Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/AG01SLASHHASH3.38.trs

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

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

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))

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:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))

The set Q consists of the following terms:

rev₁(nil)
rev₁(cons₂(x0, x1))
rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

REV2₂(x, cons₂(y, l)) -> REV2₂(y, l)
REV₁(cons₂(x, l)) -> REV2₂(x, l)
REV₁(cons₂(x, l)) -> REV1₂(x, l)
REV1₂(x, cons₂(y, l)) -> REV1₂(y, l)
REV2₂(x, cons₂(y, l)) -> REV₁(cons₂(x, rev2₂(y, l)))

The TRS R consists of the following rules:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))

The set Q consists of the following terms:

rev₁(nil)
rev₁(cons₂(x0, x1))
rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

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:

REV2₂(x, cons₂(y, l)) -> REV2₂(y, l)
REV₁(cons₂(x, l)) -> REV2₂(x, l)
REV₁(cons₂(x, l)) -> REV1₂(x, l)
REV1₂(x, cons₂(y, l)) -> REV1₂(y, l)
REV2₂(x, cons₂(y, l)) -> REV₁(cons₂(x, rev2₂(y, l)))

The TRS R consists of the following rules:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))

The set Q consists of the following terms:

rev₁(nil)
rev₁(cons₂(x0, x1))
rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

REV1₂(x, cons₂(y, l)) -> REV1₂(y, l)

The TRS R consists of the following rules:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))

The set Q consists of the following terms:

rev₁(nil)
rev₁(cons₂(x0, x1))
rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

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.

REV1₂(x, cons₂(y, l)) -> REV1₂(y, l)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
REV1₂(x1, x2) = REV1₁(x2)
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

cons₂ > REV1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))

The set Q consists of the following terms:

rev₁(nil)
rev₁(cons₂(x0, x1))
rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

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

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

REV2₂(x, cons₂(y, l)) -> REV2₂(y, l)
REV₁(cons₂(x, l)) -> REV2₂(x, l)
REV2₂(x, cons₂(y, l)) -> REV₁(cons₂(x, rev2₂(y, l)))

The TRS R consists of the following rules:

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))

The set Q consists of the following terms:

rev₁(nil)
rev₁(cons₂(x0, x1))
rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

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.

REV2₂(x, cons₂(y, l)) -> REV2₂(y, l)
REV₁(cons₂(x, l)) -> REV2₂(x, l)
REV2₂(x, cons₂(y, l)) -> REV₁(cons₂(x, rev2₂(y, l)))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
REV2₂(x1, x2) = REV2₁(x2)
cons₂(x1, x2) = cons₁(x2)
REV₁(x1) = x1
rev2₂(x1, x2) = x2
nil = nil
rev₁(x1) = x1
rev1₂(x1, x2) = rev1
0 = 0
s₁(x1) = s

Lexicographic Path Order [19].
Precedence:

cons₁ > REV2₁
nil > 0
rev1 > 0
rev1 > s

The following usable rules [14] were oriented:

rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

rev₁(nil) -> nil
rev₁(cons₂(x, l)) -> cons₂(rev1₂(x, l), rev2₂(x, l))
rev1₂(0, nil) -> 0
rev1₂(s₁(x), nil) -> s₁(x)
rev1₂(x, cons₂(y, l)) -> rev1₂(y, l)
rev2₂(x, nil) -> nil
rev2₂(x, cons₂(y, l)) -> rev₁(cons₂(x, rev2₂(y, l)))

The set Q consists of the following terms:

rev₁(nil)
rev₁(cons₂(x0, x1))
rev1₂(0, nil)
rev1₂(s₁(x0), nil)
rev1₂(x0, cons₂(x1, x2))
rev2₂(x0, nil)
rev2₂(x0, cons₂(x1, x2))

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