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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 UsableRulesProof (⇔)
    ↳12 QDP
    ↳13 QDPSizeChangeProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 UsableRulesProof (⇔)
    ↳17 QDP
    ↳18 QDPSizeChangeProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 UsableRulesProof (⇔)
    ↳22 QDP
      ↳23 Instantiation (⇔)
        ↳24 QDP
      ↳25 Instantiation (⇔)
        ↳26 QDP
        ↳27 Instantiation (⇔)
        ↳28 QDP
        ↳29 NonTerminationProof (⇔)
        ↳30 FALSE
  ↳31 QDP
    ↳32 UsableRulesProof (⇔)
    ↳33 QDP
    ↳34 QDPSizeChangeProof (⇔)
    ↳35 TRUE
  ↳36 QDP
    ↳37 UsableRulesProof (⇔)
    ↳38 QDP
    ↳39 QDPSizeChangeProof (⇔)
    ↳40 TRUE
  ↳41 QDP
    ↳42 UsableRulesProof (⇔)
    ↳43 QDP
    ↳44 QDPSizeChangeProof (⇔)
    ↳45 TRUE
  ↳46 QDP
    ↳47 UsableRulesProof (⇔)
    ↳48 QDP
    ↳49 QDPSizeChangeProof (⇔)
    ↳50 TRUE

(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)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

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:

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))
DBL1(s(X)) → DBL1(X)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
QUOTE(s(X)) → QUOTE(X)
QUOTE(dbl(X)) → DBL1(X)
QUOTE(sel(X, Y)) → SEL1(X, 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)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

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 8 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

SEL1(s(X), cons(Y, Z)) → SEL1(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)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)

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

(8) 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:

  • SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
    The graph contains the following edges 1 > 1, 2 > 2

(9) TRUE

(10) Obligation:

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

DBL1(s(X)) → DBL1(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)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

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

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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

DBL1(s(X)) → DBL1(X)

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

(13) 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:

  • DBL1(s(X)) → DBL1(X)
    The graph contains the following edges 1 > 1

(14) TRUE

(15) Obligation:

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

QUOTE(s(X)) → QUOTE(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)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

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

(16) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

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

QUOTE(s(X)) → QUOTE(X)

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

(18) 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:

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

(19) TRUE

(20) 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)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

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

(21) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(22) 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.

(23) 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)))

(24) 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.

(25) 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)))

(26) 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.

(27) 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))))

(28) 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.

(29) 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:
  • Semiunifier: [ ]
  • Matcher: [z0 / s(z0)]



Rewriting sequence

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



(30) FALSE

(31) 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)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

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

(32) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(33) 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.

(34) 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

(35) TRUE

(36) 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)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

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

(37) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(38) 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.

(39) 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

(40) TRUE

(41) 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)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

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

(42) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(43) 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.

(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:

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

(45) TRUE

(46) 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)))
dbl1(0) → 01
dbl1(s(X)) → s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) → X
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
quote(0) → 01
quote(s(X)) → s1(quote(X))
quote(dbl(X)) → dbl1(X)
quote(sel(X, Y)) → sel1(X, Y)

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

(47) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(48) 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.

(49) 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

(50) TRUE