Termination w.r.t. Q proof of /tmp/tmpDQ2BmO/Ex3_2_Luc97

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:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
ACTIVE(dbls(cons(X, Y))) → DBLS(Y)
PROPER(dbl(X)) → DBL(proper(X))
PROPER(dbl(X)) → PROPER(X)
PROPER(dbls(X)) → PROPER(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
INDX(ok(X1), ok(X2)) → INDX(X1, X2)
ACTIVE(indx(cons(X, Y), Z)) → SEL(X, Z)
DBL(mark(X)) → DBL(X)
PROPER(indx(X1, X2)) → INDX(proper(X1), proper(X2))
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(indx(X1, X2)) → PROPER(X1)
PROPER(indx(X1, X2)) → PROPER(X2)
PROPER(from(X)) → FROM(proper(X))
DBLS(ok(X)) → DBLS(X)
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
INDX(mark(X1), X2) → INDX(X1, X2)
ACTIVE(dbl(s(X))) → DBL(X)
ACTIVE(indx(X1, X2)) → INDX(active(X1), X2)
ACTIVE(dbls(cons(X, Y))) → DBL(X)
FROM(ok(X)) → FROM(X)
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(dbl(s(X))) → S(dbl(X))
DBL(ok(X)) → DBL(X)
S(ok(X)) → S(X)
SEL(mark(X1), X2) → SEL(X1, X2)
ACTIVE(dbl(X)) → ACTIVE(X)
TOP(mark(X)) → PROPER(X)
ACTIVE(dbls(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
TOP(ok(X)) → ACTIVE(X)
ACTIVE(dbl(X)) → DBL(active(X))
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
ACTIVE(indx(cons(X, Y), Z)) → INDX(Y, Z)
SEL(X1, mark(X2)) → SEL(X1, X2)
ACTIVE(dbls(cons(X, Y))) → CONS(dbl(X), dbls(Y))
PROPER(from(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
DBLS(mark(X)) → DBLS(X)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(dbl(s(X))) → S(s(dbl(X)))
ACTIVE(indx(cons(X, Y), Z)) → CONS(sel(X, Z), indx(Y, Z))
ACTIVE(indx(X1, X2)) → ACTIVE(X1)
PROPER(dbls(X)) → DBLS(proper(X))
ACTIVE(from(X)) → S(X)
PROPER(sel(X1, X2)) → PROPER(X1)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
ACTIVE(dbls(X)) → DBLS(active(X))
ACTIVE(from(X)) → CONS(X, from(s(X)))

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
ACTIVE(dbls(cons(X, Y))) → DBLS(Y)
PROPER(dbl(X)) → DBL(proper(X))
PROPER(dbl(X)) → PROPER(X)
PROPER(dbls(X)) → PROPER(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
INDX(ok(X1), ok(X2)) → INDX(X1, X2)
ACTIVE(indx(cons(X, Y), Z)) → SEL(X, Z)
DBL(mark(X)) → DBL(X)
PROPER(indx(X1, X2)) → INDX(proper(X1), proper(X2))
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(indx(X1, X2)) → PROPER(X1)
PROPER(indx(X1, X2)) → PROPER(X2)
PROPER(from(X)) → FROM(proper(X))
DBLS(ok(X)) → DBLS(X)
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
INDX(mark(X1), X2) → INDX(X1, X2)
ACTIVE(dbl(s(X))) → DBL(X)
ACTIVE(indx(X1, X2)) → INDX(active(X1), X2)
ACTIVE(dbls(cons(X, Y))) → DBL(X)
FROM(ok(X)) → FROM(X)
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(dbl(s(X))) → S(dbl(X))
DBL(ok(X)) → DBL(X)
S(ok(X)) → S(X)
SEL(mark(X1), X2) → SEL(X1, X2)
ACTIVE(dbl(X)) → ACTIVE(X)
TOP(mark(X)) → PROPER(X)
ACTIVE(dbls(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
TOP(ok(X)) → ACTIVE(X)
ACTIVE(dbl(X)) → DBL(active(X))
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
ACTIVE(indx(cons(X, Y), Z)) → INDX(Y, Z)
SEL(X1, mark(X2)) → SEL(X1, X2)
ACTIVE(dbls(cons(X, Y))) → CONS(dbl(X), dbls(Y))
PROPER(from(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
DBLS(mark(X)) → DBLS(X)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(dbl(s(X))) → S(s(dbl(X)))
ACTIVE(indx(cons(X, Y), Z)) → CONS(sel(X, Z), indx(Y, Z))
ACTIVE(indx(X1, X2)) → ACTIVE(X1)
PROPER(dbls(X)) → DBLS(proper(X))
ACTIVE(from(X)) → S(X)
PROPER(sel(X1, X2)) → PROPER(X1)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
ACTIVE(dbls(X)) → DBLS(active(X))
ACTIVE(from(X)) → CONS(X, from(s(X)))

