Termination w.r.t. Q proof of TRS/nontermin/CSREx2

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:

fst₂(0, Z) -> nil
fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, fst₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
len₁(nil) -> 0
len₁(cons₂(X, Z)) -> s₁(len₁(Z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

fst₂(0, Z) -> nil
fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, fst₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
len₁(nil) -> 0
len₁(cons₂(X, Z)) -> s₁(len₁(Z))

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:

ADD₂(s₁(X), Y) -> ADD₂(X, Y)
LEN₁(cons₂(X, Z)) -> LEN₁(Z)
FROM₁(X) -> FROM₁(s₁(X))
FST₂(s₁(X), cons₂(Y, Z)) -> FST₂(X, Z)

The TRS R consists of the following rules:

fst₂(0, Z) -> nil
fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, fst₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
len₁(nil) -> 0
len₁(cons₂(X, Z)) -> s₁(len₁(Z))

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:

ADD₂(s₁(X), Y) -> ADD₂(X, Y)
LEN₁(cons₂(X, Z)) -> LEN₁(Z)
FROM₁(X) -> FROM₁(s₁(X))
FST₂(s₁(X), cons₂(Y, Z)) -> FST₂(X, Z)

The TRS R consists of the following rules:

fst₂(0, Z) -> nil
fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, fst₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
len₁(nil) -> 0
len₁(cons₂(X, Z)) -> s₁(len₁(Z))

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:

LEN₁(cons₂(X, Z)) -> LEN₁(Z)

The TRS R consists of the following rules:

fst₂(0, Z) -> nil
fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, fst₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
len₁(nil) -> 0
len₁(cons₂(X, Z)) -> s₁(len₁(Z))

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.

LEN₁(cons₂(X, Z)) -> LEN₁(Z)

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

POL(LEN₁(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:

fst₂(0, Z) -> nil
fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, fst₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
len₁(nil) -> 0
len₁(cons₂(X, Z)) -> s₁(len₁(Z))

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:

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

The TRS R consists of the following rules:

fst₂(0, Z) -> nil
fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, fst₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
len₁(nil) -> 0
len₁(cons₂(X, Z)) -> s₁(len₁(Z))

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.

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

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

POL(ADD₂(x₁, x₂)) = 2·x₁
POL(s₁(x₁)) = 1 + 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:

fst₂(0, Z) -> nil
fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, fst₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
len₁(nil) -> 0
len₁(cons₂(X, Z)) -> s₁(len₁(Z))

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:

FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

fst₂(0, Z) -> nil
fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, fst₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
len₁(nil) -> 0
len₁(cons₂(X, Z)) -> s₁(len₁(Z))

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:

FST₂(s₁(X), cons₂(Y, Z)) -> FST₂(X, Z)

The TRS R consists of the following rules:

fst₂(0, Z) -> nil
fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, fst₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
len₁(nil) -> 0
len₁(cons₂(X, Z)) -> s₁(len₁(Z))

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.

FST₂(s₁(X), cons₂(Y, Z)) -> FST₂(X, Z)

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

POL(FST₂(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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

fst₂(0, Z) -> nil
fst₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, fst₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
len₁(nil) -> 0
len₁(cons₂(X, Z)) -> s₁(len₁(Z))

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.