Termination w.r.t. Q proof of TRS/nontermin/CSREx1

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:

eq₂(0, 0) -> true
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(length₁(L))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

eq₂(0, 0) -> true
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(length₁(L))

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:

EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)
TAKE₂(s₁(X), cons₂(Y, L)) -> TAKE₂(X, L)
LENGTH₁(cons₂(X, L)) -> LENGTH₁(L)
INF₁(X) -> INF₁(s₁(X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(length₁(L))

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:

EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)
TAKE₂(s₁(X), cons₂(Y, L)) -> TAKE₂(X, L)
LENGTH₁(cons₂(X, L)) -> LENGTH₁(L)
INF₁(X) -> INF₁(s₁(X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(length₁(L))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LENGTH₁(cons₂(X, L)) -> LENGTH₁(L)

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(length₁(L))

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.

LENGTH₁(cons₂(X, L)) -> LENGTH₁(L)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LENGTH₁(x₁)) = 2·x₁
POL(cons₂(x₁, x₂)) = 1 + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(length₁(L))

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

TAKE₂(s₁(X), cons₂(Y, L)) -> TAKE₂(X, L)

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(length₁(L))

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.

TAKE₂(s₁(X), cons₂(Y, L)) -> TAKE₂(X, L)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(TAKE₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(cons₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(length₁(L))

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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

INF₁(X) -> INF₁(s₁(X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(length₁(L))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(length₁(L))

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.

EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(EQ₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(Y, take₂(X, L))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(length₁(L))

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.