Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVE1(nats1(X)) -> NATS1(active1(X))
ACTIVE1(nats1(N)) -> S1(N)
S1(mark1(X)) -> S1(X)
PROPER1(nats1(X)) -> PROPER1(X)
ACTIVE1(zprimes) -> S1(0)
FILTER3(mark1(X1), X2, X3) -> FILTER3(X1, X2, X3)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X3)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X2)
ACTIVE1(filter3(X1, X2, X3)) -> FILTER3(X1, X2, active1(X3))
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(filter3(X1, X2, X3)) -> FILTER3(X1, active1(X2), X3)
ACTIVE1(sieve1(cons2(s1(N), Y))) -> CONS2(s1(N), sieve1(filter3(Y, N, N)))
PROPER1(filter3(X1, X2, X3)) -> PROPER1(X3)
ACTIVE1(s1(X)) -> S1(active1(X))
ACTIVE1(sieve1(cons2(0, Y))) -> CONS2(0, sieve1(Y))
ACTIVE1(filter3(X1, X2, X3)) -> FILTER3(active1(X1), X2, X3)
PROPER1(s1(X)) -> PROPER1(X)
ACTIVE1(sieve1(cons2(s1(N), Y))) -> SIEVE1(filter3(Y, N, N))
ACTIVE1(zprimes) -> SIEVE1(nats1(s1(s1(0))))
ACTIVE1(sieve1(cons2(0, Y))) -> SIEVE1(Y)
ACTIVE1(zprimes) -> S1(s1(0))
ACTIVE1(filter3(cons2(X, Y), 0, M)) -> CONS2(0, filter3(Y, M, M))
ACTIVE1(sieve1(X)) -> SIEVE1(active1(X))
TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(zprimes) -> NATS1(s1(s1(0)))
FILTER3(X1, X2, mark1(X3)) -> FILTER3(X1, X2, X3)
FILTER3(ok1(X1), ok1(X2), ok1(X3)) -> FILTER3(X1, X2, X3)
FILTER3(X1, mark1(X2), X3) -> FILTER3(X1, X2, X3)
ACTIVE1(nats1(N)) -> CONS2(N, nats1(s1(N)))
PROPER1(sieve1(X)) -> SIEVE1(proper1(X))
NATS1(mark1(X)) -> NATS1(X)
PROPER1(s1(X)) -> S1(proper1(X))
PROPER1(sieve1(X)) -> PROPER1(X)
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
SIEVE1(ok1(X)) -> SIEVE1(X)
ACTIVE1(filter3(cons2(X, Y), s1(N), M)) -> FILTER3(Y, N, M)
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X1)
PROPER1(filter3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(filter3(X1, X2, X3)) -> PROPER1(X2)
SIEVE1(mark1(X)) -> SIEVE1(X)
ACTIVE1(filter3(cons2(X, Y), 0, M)) -> FILTER3(Y, M, M)
ACTIVE1(nats1(X)) -> ACTIVE1(X)
ACTIVE1(sieve1(X)) -> ACTIVE1(X)
PROPER1(filter3(X1, X2, X3)) -> FILTER3(proper1(X1), proper1(X2), proper1(X3))
NATS1(ok1(X)) -> NATS1(X)
ACTIVE1(filter3(cons2(X, Y), s1(N), M)) -> CONS2(X, filter3(Y, N, M))
ACTIVE1(nats1(N)) -> NATS1(s1(N))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(sieve1(cons2(s1(N), Y))) -> FILTER3(Y, N, N)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(nats1(X)) -> NATS1(proper1(X))
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

