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 interpretation [21]:

POL(REV12(x1, x2)) = 2·x1 + 3·x2   
POL(cons2(x1, x2)) = 1 + x1 + 2·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(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)
REV1(cons2(x, l)) -> REV22(x, l)
REV22(x, cons2(y, l)) -> REV1(cons2(x, rev22(y, l)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(REV1(x1)) = 2 + x1   
POL(REV22(x1, x2)) = 2 + 3·x2   
POL(cons2(x1, x2)) = 2 + 3·x2   
POL(nil) = 0   
POL(rev1(x1)) = x1   
POL(rev12(x1, x2)) = 3·x2   
POL(rev22(x1, x2)) = x2   
POL(s1(x1)) = 0   

The following usable rules [14] were oriented:

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



↳ 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)))

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.