
          mark_A(x1) = x1
          sieve_A(x1) = x1 + 3
          filter_A(x1,x2,x3) = x1 + x2 + x3 + 2
          a__filter_A(x1,x2,x3) = x1 + x2 + x3 + 2
          cons_A(x1,x2) = 4
          |0|_A() = 8
          a__sieve_A(x1) = x1 + 3
          a__nats_A(x1) = x1 + 7
          a__zprimes_A() = 31
          zprimes_A() = 31
  
    precedence:
    
      mark = |0| = a__sieve = a__zprimes > a__filter > filter > mark# = a__nats# > nats = sieve = a__nats > s > zprimes > cons
    
    partial status:
  
      pi(mark#) = [1]
      pi(s) = [1]
      pi(nats) = []
      pi(a__nats#) = [1]
      pi(mark) = []
      pi(sieve) = [1]
      pi(filter) = [1]
      pi(a__filter) = [1, 2, 3]
      pi(cons) = []
      pi(|0|) = []
      pi(a__sieve) = []
      pi(a__nats) = [1]
      pi(a__zprimes) = []
      pi(zprimes) = []

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


-- 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)

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:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          s_A(x1) = x1 + 1
          filter_A(x1,x2,x3) = x1 + x2 + x3 + 1
          sieve_A(x1) = x1 + 1
          nats_A(x1) = x1 + 1
  
    precedence:
    
      mark# = s = filter = sieve = nats
    
    partial status:
  
      pi(mark#) = []
      pi(s) = [1]
      pi(filter) = [3]
      pi(sieve) = [1]
      pi(nats) = [1]

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#(filter(X1,X2,X3)) -> mark#(X1)
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)

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#(filter(X1,X2,X3)) -> mark#(X1)
p2: mark#(nats(X)) -> mark#(X)
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 set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1 + 1
          filter_A(x1,x2,x3) = x1 + x2 + x3 + 2
          nats_A(x1) = x1 + 2
          sieve_A(x1) = x1 + 2
  
    precedence:
    
      mark# = filter = nats = sieve
    
    partial status:
  
      pi(mark#) = [1]
      pi(filter) = [1, 2, 3]
      pi(nats) = [1]
      pi(sieve) = [1]

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#(filter(X1,X2,X3)) -> mark#(X1)
p2: mark#(nats(X)) -> mark#(X)
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 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#(filter(X1,X2,X3)) -> mark#(X1)
p2: mark#(filter(X1,X2,X3)) -> mark#(X2)
p3: mark#(filter(X1,X2,X3)) -> mark#(X3)
p4: mark#(nats(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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          filter_A(x1,x2,x3) = x1 + x2 + x3 + 1
          nats_A(x1) = x1 + 1
  
    precedence:
    
      mark# = filter = nats
    
    partial status:
  
      pi(mark#) = [1]
      pi(filter) = [2, 3]
      pi(nats) = [1]

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


-- Reduction pair.

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

p1: mark#(filter(X1,X2,X3)) -> mark#(X2)
p2: mark#(nats(X)) -> mark#(X)
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:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          filter_A(x1,x2,x3) = x1 + x2 + x3 + 1
          nats_A(x1) = x1 + 1
  
    precedence:
    
      mark# = filter = nats
    
    partial status:
  
      pi(mark#) = [1]
      pi(filter) = [2, 3]
      pi(nats) = [1]

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#(nats(X)) -> mark#(X)
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#(nats(X)) -> mark#(X)
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:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          nats_A(x1) = x1 + 1
          filter_A(x1,x2,x3) = x1 + x2 + x3 + 1
  
    precedence:
    
      mark# = filter > nats
    
    partial status:
  
      pi(mark#) = []
      pi(nats) = []
      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#(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}


-- Reduction pair.

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

p1: 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:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          mark#_A(x1) = x1
          filter_A(x1,x2,x3) = x1 + x2 + x3 + 1
  
    precedence:
    
      mark# = filter
    
    partial status:
  
      pi(mark#) = [1]
      pi(filter) = [3]

The next rules are strictly ordered:

  p1

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