ACTIVE1(nats1(X)) -> NATS1(active1(X))
ACTIVE1(nats1(N)) -> S1(N)
S1(mark1(X)) -> S1(X)
PROPER1(nats1(X)) -> PROPER1(X)
ACTIVE1(zprimes) -> S1(0)
FILTER3(mark1(X1), X2, X3) -> FILTER3(X1, X2, X3)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X3)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X2)
ACTIVE1(filter3(X1, X2, X3)) -> FILTER3(X1, X2, active1(X3))
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(filter3(X1, X2, X3)) -> FILTER3(X1, active1(X2), X3)
ACTIVE1(sieve1(cons2(s1(N), Y))) -> CONS2(s1(N), sieve1(filter3(Y, N, N)))
PROPER1(filter3(X1, X2, X3)) -> PROPER1(X3)
ACTIVE1(s1(X)) -> S1(active1(X))
ACTIVE1(sieve1(cons2(0, Y))) -> CONS2(0, sieve1(Y))
ACTIVE1(filter3(X1, X2, X3)) -> FILTER3(active1(X1), X2, X3)
PROPER1(s1(X)) -> PROPER1(X)
ACTIVE1(sieve1(cons2(s1(N), Y))) -> SIEVE1(filter3(Y, N, N))
ACTIVE1(zprimes) -> SIEVE1(nats1(s1(s1(0))))
ACTIVE1(sieve1(cons2(0, Y))) -> SIEVE1(Y)
ACTIVE1(zprimes) -> S1(s1(0))
ACTIVE1(filter3(cons2(X, Y), 0, M)) -> CONS2(0, filter3(Y, M, M))
ACTIVE1(sieve1(X)) -> SIEVE1(active1(X))
TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(zprimes) -> NATS1(s1(s1(0)))
FILTER3(X1, X2, mark1(X3)) -> FILTER3(X1, X2, X3)
FILTER3(ok1(X1), ok1(X2), ok1(X3)) -> FILTER3(X1, X2, X3)
FILTER3(X1, mark1(X2), X3) -> FILTER3(X1, X2, X3)
ACTIVE1(nats1(N)) -> CONS2(N, nats1(s1(N)))
PROPER1(sieve1(X)) -> SIEVE1(proper1(X))
NATS1(mark1(X)) -> NATS1(X)
PROPER1(s1(X)) -> S1(proper1(X))
PROPER1(sieve1(X)) -> PROPER1(X)
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
SIEVE1(ok1(X)) -> SIEVE1(X)
ACTIVE1(filter3(cons2(X, Y), s1(N), M)) -> FILTER3(Y, N, M)
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X1)
PROPER1(filter3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(filter3(X1, X2, X3)) -> PROPER1(X2)
SIEVE1(mark1(X)) -> SIEVE1(X)
ACTIVE1(filter3(cons2(X, Y), 0, M)) -> FILTER3(Y, M, M)
ACTIVE1(nats1(X)) -> ACTIVE1(X)
ACTIVE1(sieve1(X)) -> ACTIVE1(X)
PROPER1(filter3(X1, X2, X3)) -> FILTER3(proper1(X1), proper1(X2), proper1(X3))
NATS1(ok1(X)) -> NATS1(X)
ACTIVE1(filter3(cons2(X, Y), s1(N), M)) -> CONS2(X, filter3(Y, N, M))
ACTIVE1(nats1(N)) -> NATS1(s1(N))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(sieve1(cons2(s1(N), Y))) -> FILTER3(Y, N, N)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(nats1(X)) -> NATS1(proper1(X))
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 8 SCCs with 30 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

NATS1(ok1(X)) -> NATS1(X)
NATS1(mark1(X)) -> NATS1(X)

The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


NATS1(ok1(X)) -> NATS1(X)
NATS1(mark1(X)) -> NATS1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(NATS1(x1)) = 3·x1   
POL(mark1(x1)) = 3 + 2·x1   
POL(ok1(x1)) = 3 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

SIEVE1(mark1(X)) -> SIEVE1(X)
SIEVE1(ok1(X)) -> SIEVE1(X)

