Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx9_BLR02

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.

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 contains 8 SCCs with 30 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

NATS₁(mark₁(X)) -> NATS₁(X)

Used argument filtering: NATS₁(x1) = x1
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

NATS₁(ok₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

NATS₁(ok₁(X)) -> NATS₁(X)

Used argument filtering: NATS₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SIEVE₁(ok₁(X)) -> SIEVE₁(X)

Used argument filtering: SIEVE₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SIEVE₁(mark₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SIEVE₁(mark₁(X)) -> SIEVE₁(X)

Used argument filtering: SIEVE₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(mark₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S₁(ok₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(ok₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: CONS₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

CONS₂(mark₁(X1), 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)

Used argument filtering: CONS₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: FILTER₃(x1, x2, x3) = x3
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: FILTER₃(x1, x2, x3) = x3
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ 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, 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: FILTER₃(x1, x2, x3) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FILTER₃(mark₁(X1), 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: FILTER₃(x1, x2, x3) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(filter₃(X1, X2, X3)) -> PROPER₁(X3)
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)

Used argument filtering: PROPER₁(x1) = x1
filter₃(x1, x2, x3) = filter₃(x1, x2, x3)
s₁(x1) = x1
cons₂(x1, x2) = cons₂(x1, x2)
sieve₁(x1) = x1
nats₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(s₁(X)) -> PROPER₁(X)
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(nats₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
sieve₁(x1) = x1
nats₁(x1) = nats₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(sieve₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(sieve₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
sieve₁(x1) = sieve₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(s₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(s₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(filter₃(X1, X2, X3)) -> ACTIVE₁(X3)
ACTIVE₁(filter₃(X1, X2, X3)) -> ACTIVE₁(X1)
ACTIVE₁(filter₃(X1, X2, X3)) -> ACTIVE₁(X2)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = x1
filter₃(x1, x2, x3) = filter₃(x1, x2, x3)
cons₂(x1, x2) = x1
s₁(x1) = x1
nats₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(nats₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = x1
cons₂(x1, x2) = x1
s₁(x1) = x1
nats₁(x1) = nats₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(s₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = x1
cons₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = x1
cons₂(x1, x2) = cons₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(sieve₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(sieve₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
sieve₁(x1) = sieve₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ 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

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.