Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 DependencyPairsProof (⇔, 3 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 NonLoopProof (⇔, 0 ms)
        ↳7 NO
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔, 0 ms)
        ↳10 YES
    ↳11 QDP
        ↳12 MNOCProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 UsableRulesProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 QReductionProof (⇔, 0 ms)
        ↳17 QDP
        ↳18 QDPOrderProof (⇔, 19 ms)
        ↳19 QDP
        ↳20 PisEmptyProof (⇔, 0 ms)
        ↳21 YES

(0) Obligation:

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

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
SIEVE(cons(0, Y)) → SIEVE(Y)
SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))
SIEVE(cons(s(N), Y)) → FILTER(Y, N, N)
NATS(N) → NATS(s(N))
ZPRIMES → SIEVE(nats(s(s(0))))
ZPRIMES → NATS(s(s(0)))

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

NATS(N) → NATS(s(N))

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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

(6) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [x0 / s(x0)] on the rule
NATS(s(x0))[ ]n[ ] → NATS(s(x0))[ ]n[x0 / s(x0)]
This rule is correct for the QDP as the following derivation shows:

intermediate steps: Equivalent (Simplify mu) - Instantiate mu - Instantiation
NATS(N)[ ]n[ ] → NATS(s(N))[ ]n[ ]
    by OriginalRule from TRS P

(7) NO

(8) Obligation:

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

FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • FILTER(cons(X, Y), s(N), M) → FILTER(Y, N, M)
    The graph contains the following edges 1 > 1, 2 > 2, 3 >= 3

  • FILTER(cons(X, Y), 0, M) → FILTER(Y, M, M)
    The graph contains the following edges 1 > 1, 3 >= 2, 3 >= 3

(10) YES

(11) Obligation:

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

SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))
SIEVE(cons(0, Y)) → SIEVE(Y)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

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

(12) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(13) Obligation:

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

SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))
SIEVE(cons(0, Y)) → SIEVE(Y)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))
sieve(cons(0, Y)) → cons(0, sieve(Y))
sieve(cons(s(N), Y)) → cons(s(N), sieve(filter(Y, N, N)))
nats(N) → cons(N, nats(s(N)))
zprimes → sieve(nats(s(s(0))))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)
sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes

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

(14) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(15) Obligation:

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

SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))
SIEVE(cons(0, Y)) → SIEVE(Y)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)
sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes

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

(16) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

sieve(cons(0, x0))
sieve(cons(s(x0), x1))
nats(x0)
zprimes

(17) Obligation:

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

SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))
SIEVE(cons(0, Y)) → SIEVE(Y)

The TRS R consists of the following rules:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


SIEVE(cons(s(N), Y)) → SIEVE(filter(Y, N, N))
SIEVE(cons(0, Y)) → SIEVE(Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(SIEVE(x1)) = x1   
POL(cons(x1, x2)) = 1 + x2   
POL(filter(x1, x2, x3)) = x1   
POL(s(x1)) = 0   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))

(19) Obligation:

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

filter(cons(X, Y), 0, M) → cons(0, filter(Y, M, M))
filter(cons(X, Y), s(N), M) → cons(X, filter(Y, N, M))

The set Q consists of the following terms:

filter(cons(x0, x1), 0, x2)
filter(cons(x0, x1), s(x2), x3)

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

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) YES