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
  ↳ 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))))

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:

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))))

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:

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))))

Q is empty.
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
  ↳ 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))))

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.


REV12(x, ++2(y, z)) -> REV12(y, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(++2(x1, x2)) = 1 + x2   
POL(REV12(x1, x2)) = x2   

The following usable rules [14] were oriented: none



↳ 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))))

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

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))))

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.


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 be oriented weakly.

REV22(x, ++2(y, z)) -> REV1(++2(x, rev1(rev22(y, z))))
Used ordering: Polynomial interpretation [21]:

POL(++2(x1, x2)) = 1 + x2   
POL(REV1(x1)) = 1 + x1   
POL(REV22(x1, x2)) = 1 + x2   
POL(nil) = 0   
POL(rev1(x1)) = x1   
POL(rev12(x1, x2)) = 0   
POL(rev22(x1, x2)) = x2   

The following usable rules [14] were oriented:

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



↳ 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))))

Q is empty.
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.