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 NonTerminationProof (⇔)
        ↳17 NO
    ↳18 QDP
        ↳19 UsableRulesProof (⇔)
        ↳20 QDP
        ↳21 QReductionProof (⇔)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔)
        ↳27 QDP
        ↳28 QReductionProof (⇔)
        ↳29 QDP
        ↳30 QDPSizeChangeProof (⇔)
        ↳31 YES
    ↳32 QDP
        ↳33 UsableRulesProof (⇔)
        ↳34 QDP
        ↳35 QReductionProof (⇔)
        ↳36 QDP
        ↳37 QDPSizeChangeProof (⇔)
        ↳38 YES
    ↳39 QDP
        ↳40 UsableRulesProof (⇔)
        ↳41 QDP
        ↳42 QReductionProof (⇔)
        ↳43 QDP
        ↳44 QDPSizeChangeProof (⇔)
        ↳45 YES

(0) Obligation:

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

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

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:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

(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:

DBL(s(X)) → DBL(X)
DBLS(cons(X, Y)) → DBL(X)
DBLS(cons(X, Y)) → DBLS(Y)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
INDX(cons(X, Y), Z) → SEL(X, Z)
INDX(cons(X, Y), Z) → INDX(Y, Z)
FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 2 less nodes.

(6) Complex Obligation (AND)

(7) 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:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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:

FROM(X) → FROM(s(X))

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

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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].

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

(11) 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.

(12) 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)))

(13) 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.

(14) 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))))

(15) 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.

(16) 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)))).



(17) NO

(18) Obligation:

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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:

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

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

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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].

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

(22) Obligation:

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

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:

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

(24) YES

(25) Obligation:

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

INDX(cons(X, Y), Z) → INDX(Y, Z)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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:

INDX(cons(X, Y), Z) → INDX(Y, Z)

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

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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].

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

(29) Obligation:

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

INDX(cons(X, Y), Z) → INDX(Y, Z)

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

(30) 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:

  • INDX(cons(X, Y), Z) → INDX(Y, Z)
    The graph contains the following edges 1 > 1, 2 >= 2

(31) YES

(32) Obligation:

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

DBL(s(X)) → DBL(X)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(33) 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.

(34) Obligation:

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

DBL(s(X)) → DBL(X)

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

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(35) 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].

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

(36) Obligation:

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

DBL(s(X)) → DBL(X)

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

(37) 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:

  • DBL(s(X)) → DBL(X)
    The graph contains the following edges 1 > 1

(38) YES

(39) Obligation:

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

DBLS(cons(X, Y)) → DBLS(Y)

The TRS R consists of the following rules:

dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
dbls(nil) → nil
dbls(cons(X, Y)) → cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, Z)
indx(nil, X) → nil
indx(cons(X, Y), Z) → cons(sel(X, Z), indx(Y, Z))
from(X) → cons(X, from(s(X)))

The set Q consists of the following terms:

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(40) 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.

(41) Obligation:

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

DBLS(cons(X, Y)) → DBLS(Y)

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

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

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

(42) 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].

dbl(0)
dbl(s(x0))
dbls(nil)
dbls(cons(x0, x1))
sel(0, cons(x0, x1))
sel(s(x0), cons(x1, x2))
indx(nil, x0)
indx(cons(x0, x1), x2)
from(x0)

(43) Obligation:

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

DBLS(cons(X, Y)) → DBLS(Y)

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

(44) 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:

  • DBLS(cons(X, Y)) → DBLS(Y)
    The graph contains the following edges 1 > 1

(45) YES