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
  ↳ 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)))

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:

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)))

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:

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)))

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, 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)))

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, cons2(y, l)) -> REV12(y, l)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( REV12(x1, x2) ) = max{0, x2 - 2}


POL( cons2(x1, x2) ) = x2 + 3



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(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.
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, 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)))

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, cons2(y, l)) -> REV22(y, l)
REV22(x, cons2(y, l)) -> REV1(cons2(x, rev22(y, l)))
The remaining pairs can at least be oriented weakly.

REV1(cons2(x, l)) -> REV22(x, l)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( REV22(x1, x2) ) = max{0, x2 - 2}


POL( cons2(x1, x2) ) = x2 + 3


POL( REV1(x1) ) = max{0, x1 - 3}


POL( rev22(x1, x2) ) = x2


POL( nil ) = 0


POL( rev1(x1) ) = x1



The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof

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

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

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.