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:

rev1(nil) -> nil
rev1(++2(x, y)) -> ++2(rev12(x, y), rev22(x, y))
rev12(x, nil) -> x
rev12(x, ++2(y, z)) -> rev12(y, z)
rev22(x, nil) -> nil
rev22(x, ++2(y, z)) -> rev1(++2(x, rev1(rev22(y, z))))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

rev1(nil) -> nil
rev1(++2(x, y)) -> ++2(rev12(x, y), rev22(x, y))
rev12(x, nil) -> x
rev12(x, ++2(y, z)) -> rev12(y, z)
rev22(x, nil) -> nil
rev22(x, ++2(y, z)) -> rev1(++2(x, rev1(rev22(y, 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:

rev1(nil) -> nil
rev1(++2(x, y)) -> ++2(rev12(x, y), rev22(x, y))
rev12(x, nil) -> x
rev12(x, ++2(y, z)) -> rev12(y, z)
rev22(x, nil) -> nil
rev22(x, ++2(y, z)) -> rev1(++2(x, rev1(rev22(y, z))))

The set Q consists of the following terms:

rev1(nil)
rev1(++2(x0, x1))
rev12(x0, nil)
rev12(x0, ++2(x1, x2))
rev22(x0, nil)
rev22(x0, ++2(x1, x2))


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

REV12(x, ++2(y, z)) -> REV12(y, z)
REV1(++2(x, y)) -> REV12(x, y)
REV22(x, ++2(y, z)) -> REV22(y, z)
REV22(x, ++2(y, z)) -> REV1(rev22(y, z))
REV22(x, ++2(y, z)) -> REV1(++2(x, rev1(rev22(y, z))))
REV1(++2(x, y)) -> REV22(x, y)

The TRS R consists of the following rules:

rev1(nil) -> nil
rev1(++2(x, y)) -> ++2(rev12(x, y), rev22(x, y))
rev12(x, nil) -> x
rev12(x, ++2(y, z)) -> rev12(y, z)
rev22(x, nil) -> nil
rev22(x, ++2(y, z)) -> rev1(++2(x, rev1(rev22(y, z))))

The set Q consists of the following terms:

rev1(nil)
rev1(++2(x0, x1))
rev12(x0, nil)
rev12(x0, ++2(x1, x2))
rev22(x0, nil)
rev22(x0, ++2(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:

REV12(x, ++2(y, z)) -> REV12(y, z)
REV1(++2(x, y)) -> REV12(x, y)
REV22(x, ++2(y, z)) -> REV22(y, z)
REV22(x, ++2(y, z)) -> REV1(rev22(y, z))
REV22(x, ++2(y, z)) -> REV1(++2(x, rev1(rev22(y, z))))
REV1(++2(x, y)) -> REV22(x, y)

The TRS R consists of the following rules:

rev1(nil) -> nil
rev1(++2(x, y)) -> ++2(rev12(x, y), rev22(x, y))
rev12(x, nil) -> x
rev12(x, ++2(y, z)) -> rev12(y, z)
rev22(x, nil) -> nil
rev22(x, ++2(y, z)) -> rev1(++2(x, rev1(rev22(y, z))))

The set Q consists of the following terms:

rev1(nil)
rev1(++2(x0, x1))
rev12(x0, nil)
rev12(x0, ++2(x1, x2))
rev22(x0, nil)
rev22(x0, ++2(x1, x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP

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

REV12(x, ++2(y, z)) -> REV12(y, z)

The TRS R consists of the following rules:

rev1(nil) -> nil
rev1(++2(x, y)) -> ++2(rev12(x, y), rev22(x, y))
rev12(x, nil) -> x
rev12(x, ++2(y, z)) -> rev12(y, z)
rev22(x, nil) -> nil
rev22(x, ++2(y, z)) -> rev1(++2(x, rev1(rev22(y, z))))

The set Q consists of the following terms:

rev1(nil)
rev1(++2(x0, x1))
rev12(x0, nil)
rev12(x0, ++2(x1, x2))
rev22(x0, nil)
rev22(x0, ++2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

REV12(x, ++2(y, z)) -> REV12(y, z)
Used argument filtering: REV12(x1, x2)  =  x2
++2(x1, x2)  =  ++1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

rev1(nil) -> nil
rev1(++2(x, y)) -> ++2(rev12(x, y), rev22(x, y))
rev12(x, nil) -> x
rev12(x, ++2(y, z)) -> rev12(y, z)
rev22(x, nil) -> nil
rev22(x, ++2(y, z)) -> rev1(++2(x, rev1(rev22(y, z))))

The set Q consists of the following terms:

rev1(nil)
rev1(++2(x0, x1))
rev12(x0, nil)
rev12(x0, ++2(x1, x2))
rev22(x0, nil)
rev22(x0, ++2(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

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

REV22(x, ++2(y, z)) -> REV22(y, z)
REV22(x, ++2(y, z)) -> REV1(rev22(y, z))
REV22(x, ++2(y, z)) -> REV1(++2(x, rev1(rev22(y, z))))
REV1(++2(x, y)) -> REV22(x, y)

The TRS R consists of the following rules:

rev1(nil) -> nil
rev1(++2(x, y)) -> ++2(rev12(x, y), rev22(x, y))
rev12(x, nil) -> x
rev12(x, ++2(y, z)) -> rev12(y, z)
rev22(x, nil) -> nil
rev22(x, ++2(y, z)) -> rev1(++2(x, rev1(rev22(y, z))))

The set Q consists of the following terms:

rev1(nil)
rev1(++2(x0, x1))
rev12(x0, nil)
rev12(x0, ++2(x1, x2))
rev22(x0, nil)
rev22(x0, ++2(x1, x2))

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