Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/nonterminSLASHCSRSLASHEx2

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

  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s₁(x0), cons₂(x1, x2))
from₁(x0)
add₂(0, x0)
add₂(s₁(x0), x1)
len₁(nil)
len₁(cons₂(x0, x1))

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))

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s₁(x0), cons₂(x1, x2))
from₁(x0)
add₂(0, x0)
add₂(s₁(x0), x1)
len₁(nil)
len₁(cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s₁(x0), cons₂(x1, x2))
from₁(x0)
add₂(0, x0)
add₂(s₁(x0), x1)
len₁(nil)
len₁(cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s₁(x0), cons₂(x1, x2))
from₁(x0)
add₂(0, x0)
add₂(s₁(x0), x1)
len₁(nil)
len₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

LEN₁(cons₂(X, Z)) -> LEN₁(Z)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LEN₁(x1) = LEN₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

cons₂ > LEN₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s₁(x0), cons₂(x1, x2))
from₁(x0)
add₂(0, x0)
add₂(s₁(x0), x1)
len₁(nil)
len₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s₁(x0), cons₂(x1, x2))
from₁(x0)
add₂(0, x0)
add₂(s₁(x0), x1)
len₁(nil)
len₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ADD₂(x1, x2) = ADD₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > ADD₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s₁(x0), cons₂(x1, x2))
from₁(x0)
add₂(0, x0)
add₂(s₁(x0), x1)
len₁(nil)
len₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s₁(x0), cons₂(x1, x2))
from₁(x0)
add₂(0, x0)
add₂(s₁(x0), x1)
len₁(nil)
len₁(cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s₁(x0), cons₂(x1, x2))
from₁(x0)
add₂(0, x0)
add₂(s₁(x0), x1)
len₁(nil)
len₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
FST₂(x1, x2) = FST₁(x1)
s₁(x1) = s₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

s₁ > FST₁
cons₂ > FST₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

fst₂(0, x0)
fst₂(s₁(x0), cons₂(x1, x2))
from₁(x0)
add₂(0, x0)
add₂(s₁(x0), x1)
len₁(nil)
len₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.