NO Termination w.r.t. Q proof of Strategy_removed_CSR_05_Ex9_BLR02.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 11 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 TransformationProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 TransformationProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 NonTerminationLoopProof (⇒, 0 ms)
        ↳17 NO
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QReductionProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QReductionProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QDPOrderProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 PisEmptyProof (⇔, 0 ms)
        ↳33 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) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) 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))))

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

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) 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))))

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.

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) 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))))

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.

(8) 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.

(9) Obligation:

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

NATS(N) → NATS(s(N))

R is empty.
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.

(10) 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].

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

(11) Obligation:

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

NATS(N) → NATS(s(N))

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

(12) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule NATS(N) → NATS(s(N)) we obtained the following new rules [LPAR04]:

NATS(s(z0)) → NATS(s(s(z0))) → NATS(s(z0)) → NATS(s(s(z0)))

(13) Obligation:

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

NATS(s(z0)) → NATS(s(s(z0)))

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

(14) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule NATS(s(z0)) → NATS(s(s(z0))) we obtained the following new rules [LPAR04]:

NATS(s(s(z0))) → NATS(s(s(s(z0)))) → NATS(s(s(z0))) → NATS(s(s(s(z0))))

(15) Obligation:

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

NATS(s(s(z0))) → NATS(s(s(s(z0))))

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

(16) NonTerminationLoopProof (COMPLETE transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = NATS(s(s(z0))) evaluates to t =NATS(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [z0 / s(z0)]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from NATS(s(s(z0))) to NATS(s(s(s(z0)))).



(17) NO

(18) 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))))

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.

(19) 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.

(20) 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)

R is empty.
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.

(21) 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].

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

(22) 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)

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

(23) 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

(24) YES

(25) 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.

(26) 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.

(27) 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.

(28) 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

(29) 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.

(30) 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: Combined order from the following AFS and order.
SIEVE(x1)  =  x1
cons(x1, x2)  =  cons(x2)
filter(x1, x2, x3)  =  x1

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

dummyConstant=1
cons_1=1

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))

(31) 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.

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) YES