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 QDPSizeChangeProof (⇔)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔)
        ↳16 QDP
        ↳17 QReductionProof (⇔)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔)
        ↳23 QDP
        ↳24 QReductionProof (⇔)
        ↳25 QDP
        ↳26 Instantiation (⇔)
        ↳27 QDP
        ↳28 Instantiation (⇔)
        ↳29 QDP
        ↳30 NonTerminationProof (⇔)
        ↳31 NO

(0) Obligation:

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

from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

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:

from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

(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:

FROM(X) → FROM(s(X))
2ND(cons(X, XS)) → HEAD(XS)
TAKE(s(N), cons(X, XS)) → TAKE(N, XS)
SEL(s(N), cons(X, XS)) → SEL(N, XS)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

SEL(s(N), cons(X, XS)) → SEL(N, XS)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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:

SEL(s(N), cons(X, XS)) → SEL(N, XS)

R is empty.
The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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].

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

(11) Obligation:

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

SEL(s(N), cons(X, XS)) → SEL(N, XS)

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

(12) 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:

  • SEL(s(N), cons(X, XS)) → SEL(N, XS)
    The graph contains the following edges 1 > 1, 2 > 2

(13) YES

(14) Obligation:

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

TAKE(s(N), cons(X, XS)) → TAKE(N, XS)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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

(15) 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.

(16) Obligation:

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

TAKE(s(N), cons(X, XS)) → TAKE(N, XS)

R is empty.
The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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

(17) 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].

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

(18) Obligation:

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

TAKE(s(N), cons(X, XS)) → TAKE(N, XS)

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

(19) 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:

  • TAKE(s(N), cons(X, XS)) → TAKE(N, XS)
    The graph contains the following edges 1 > 1, 2 > 2

(20) YES

(21) Obligation:

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(XS)
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, take(N, XS))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, XS)

The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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

(22) 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.

(23) Obligation:

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

FROM(X) → FROM(s(X))

R is empty.
The set Q consists of the following terms:

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

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

(24) 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].

from(x0)
head(cons(x0, x1))
2nd(cons(x0, x1))
take(0, x0)
take(s(x0), cons(x1, x2))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))

(25) Obligation:

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

FROM(X) → FROM(s(X))

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

(26) Instantiation (EQUIVALENT transformation)

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

FROM(s(z0)) → FROM(s(s(z0)))

(27) Obligation:

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

FROM(s(z0)) → FROM(s(s(z0)))

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

(28) Instantiation (EQUIVALENT transformation)

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

FROM(s(s(z0))) → FROM(s(s(s(z0))))

(29) Obligation:

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

FROM(s(s(z0))) → FROM(s(s(s(z0))))

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

(30) 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 = FROM(s(s(z0))) evaluates to t =FROM(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 FROM(s(s(z0))) to FROM(s(s(s(z0)))).



(31) NO