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:
filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(n__filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(n__s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
s(X) → n__s(X)
activate(n__filter(X1, X2, X3)) → filter(activate(X1), activate(X2), activate(X3))
activate(n__sieve(X)) → sieve(activate(X))
activate(n__nats(X)) → nats(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(n__filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(n__s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
s(X) → n__s(X)
activate(n__filter(X1, X2, X3)) → filter(activate(X1), activate(X2), activate(X3))
activate(n__sieve(X)) → sieve(activate(X))
activate(n__nats(X)) → nats(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → 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:
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X3)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
ZPRIMES → SIEVE(nats(s(s(0))))
ACTIVATE(n__nats(X)) → ACTIVATE(X)
ACTIVATE(n__nats(X)) → NATS(activate(X))
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X1)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ZPRIMES → S(0)
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X2)
ZPRIMES → S(s(0))
ACTIVATE(n__s(X)) → S(activate(X))
SIEVE(cons(0, Y)) → ACTIVATE(Y)
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(activate(X1), activate(X2), activate(X3))
ACTIVATE(n__sieve(X)) → ACTIVATE(X)
ZPRIMES → NATS(s(s(0)))
ACTIVATE(n__sieve(X)) → SIEVE(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
The TRS R consists of the following rules:
filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(n__filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(n__s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
s(X) → n__s(X)
activate(n__filter(X1, X2, X3)) → filter(activate(X1), activate(X2), activate(X3))
activate(n__sieve(X)) → sieve(activate(X))
activate(n__nats(X)) → nats(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X3)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
ZPRIMES → SIEVE(nats(s(s(0))))
ACTIVATE(n__nats(X)) → ACTIVATE(X)
ACTIVATE(n__nats(X)) → NATS(activate(X))
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X1)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ZPRIMES → S(0)
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X2)
ZPRIMES → S(s(0))
ACTIVATE(n__s(X)) → S(activate(X))
SIEVE(cons(0, Y)) → ACTIVATE(Y)
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(activate(X1), activate(X2), activate(X3))
ACTIVATE(n__sieve(X)) → ACTIVATE(X)
ZPRIMES → NATS(s(s(0)))
ACTIVATE(n__sieve(X)) → SIEVE(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
The TRS R consists of the following rules:
filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(n__filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(n__s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
s(X) → n__s(X)
activate(n__filter(X1, X2, X3)) → filter(activate(X1), activate(X2), activate(X3))
activate(n__sieve(X)) → sieve(activate(X))
activate(n__nats(X)) → nats(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X3)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
ZPRIMES → SIEVE(nats(s(s(0))))
ACTIVATE(n__nats(X)) → NATS(activate(X))
ACTIVATE(n__nats(X)) → ACTIVATE(X)
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X1)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ZPRIMES → S(0)
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X2)
ZPRIMES → S(s(0))
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(activate(X1), activate(X2), activate(X3))
SIEVE(cons(0, Y)) → ACTIVATE(Y)
ACTIVATE(n__sieve(X)) → ACTIVATE(X)
ZPRIMES → NATS(s(s(0)))
ACTIVATE(n__sieve(X)) → SIEVE(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
The TRS R consists of the following rules:
filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(n__filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(n__s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
s(X) → n__s(X)
activate(n__filter(X1, X2, X3)) → filter(activate(X1), activate(X2), activate(X3))
activate(n__sieve(X)) → sieve(activate(X))
activate(n__nats(X)) → nats(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 6 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X3)
SIEVE(cons(s(N), Y)) → ACTIVATE(Y)
FILTER(cons(X, Y), s(N), M) → ACTIVATE(Y)
SIEVE(cons(0, Y)) → ACTIVATE(Y)
ACTIVATE(n__filter(X1, X2, X3)) → FILTER(activate(X1), activate(X2), activate(X3))
ACTIVATE(n__nats(X)) → ACTIVATE(X)
ACTIVATE(n__sieve(X)) → ACTIVATE(X)
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X1)
FILTER(cons(X, Y), 0, M) → ACTIVATE(Y)
ACTIVATE(n__filter(X1, X2, X3)) → ACTIVATE(X2)
ACTIVATE(n__sieve(X)) → SIEVE(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
The TRS R consists of the following rules:
filter(cons(X, Y), 0, M) → cons(0, n__filter(activate(Y), M, M))
filter(cons(X, Y), s(N), M) → cons(X, n__filter(activate(Y), N, M))
sieve(cons(0, Y)) → cons(0, n__sieve(activate(Y)))
sieve(cons(s(N), Y)) → cons(s(N), n__sieve(n__filter(activate(Y), N, N)))
nats(N) → cons(N, n__nats(n__s(N)))
zprimes → sieve(nats(s(s(0))))
filter(X1, X2, X3) → n__filter(X1, X2, X3)
sieve(X) → n__sieve(X)
nats(X) → n__nats(X)
s(X) → n__s(X)
activate(n__filter(X1, X2, X3)) → filter(activate(X1), activate(X2), activate(X3))
activate(n__sieve(X)) → sieve(activate(X))
activate(n__nats(X)) → nats(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.