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 10 SCCs with 27 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FROM(ok(X)) → FROM(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:

FROM(ok(X)) → FROM(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

From the DPs we obtained the following set of size-change graphs:

CONS(ok(X1), ok(X2)) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(ok(X)) → S(X)

From the DPs we obtained the following set of size-change graphs:

S(ok(X)) → S(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

INDX(ok(X1), ok(X2)) → INDX(X1, X2)
INDX(mark(X1), X2) → INDX(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

INDX(ok(X1), ok(X2)) → INDX(X1, X2)
INDX(mark(X1), X2) → INDX(X1, X2)

From the DPs we obtained the following set of size-change graphs:

INDX(ok(X1), ok(X2)) → INDX(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2
INDX(mark(X1), X2) → INDX(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL(mark(X1), X2) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL(mark(X1), X2) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)

From the DPs we obtained the following set of size-change graphs:

SEL(mark(X1), X2) → SEL(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2
SEL(X1, mark(X2)) → SEL(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DBLS(mark(X)) → DBLS(X)
DBLS(ok(X)) → DBLS(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DBLS(mark(X)) → DBLS(X)
DBLS(ok(X)) → DBLS(X)

From the DPs we obtained the following set of size-change graphs:

DBLS(mark(X)) → DBLS(X)
The graph contains the following edges 1 > 1
DBLS(ok(X)) → DBLS(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DBL(mark(X)) → DBL(X)
DBL(ok(X)) → DBL(X)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DBL(ok(X)) → DBL(X)
DBL(mark(X)) → DBL(X)

From the DPs we obtained the following set of size-change graphs:

DBL(mark(X)) → DBL(X)
The graph contains the following edges 1 > 1
DBL(ok(X)) → DBL(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

PROPER(indx(X1, X2)) → PROPER(X2)
PROPER(dbl(X)) → PROPER(X)
PROPER(dbls(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
PROPER(indx(X1, X2)) → PROPER(X1)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

PROPER(dbl(X)) → PROPER(X)
PROPER(indx(X1, X2)) → PROPER(X2)
PROPER(dbls(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
PROPER(indx(X1, X2)) → PROPER(X1)

From the DPs we obtained the following set of size-change graphs:

PROPER(indx(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(dbl(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(dbls(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(cons(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(s(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(sel(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(sel(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(cons(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(from(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(indx(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

ACTIVE(dbl(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(dbls(X)) → ACTIVE(X)
ACTIVE(indx(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

ACTIVE(dbl(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(dbls(X)) → ACTIVE(X)
ACTIVE(indx(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)

From the DPs we obtained the following set of size-change graphs:

ACTIVE(dbl(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
ACTIVE(dbls(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
ACTIVE(indx(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)
dbl(mark(X)) → mark(dbl(X))
dbls(mark(X)) → mark(dbls(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
dbl(ok(X)) → ok(dbl(X))
s(ok(X)) → ok(s(X))
dbls(ok(X)) → ok(dbls(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(TOP(x₁)) = x₁
POL(active(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(dbl(x₁)) = 2·x₁
POL(dbls(x₁)) = x₁
POL(from(x₁)) = x₁
POL(indx(x₁, x₂)) = x₁ + x₂
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(ok(x₁)) = 2·x₁
POL(proper(x₁)) = x₁
POL(s(x₁)) = x₁
POL(sel(x₁, x₂)) = x₁ + x₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
from(ok(X)) → ok(from(X))
indx(mark(X1), X2) → mark(indx(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
dbls(mark(X)) → mark(dbls(X))
dbls(ok(X)) → ok(dbls(X))
s(ok(X)) → ok(s(X))
dbl(mark(X)) → mark(dbl(X))
dbl(ok(X)) → ok(dbl(X))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(mark(X)) → TOP(proper(X)) at position [0] we obtained the following new rules:

TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(indx(x0, x1))) → TOP(indx(proper(x0), proper(x1)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(dbl(x0))) → TOP(dbl(proper(x0)))
TOP(mark(dbls(x0))) → TOP(dbls(proper(x0)))
TOP(mark(nil)) → TOP(ok(nil))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(0)) → TOP(ok(0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

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

TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(indx(x0, x1))) → TOP(indx(proper(x0), proper(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(nil)) → TOP(ok(nil))
TOP(mark(dbls(x0))) → TOP(dbls(proper(x0)))
TOP(mark(dbl(x0))) → TOP(dbl(proper(x0)))
TOP(mark(0)) → TOP(ok(0))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
from(ok(X)) → ok(from(X))
indx(mark(X1), X2) → mark(indx(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
dbls(mark(X)) → mark(dbls(X))
dbls(ok(X)) → ok(dbls(X))
s(ok(X)) → ok(s(X))
dbl(mark(X)) → mark(dbl(X))
dbl(ok(X)) → ok(dbl(X))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(ok(X)) → TOP(active(X)) at position [0] we obtained the following new rules:

TOP(ok(indx(nil, x0))) → TOP(mark(nil))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(ok(dbls(nil))) → TOP(mark(nil))
TOP(ok(dbls(x0))) → TOP(dbls(active(x0)))
TOP(ok(dbls(cons(x0, x1)))) → TOP(mark(cons(dbl(x0), dbls(x1))))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(indx(x0, x1))) → TOP(indx(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(indx(cons(x0, x1), x2))) → TOP(mark(cons(sel(x0, x2), indx(x1, x2))))
TOP(ok(dbl(s(x0)))) → TOP(mark(s(s(dbl(x0)))))
TOP(ok(dbl(x0))) → TOP(dbl(active(x0)))
TOP(ok(dbl(0))) → TOP(mark(0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

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

TOP(ok(indx(nil, x0))) → TOP(mark(nil))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(ok(dbls(x0))) → TOP(dbls(active(x0)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(dbls(nil))) → TOP(mark(nil))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(dbls(cons(x0, x1)))) → TOP(mark(cons(dbl(x0), dbls(x1))))
TOP(mark(0)) → TOP(ok(0))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(indx(x0, x1))) → TOP(indx(proper(x0), proper(x1)))
TOP(ok(indx(x0, x1))) → TOP(indx(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(indx(cons(x0, x1), x2))) → TOP(mark(cons(sel(x0, x2), indx(x1, x2))))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(nil)) → TOP(ok(nil))
TOP(mark(dbls(x0))) → TOP(dbls(proper(x0)))
TOP(mark(dbl(x0))) → TOP(dbl(proper(x0)))
TOP(ok(dbl(x0))) → TOP(dbl(active(x0)))
TOP(ok(dbl(s(x0)))) → TOP(mark(s(s(dbl(x0)))))
TOP(ok(dbl(0))) → TOP(mark(0))

The TRS R consists of the following rules:

proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
from(ok(X)) → ok(from(X))
indx(mark(X1), X2) → mark(indx(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
dbls(mark(X)) → mark(dbls(X))
dbls(ok(X)) → ok(dbls(X))
s(ok(X)) → ok(s(X))
dbl(mark(X)) → mark(dbl(X))
dbl(ok(X)) → ok(dbl(X))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