The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


SIEVE1(mark1(X)) -> SIEVE1(X)
SIEVE1(ok1(X)) -> SIEVE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SIEVE1(x1)) = 3·x1   
POL(mark1(x1)) = 3 + x1   
POL(ok1(x1)) = 3 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

S1(ok1(X)) -> S1(X)
S1(mark1(X)) -> S1(X)

The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


S1(ok1(X)) -> S1(X)
S1(mark1(X)) -> S1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(S1(x1)) = 3·x1   
POL(mark1(x1)) = 3 + 2·x1   
POL(ok1(x1)) = 3 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


CONS2(mark1(X1), X2) -> CONS2(X1, X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CONS2(x1, x2)) = 3·x1 + 3·x2   
POL(mark1(x1)) = 3 + x1   
POL(ok1(x1)) = 3 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

FILTER3(mark1(X1), X2, X3) -> FILTER3(X1, X2, X3)
FILTER3(X1, X2, mark1(X3)) -> FILTER3(X1, X2, X3)
FILTER3(ok1(X1), ok1(X2), ok1(X3)) -> FILTER3(X1, X2, X3)
FILTER3(X1, mark1(X2), X3) -> FILTER3(X1, X2, X3)

The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


FILTER3(mark1(X1), X2, X3) -> FILTER3(X1, X2, X3)
FILTER3(X1, X2, mark1(X3)) -> FILTER3(X1, X2, X3)
FILTER3(ok1(X1), ok1(X2), ok1(X3)) -> FILTER3(X1, X2, X3)
FILTER3(X1, mark1(X2), X3) -> FILTER3(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FILTER3(x1, x2, x3)) = 3·x1 + 3·x2 + 3·x3   
POL(mark1(x1)) = 3 + x1   
POL(ok1(x1)) = 3 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER1(filter3(X1, X2, X3)) -> PROPER1(X3)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(filter3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(filter3(X1, X2, X3)) -> PROPER1(X2)
PROPER1(sieve1(X)) -> PROPER1(X)
PROPER1(nats1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(filter3(X1, X2, X3)) -> PROPER1(X3)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(filter3(X1, X2, X3)) -> PROPER1(X1)
PROPER1(filter3(X1, X2, X3)) -> PROPER1(X2)
PROPER1(sieve1(X)) -> PROPER1(X)
PROPER1(nats1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = 3·x1   
POL(cons2(x1, x2)) = 3 + 2·x1 + 2·x2   
POL(filter3(x1, x2, x3)) = 3 + 2·x1 + 2·x2 + x3   
POL(nats1(x1)) = 3 + 2·x1   
POL(s1(x1)) = 3 + 2·x1   
POL(sieve1(x1)) = 3 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

ACTIVE1(sieve1(X)) -> ACTIVE1(X)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X3)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X1)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X2)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(nats1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVE1(sieve1(X)) -> ACTIVE1(X)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X3)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X1)
ACTIVE1(filter3(X1, X2, X3)) -> ACTIVE1(X2)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(nats1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = 3·x1   
POL(cons2(x1, x2)) = 3 + 2·x1   
POL(filter3(x1, x2, x3)) = 3 + 2·x1 + 2·x2 + 2·x3   
POL(nats1(x1)) = 3 + 2·x1   
POL(s1(x1)) = 3 + 2·x1   
POL(sieve1(x1)) = 3 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))

The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TOP1(mark1(X)) -> TOP1(proper1(X))
The remaining pairs can at least be oriented weakly.

