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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
      ↳12 Instantiation (⇔)
        ↳13 QDP
      ↳14 Instantiation (⇔)
        ↳15 QDP
        ↳16 Instantiation (⇔)
        ↳17 QDP
        ↳18 NonTerminationProof (⇔)
        ↳19 FALSE
  ↳20 QDP
    ↳21 UsableRulesProof (⇔)
    ↳22 QDP
    ↳23 QReductionProof (⇔)
    ↳24 QDP
    ↳25 QDPSizeChangeProof (⇔)
    ↳26 TRUE
  ↳27 QDP
    ↳28 UsableRulesProof (⇔)
    ↳29 QDP
    ↳30 QReductionProof (⇔)
    ↳31 QDP
    ↳32 QDPOrderProof (⇔)
    ↳33 QDP
    ↳34 PisEmptyProof (⇔)
    ↳35 TRUE

(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) Instantiation (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)))

(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) Instantiation (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)))

(15) 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.

(16) Instantiation (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))))

(17) 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.

(18) NonTerminationProof (EQUIVALENT 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)))).



(19) FALSE

(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)

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.

(21) 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.

(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.
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.

(23) 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

(24) 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.

(25) 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

(26) TRUE

(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))
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.

(28) 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.

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

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

(30) 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

(31) 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.

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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 + x1 + x2   
POL(filter(x1, x2, x3)) = x1   
POL(s(x1)) = x1   

The following usable rules [FROCOS05] were oriented:

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

(33) 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.

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE