Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.


QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

QUOTE(sel(X, Y)) → SEL1(X, Y)
DBL1(s(X)) → DBL1(X)
QUOTE(s(X)) → QUOTE(X)
INDX(cons(X, Y), Z) → SEL(X, Z)
QUOTE(dbl(X)) → DBL1(X)
DBLS(cons(X, Y)) → DBL(X)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
DBLS(cons(X, Y)) → DBLS(Y)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
DBL(s(X)) → DBL(X)
FROM(X) → FROM(s(X))
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.

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

QUOTE(sel(X, Y)) → SEL1(X, Y)
DBL1(s(X)) → DBL1(X)
QUOTE(s(X)) → QUOTE(X)
INDX(cons(X, Y), Z) → SEL(X, Z)
QUOTE(dbl(X)) → DBL1(X)
DBLS(cons(X, Y)) → DBL(X)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
DBLS(cons(X, Y)) → DBLS(Y)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
DBL(s(X)) → DBL(X)
FROM(X) → FROM(s(X))
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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

QUOTE(sel(X, Y)) → SEL1(X, Y)
DBLS(cons(X, Y)) → DBL(X)
DBLS(cons(X, Y)) → DBLS(Y)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
FROM(X) → FROM(s(X))
DBL1(s(X)) → DBL1(X)
QUOTE(s(X)) → QUOTE(X)
INDX(cons(X, Y), Z) → SEL(X, Z)
QUOTE(dbl(X)) → DBL1(X)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
DBL(s(X)) → DBL(X)
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.
The approximation of the Dependency Graph [13,14,18] contains 8 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SEL1(x1, x2)  =  SEL1(x1)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [19].
Precedence:
s1 > SEL11
cons2 > SEL11

Status:
s1: multiset
SEL11: [1]
cons2: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


DBL1(s(X)) → DBL1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DBL1(x1)  =  DBL1(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
s1 > DBL11

Status:
DBL11: [1]
s1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


QUOTE(s(X)) → QUOTE(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
QUOTE(x1)  =  QUOTE(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
s1 > QUOTE1

Status:
QUOTE1: [1]
s1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


SEL(s(X), cons(Y, Z)) → SEL(X, Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x1)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [19].
Precedence:
s1 > SEL1
cons2 > SEL1

Status:
SEL1: [1]
s1: multiset
cons2: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


INDX(cons(X, Y), Z) → INDX(Y, Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
INDX(x1, x2)  =  INDX(x1)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [19].
Precedence:
cons2 > INDX1

Status:
INDX1: [1]
cons2: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


DBL(s(X)) → DBL(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DBL(x1)  =  DBL(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
s1 > DBL1

Status:
s1: multiset
DBL1: [1]

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


DBLS(cons(X, Y)) → DBLS(Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DBLS(x1)  =  DBLS(x1)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [19].
Precedence:
cons2 > DBLS1

Status:
DBLS1: [1]
cons2: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof

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