Termination w.r.t. Q proof of /tmp/tmpR5plyf/Ex14_AEGL02

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → 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:

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH(n__cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__nil) → NIL
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
LENGTH(n__cons(X, Y)) → S(length1(activate(Y)))
LENGTH1(X) → LENGTH(activate(X))
LENGTH1(X) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
FROM(X) → CONS(X, n__from(n__s(X)))
ACTIVATE(n__from(X)) → FROM(activate(X))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → 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:

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH(n__cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__nil) → NIL
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
LENGTH(n__cons(X, Y)) → S(length1(activate(Y)))
LENGTH1(X) → LENGTH(activate(X))
LENGTH1(X) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
FROM(X) → CONS(X, n__from(n__s(X)))
ACTIVATE(n__from(X)) → FROM(activate(X))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 8 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → 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

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

ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(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:

ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
The graph contains the following edges 1 > 1
ACTIVATE(n__from(X)) → ACTIVATE(X)
The graph contains the following edges 1 > 1
ACTIVATE(n__s(X)) → ACTIVATE(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ Narrowing

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(X) → LENGTH(activate(X))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LENGTH1(X) → LENGTH(activate(X)) at position [0] we obtained the following new rules:

LENGTH1(n__cons(x0, x1)) → LENGTH(cons(activate(x0), x1))
LENGTH1(n__s(x0)) → LENGTH(s(activate(x0)))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__from(x0)) → LENGTH(from(activate(x0)))
LENGTH1(n__nil) → LENGTH(nil)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ Narrowing

              ↳ QDP

                ↳ Narrowing

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(n__from(x0)) → LENGTH(from(activate(x0)))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__nil) → LENGTH(nil)
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(activate(x0), x1))
LENGTH1(n__s(x0)) → LENGTH(s(activate(x0)))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LENGTH1(n__nil) → LENGTH(nil) at position [0] we obtained the following new rules:

LENGTH1(n__nil) → LENGTH(n__nil)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ Narrowing

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__from(x0)) → LENGTH(from(activate(x0)))
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(activate(x0), x1))
LENGTH1(n__s(x0)) → LENGTH(s(activate(x0)))
LENGTH1(n__nil) → LENGTH(n__nil)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ Narrowing

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(n__from(x0)) → LENGTH(from(activate(x0)))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(activate(x0), x1))
LENGTH1(n__s(x0)) → LENGTH(s(activate(x0)))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

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.

LENGTH1(n__s(x0)) → LENGTH(s(activate(x0)))

The remaining pairs can at least be oriented weakly.

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(n__from(x0)) → LENGTH(from(activate(x0)))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(activate(x0), x1))

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( from(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( cons(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	1	1	\
\	0	0	/

x₂

M( n__from(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

M( n__cons(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	1	1	\
\	0	0	/

x₂

M( activate(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( s(x₁) ) =

/	0	\
\	1	/

/	0	0	\
\	0	0	/

x₁

M( n__s(x₁) ) =

/	0	\
\	1	/

/	0	0	\
\	0	0	/

x₁

M( n__nil ) =

/	0	\
\	0	/

M( nil ) =

/	0	\
\	0	/

Tuple symbols:

M( LENGTH(x₁) ) =

[

]

x₁

M( LENGTH1(x₁) ) =

[

]

x₁

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

activate(X) → X
from(X) → cons(X, n__from(n__s(X)))
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ Narrowing

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ NonTerminationProof

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__from(x0)) → LENGTH(from(activate(x0)))
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(activate(x0), x1))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__from(x0)) → LENGTH(from(activate(x0)))
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(activate(x0), x1))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

s = LENGTH1(activate(n__from(X'))) evaluates to t =LENGTH1(activate(n__from(n__s(activate(X')))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [X' / n__s(activate(X'))]

Rewriting sequence

LENGTH1(activate(n__from(X'))) → LENGTH1(from(activate(X')))
with rule activate(n__from(X'')) → from(activate(X'')) at position [0] and matcher [X'' / X']

LENGTH1(from(activate(X'))) → LENGTH1(cons(activate(X'), n__from(n__s(activate(X')))))
with rule from(X) → cons(X, n__from(n__s(X))) at position [0] and matcher [X / activate(X')]

LENGTH1(cons(activate(X'), n__from(n__s(activate(X'))))) → LENGTH1(n__cons(activate(X'), n__from(n__s(activate(X')))))
with rule cons(X1, X2) → n__cons(X1, X2) at position [0] and matcher [X2 / n__from(n__s(activate(X'))), X1 / activate(X')]

LENGTH1(n__cons(activate(X'), n__from(n__s(activate(X'))))) → LENGTH(n__cons(activate(X'), n__from(n__s(activate(X')))))
with rule LENGTH1(x0) → LENGTH(x0) at position [] and matcher [x0 / n__cons(activate(X'), n__from(n__s(activate(X'))))]

LENGTH(n__cons(activate(X'), n__from(n__s(activate(X'))))) → LENGTH1(activate(n__from(n__s(activate(X')))))
with rule LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.