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(++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))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [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, ++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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


REV12(x, ++2(y, z)) -> REV12(y, z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
REV12(x1, x2)  =  REV12(x1, x2)
++2(x1, x2)  =  ++2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[REV12, ++2]


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(++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
                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


REV22(x, ++2(y, z)) -> REV22(y, z)
REV22(x, ++2(y, z)) -> REV1(rev22(y, z))
REV1(++2(x, y)) -> REV22(x, y)
The remaining pairs can at least by weakly be oriented.

REV22(x, ++2(y, z)) -> REV1(++2(x, rev1(rev22(y, z))))
Used ordering: Combined order from the following AFS and order.
REV22(x1, x2)  =  x2
++2(x1, x2)  =  ++1(x2)
REV1(x1)  =  x1
rev22(x1, x2)  =  x2
rev1(x1)  =  x1
nil  =  nil
rev12(x1, x2)  =  rev11(x2)

Lexicographic Path Order [19].
Precedence:
[++1, rev11]


The following usable rules [14] were oriented:

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



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

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

REV22(x, ++2(y, z)) -> REV1(++2(x, rev1(rev22(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.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.