Termination w.r.t. Q proof of TRS/TRCSREx9_BLR02

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 Order [17,21] with Interpretation:

POL( mark₁(x₁) ) = 2x₁ + 2

POL( ok₁(x₁) ) = 3x₁ + 1

POL( NATS₁(x₁) ) = 2x₁

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 Order [17,21] with Interpretation:

POL( SIEVE₁(x₁) ) = 2x₁

POL( mark₁(x₁) ) = 3x₁ + 1

POL( ok₁(x₁) ) = 2x₁ + 2

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 Order [17,21] with Interpretation:

POL( S₁(x₁) ) = 2x₁

POL( mark₁(x₁) ) = 2x₁ + 2

POL( ok₁(x₁) ) = 3x₁ + 1

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 Order [17,21] with Interpretation:

POL( CONS₂(x₁, x₂) ) = max{0, 2x₁ + 2x₂ - 1}

POL( mark₁(x₁) ) = 2x₁ + 1

POL( ok₁(x₁) ) = x₁ + 3

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(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.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( mark₁(x₁) ) = 2x₁ + 1

POL( FILTER₃(x₁, ..., x₃) ) = 2x₂

POL( ok₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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)

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)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( mark₁(x₁) ) = 2x₁ + 2

POL( FILTER₃(x₁, ..., x₃) ) = 2x₁ + 3x₂ + 2x₃

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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(s₁(X)) -> PROPER₁(X)
PROPER₁(sieve₁(X)) -> PROPER₁(X)
PROPER₁(nats₁(X)) -> PROPER₁(X)

The remaining pairs can at least be oriented weakly.

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 ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = 3x₁ + 1

POL( filter₃(x₁, ..., x₃) ) = x₁ + 2x₂ + 2x₃

POL( sieve₁(x₁) ) = 2x₁ + 1

POL( cons₂(x₁, x₂) ) = 2x₁ + 2x₂

POL( nats₁(x₁) ) = 2x₁ + 1

POL( PROPER₁(x₁) ) = 3x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

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)

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(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)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( cons₂(x₁, x₂) ) = 2x₁ + 2x₂ + 1

POL( PROPER₁(x₁) ) = x₁ + 3

POL( filter₃(x₁, ..., x₃) ) = 2x₁ + 2x₂ + 3x₃ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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)

The remaining pairs can at least be oriented weakly.

ACTIVE₁(nats₁(X)) -> ACTIVE₁(X)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = 2x₁ + 1

POL( filter₃(x₁, ..., x₃) ) = 2x₁ + 3x₂ + x₃ + 1

POL( sieve₁(x₁) ) = 2x₁ + 1

POL( ACTIVE₁(x₁) ) = 3x₁

POL( cons₂(x₁, x₂) ) = 2x₁ + 3x₂ + 3

POL( nats₁(x₁) ) = 2x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

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(nats₁(X)) -> ACTIVE₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ACTIVE₁(x₁) ) = 2x₁ + 3

POL( nats₁(x₁) ) = 3x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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 Order [17,21] with Interpretation:

POL( mark₁(x₁) ) = x₁ + 1

POL( filter₃(x₁, ..., x₃) ) = 2x₁ + x₂ + 2x₃ + 1

POL( sieve₁(x₁) ) = x₁ + 1

POL( zprimes ) = 3

POL( active₁(x₁) ) = x₁

POL( s₁(x₁) ) = 2x₁

POL( 0 ) = max{0, -3}

POL( TOP₁(x₁) ) = x₁ + 1

POL( ok₁(x₁) ) = x₁

POL( nats₁(x₁) ) = 2x₁ + 1

POL( cons₂(x₁, x₂) ) = x₁

POL( proper₁(x₁) ) = x₁

The following usable rules [14] were oriented:

s₁(mark₁(X)) -> mark₁(s₁(X))
active₁(nats₁(X)) -> nats₁(active₁(X))
active₁(zprimes) -> mark₁(sieve₁(nats₁(s₁(s₁(0)))))
proper₁(sieve₁(X)) -> sieve₁(proper₁(X))
sieve₁(ok₁(X)) -> ok₁(sieve₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
active₁(filter₃(X1, X2, X3)) -> filter₃(active₁(X1), X2, X3)
active₁(sieve₁(X)) -> sieve₁(active₁(X))
proper₁(nats₁(X)) -> nats₁(proper₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
active₁(filter₃(X1, X2, X3)) -> filter₃(X1, active₁(X2), X3)
proper₁(0) -> ok₁(0)
filter₃(X1, mark₁(X2), X3) -> mark₁(filter₃(X1, X2, X3))
nats₁(ok₁(X)) -> ok₁(nats₁(X))
active₁(nats₁(N)) -> mark₁(cons₂(N, nats₁(s₁(N))))
nats₁(mark₁(X)) -> mark₁(nats₁(X))
filter₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(filter₃(X1, X2, X3))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(sieve₁(cons₂(0, Y))) -> mark₁(cons₂(0, sieve₁(Y)))
proper₁(filter₃(X1, X2, X3)) -> filter₃(proper₁(X1), proper₁(X2), proper₁(X3))
active₁(s₁(X)) -> s₁(active₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
active₁(filter₃(cons₂(X, Y), s₁(N), M)) -> mark₁(cons₂(X, filter₃(Y, N, M)))
active₁(sieve₁(cons₂(s₁(N), Y))) -> mark₁(cons₂(s₁(N), sieve₁(filter₃(Y, N, N))))
filter₃(X1, X2, mark₁(X3)) -> mark₁(filter₃(X1, X2, X3))
active₁(filter₃(cons₂(X, Y), 0, M)) -> mark₁(cons₂(0, filter₃(Y, M, M)))
proper₁(zprimes) -> ok₁(zprimes)
sieve₁(mark₁(X)) -> mark₁(sieve₁(X))
active₁(filter₃(X1, X2, X3)) -> filter₃(X1, X2, active₁(X3))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
filter₃(mark₁(X1), X2, X3) -> mark₁(filter₃(X1, X2, 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 Order [17,21] with Interpretation:

POL( mark₁(x₁) ) = 1

POL( filter₃(x₁, ..., x₃) ) = 3x₃ + 2

POL( sieve₁(x₁) ) = 3x₁ + 2

POL( zprimes ) = 3

POL( active₁(x₁) ) = max{0, 3x₁ - 3}

POL( s₁(x₁) ) = 3x₁ + 2

POL( 0 ) = 2

POL( TOP₁(x₁) ) = 2x₁ + 1

POL( ok₁(x₁) ) = 3x₁ + 2

POL( nats₁(x₁) ) = 3x₁ + 2

POL( cons₂(x₁, x₂) ) = 3x₁ + 3x₂ + 3

The following usable rules [14] were oriented:

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

↳ 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.