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

TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(dbls(x0))) → TOP(dbls(active(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(dbls(cons(x0, x1)))) → TOP(mark(cons(dbl(x0), dbls(x1))))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(indx(x0, x1))) → TOP(indx(proper(x0), proper(x1)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(indx(x0, x1))) → TOP(indx(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(indx(cons(x0, x1), x2))) → TOP(mark(cons(sel(x0, x2), indx(x1, x2))))
TOP(mark(dbl(x0))) → TOP(dbl(proper(x0)))
TOP(mark(dbls(x0))) → TOP(dbls(proper(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(dbl(x0))) → TOP(dbl(active(x0)))
TOP(ok(dbl(s(x0)))) → TOP(mark(s(s(dbl(x0)))))

The TRS R consists of the following rules:

proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
from(ok(X)) → ok(from(X))
indx(mark(X1), X2) → mark(indx(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
dbls(mark(X)) → mark(dbls(X))
dbls(ok(X)) → ok(dbls(X))
s(ok(X)) → ok(s(X))
dbl(mark(X)) → mark(dbl(X))
dbl(ok(X)) → ok(dbl(X))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

TOP(mark(s(x0))) → TOP(s(proper(x0)))

The remaining pairs can at least be oriented weakly.

TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(dbls(x0))) → TOP(dbls(active(x0)))
TOP(ok(dbls(cons(x0, x1)))) → TOP(mark(cons(dbl(x0), dbls(x1))))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(indx(x0, x1))) → TOP(indx(proper(x0), proper(x1)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(indx(x0, x1))) → TOP(indx(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(indx(cons(x0, x1), x2))) → TOP(mark(cons(sel(x0, x2), indx(x1, x2))))
TOP(mark(dbl(x0))) → TOP(dbl(proper(x0)))
TOP(mark(dbls(x0))) → TOP(dbls(proper(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(dbl(x0))) → TOP(dbl(active(x0)))
TOP(ok(dbl(s(x0)))) → TOP(mark(s(s(dbl(x0)))))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(TOP(x₁)) = x₁
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = 1
POL(dbl(x₁)) = 1
POL(dbls(x₁)) = x₁
POL(from(x₁)) = 1
POL(indx(x₁, x₂)) = x₁
POL(mark(x₁)) = 1
POL(nil) = 1
POL(ok(x₁)) = x₁
POL(proper(x₁)) = 1
POL(s(x₁)) = 0
POL(sel(x₁, x₂)) = 1

The following usable rules [17] were oriented:

dbl(mark(X)) → mark(dbl(X))
s(ok(X)) → ok(s(X))
active(dbl(0)) → mark(0)
dbl(ok(X)) → ok(dbl(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
dbls(ok(X)) → ok(dbls(X))
dbls(mark(X)) → mark(dbls(X))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

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

TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(dbls(x0))) → TOP(dbls(active(x0)))
TOP(ok(dbls(cons(x0, x1)))) → TOP(mark(cons(dbl(x0), dbls(x1))))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(indx(x0, x1))) → TOP(indx(proper(x0), proper(x1)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(indx(x0, x1))) → TOP(indx(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(dbls(x0))) → TOP(dbls(proper(x0)))
TOP(ok(indx(cons(x0, x1), x2))) → TOP(mark(cons(sel(x0, x2), indx(x1, x2))))
TOP(mark(dbl(x0))) → TOP(dbl(proper(x0)))
TOP(ok(dbl(s(x0)))) → TOP(mark(s(s(dbl(x0)))))
TOP(ok(dbl(x0))) → TOP(dbl(active(x0)))

The TRS R consists of the following rules:

proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
from(ok(X)) → ok(from(X))
indx(mark(X1), X2) → mark(indx(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
dbls(mark(X)) → mark(dbls(X))
dbls(ok(X)) → ok(dbls(X))
s(ok(X)) → ok(s(X))
dbl(mark(X)) → mark(dbl(X))
dbl(ok(X)) → ok(dbl(X))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))

The remaining pairs can at least be oriented weakly.

TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(dbls(x0))) → TOP(dbls(active(x0)))
TOP(ok(dbls(cons(x0, x1)))) → TOP(mark(cons(dbl(x0), dbls(x1))))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(indx(x0, x1))) → TOP(indx(proper(x0), proper(x1)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(indx(x0, x1))) → TOP(indx(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(mark(dbls(x0))) → TOP(dbls(proper(x0)))
TOP(ok(indx(cons(x0, x1), x2))) → TOP(mark(cons(sel(x0, x2), indx(x1, x2))))
TOP(mark(dbl(x0))) → TOP(dbl(proper(x0)))
TOP(ok(dbl(s(x0)))) → TOP(mark(s(s(dbl(x0)))))
TOP(ok(dbl(x0))) → TOP(dbl(active(x0)))

Used ordering: Combined order from the following AFS and order.
TOP(x1) = x1
ok(x1) = x1
sel(x1, x2) = sel
active(x1) = x1
mark(x1) = mark
from(x1) = from
proper(x1) = proper
dbls(x1) = dbls
cons(x1, x2) = cons
dbl(x1) = x1
s(x1) = s
0 = 0
indx(x1, x2) = indx
nil = nil

Recursive path order with status [2].
Quasi-Precedence:

[sel, mark, from, proper, dbls, s, 0, indx] > cons
[sel, mark, from, proper, dbls, s, 0, indx] > nil

Status:

from: []
sel: []
cons: multiset
mark: []
proper: []
dbls: []
s: []
0: multiset
nil: multiset
indx: []

The following usable rules [17] were oriented:

dbl(mark(X)) → mark(dbl(X))
s(ok(X)) → ok(s(X))
active(dbl(0)) → mark(0)
dbl(ok(X)) → ok(dbl(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
dbls(ok(X)) → ok(dbls(X))
dbls(mark(X)) → mark(dbls(X))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
indx(mark(X1), X2) → mark(indx(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
active(indx(X1, X2)) → indx(active(X1), X2)
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

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

TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(dbls(x0))) → TOP(dbls(active(x0)))
TOP(ok(dbls(cons(x0, x1)))) → TOP(mark(cons(dbl(x0), dbls(x1))))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(indx(x0, x1))) → TOP(indx(proper(x0), proper(x1)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(indx(x0, x1))) → TOP(indx(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(mark(dbl(x0))) → TOP(dbl(proper(x0)))
TOP(mark(dbls(x0))) → TOP(dbls(proper(x0)))
TOP(ok(indx(cons(x0, x1), x2))) → TOP(mark(cons(sel(x0, x2), indx(x1, x2))))
TOP(ok(dbl(x0))) → TOP(dbl(active(x0)))
TOP(ok(dbl(s(x0)))) → TOP(mark(s(s(dbl(x0)))))

The TRS R consists of the following rules:

proper(dbl(X)) → dbl(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(dbls(X)) → dbls(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(indx(X1, X2)) → indx(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
from(ok(X)) → ok(from(X))
indx(mark(X1), X2) → mark(indx(X1, X2))
indx(ok(X1), ok(X2)) → ok(indx(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
dbls(mark(X)) → mark(dbls(X))
dbls(ok(X)) → ok(dbls(X))
s(ok(X)) → ok(s(X))
dbl(mark(X)) → mark(dbl(X))
dbl(ok(X)) → ok(dbl(X))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(dbls(nil)) → mark(nil)
active(dbls(cons(X, Y))) → mark(cons(dbl(X), dbls(Y)))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(indx(nil, X)) → mark(nil)
active(indx(cons(X, Y), Z)) → mark(cons(sel(X, Z), indx(Y, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(dbl(X)) → dbl(active(X))
active(dbls(X)) → dbls(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(indx(X1, X2)) → indx(active(X1), X2)

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