TOP1(ok1(X)) -> TOP1(active1(X))
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(TOP1(x1)) = x1   
POL(active1(x1)) = x1   
POL(cons2(x1, x2)) = x1   
POL(filter3(x1, x2, x3)) = 2 + x1 + 2·x2 + x3   
POL(mark1(x1)) = 1 + x1   
POL(nats1(x1)) = 1 + x1   
POL(ok1(x1)) = x1   
POL(proper1(x1)) = x1   
POL(s1(x1)) = 2·x1   
POL(sieve1(x1)) = 1 + x1   
POL(zprimes) = 3   

The following usable rules [14] were oriented:

active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
proper1(s1(X)) -> s1(proper1(X))
nats1(mark1(X)) -> mark1(nats1(X))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
proper1(zprimes) -> ok1(zprimes)
active1(sieve1(X)) -> sieve1(active1(X))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(sieve1(X)) -> sieve1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
proper1(nats1(X)) -> nats1(proper1(X))
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
s1(mark1(X)) -> mark1(s1(X))
active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
proper1(0) -> ok1(0)
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
sieve1(ok1(X)) -> ok1(sieve1(X))
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(s1(X)) -> s1(active1(X))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
sieve1(mark1(X)) -> mark1(sieve1(X))
active1(nats1(X)) -> nats1(active1(X))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
nats1(ok1(X)) -> ok1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

TOP1(ok1(X)) -> TOP1(active1(X))

The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TOP1(ok1(X)) -> TOP1(active1(X))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(TOP1(x1)) = 2·x1   
POL(active1(x1)) = 2·x1   
POL(cons2(x1, x2)) = 2·x1   
POL(filter3(x1, x2, x3)) = x1   
POL(mark1(x1)) = 0   
POL(nats1(x1)) = x1   
POL(ok1(x1)) = 1 + 2·x1   
POL(s1(x1)) = 2·x1   
POL(sieve1(x1)) = x1   
POL(zprimes) = 0   

The following usable rules [14] were oriented:

active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
nats1(mark1(X)) -> mark1(nats1(X))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
active1(sieve1(X)) -> sieve1(active1(X))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
s1(ok1(X)) -> ok1(s1(X))
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
s1(mark1(X)) -> mark1(s1(X))
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
sieve1(ok1(X)) -> ok1(sieve1(X))
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(s1(X)) -> s1(active1(X))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
sieve1(mark1(X)) -> mark1(sieve1(X))
active1(nats1(X)) -> nats1(active1(X))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
nats1(ok1(X)) -> ok1(nats1(X))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(filter3(cons2(X, Y), 0, M)) -> mark1(cons2(0, filter3(Y, M, M)))
active1(filter3(cons2(X, Y), s1(N), M)) -> mark1(cons2(X, filter3(Y, N, M)))
active1(sieve1(cons2(0, Y))) -> mark1(cons2(0, sieve1(Y)))
active1(sieve1(cons2(s1(N), Y))) -> mark1(cons2(s1(N), sieve1(filter3(Y, N, N))))
active1(nats1(N)) -> mark1(cons2(N, nats1(s1(N))))
active1(zprimes) -> mark1(sieve1(nats1(s1(s1(0)))))
active1(filter3(X1, X2, X3)) -> filter3(active1(X1), X2, X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, active1(X2), X3)
active1(filter3(X1, X2, X3)) -> filter3(X1, X2, active1(X3))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(sieve1(X)) -> sieve1(active1(X))
active1(nats1(X)) -> nats1(active1(X))
filter3(mark1(X1), X2, X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, mark1(X2), X3) -> mark1(filter3(X1, X2, X3))
filter3(X1, X2, mark1(X3)) -> mark1(filter3(X1, X2, X3))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
sieve1(mark1(X)) -> mark1(sieve1(X))
nats1(mark1(X)) -> mark1(nats1(X))
proper1(filter3(X1, X2, X3)) -> filter3(proper1(X1), proper1(X2), proper1(X3))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(sieve1(X)) -> sieve1(proper1(X))
proper1(nats1(X)) -> nats1(proper1(X))
proper1(zprimes) -> ok1(zprimes)
filter3(ok1(X1), ok1(X2), ok1(X3)) -> ok1(filter3(X1, X2, X3))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
sieve1(ok1(X)) -> ok1(sieve1(X))
nats1(ok1(X)) -> ok1(nats1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.