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:

rev12(0, nil) -> 0
rev12(s1(X), nil) -> s1(X)
rev12(X, cons2(Y, L)) -> rev12(Y, L)
rev1(nil) -> nil
rev1(cons2(X, L)) -> cons2(rev12(X, L), rev22(X, L))
rev22(X, nil) -> nil
rev22(X, cons2(Y, L)) -> rev1(cons2(X, rev1(rev22(Y, L))))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

rev12(0, nil) -> 0
rev12(s1(X), nil) -> s1(X)
rev12(X, cons2(Y, L)) -> rev12(Y, L)
rev1(nil) -> nil
rev1(cons2(X, L)) -> cons2(rev12(X, L), rev22(X, L))
rev22(X, nil) -> nil
rev22(X, cons2(Y, L)) -> rev1(cons2(X, rev1(rev22(Y, L))))

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:

rev12(0, nil) -> 0
rev12(s1(X), nil) -> s1(X)
rev12(X, cons2(Y, L)) -> rev12(Y, L)
rev1(nil) -> nil
rev1(cons2(X, L)) -> cons2(rev12(X, L), rev22(X, L))
rev22(X, nil) -> nil
rev22(X, cons2(Y, L)) -> rev1(cons2(X, rev1(rev22(Y, L))))

The set Q consists of the following terms:

rev12(0, nil)
rev12(s1(x0), nil)
rev12(x0, cons2(x1, x2))
rev1(nil)
rev1(cons2(x0, x1))
rev22(x0, nil)
rev22(x0, cons2(x1, x2))


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

REV22(X, cons2(Y, L)) -> REV1(cons2(X, rev1(rev22(Y, L))))
REV22(X, cons2(Y, L)) -> REV22(Y, L)
REV1(cons2(X, L)) -> REV22(X, L)
REV1(cons2(X, L)) -> REV12(X, L)
REV22(X, cons2(Y, L)) -> REV1(rev22(Y, L))
REV12(X, cons2(Y, L)) -> REV12(Y, L)

The TRS R consists of the following rules:

rev12(0, nil) -> 0
rev12(s1(X), nil) -> s1(X)
rev12(X, cons2(Y, L)) -> rev12(Y, L)
rev1(nil) -> nil
rev1(cons2(X, L)) -> cons2(rev12(X, L), rev22(X, L))
rev22(X, nil) -> nil
rev22(X, cons2(Y, L)) -> rev1(cons2(X, rev1(rev22(Y, L))))

The set Q consists of the following terms:

rev12(0, nil)
rev12(s1(x0), nil)
rev12(x0, cons2(x1, x2))
rev1(nil)
rev1(cons2(x0, x1))
rev22(x0, nil)
rev22(x0, cons2(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:

REV22(X, cons2(Y, L)) -> REV1(cons2(X, rev1(rev22(Y, L))))
REV22(X, cons2(Y, L)) -> REV22(Y, L)
REV1(cons2(X, L)) -> REV22(X, L)
REV1(cons2(X, L)) -> REV12(X, L)
REV22(X, cons2(Y, L)) -> REV1(rev22(Y, L))
REV12(X, cons2(Y, L)) -> REV12(Y, L)

The TRS R consists of the following rules:

rev12(0, nil) -> 0
rev12(s1(X), nil) -> s1(X)
rev12(X, cons2(Y, L)) -> rev12(Y, L)
rev1(nil) -> nil
rev1(cons2(X, L)) -> cons2(rev12(X, L), rev22(X, L))
rev22(X, nil) -> nil
rev22(X, cons2(Y, L)) -> rev1(cons2(X, rev1(rev22(Y, L))))

The set Q consists of the following terms:

rev12(0, nil)
rev12(s1(x0), nil)
rev12(x0, cons2(x1, x2))
rev1(nil)
rev1(cons2(x0, x1))
rev22(x0, nil)
rev22(x0, cons2(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, cons2(Y, L)) -> REV12(Y, L)

The TRS R consists of the following rules:

rev12(0, nil) -> 0
rev12(s1(X), nil) -> s1(X)
rev12(X, cons2(Y, L)) -> rev12(Y, L)
rev1(nil) -> nil
rev1(cons2(X, L)) -> cons2(rev12(X, L), rev22(X, L))
rev22(X, nil) -> nil
rev22(X, cons2(Y, L)) -> rev1(cons2(X, rev1(rev22(Y, L))))

The set Q consists of the following terms:

rev12(0, nil)
rev12(s1(x0), nil)
rev12(x0, cons2(x1, x2))
rev1(nil)
rev1(cons2(x0, x1))
rev22(x0, nil)
rev22(x0, cons2(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, cons2(Y, L)) -> REV12(Y, L)
Used argument filtering: REV12(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi 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:

rev12(0, nil) -> 0
rev12(s1(X), nil) -> s1(X)
rev12(X, cons2(Y, L)) -> rev12(Y, L)
rev1(nil) -> nil
rev1(cons2(X, L)) -> cons2(rev12(X, L), rev22(X, L))
rev22(X, nil) -> nil
rev22(X, cons2(Y, L)) -> rev1(cons2(X, rev1(rev22(Y, L))))

The set Q consists of the following terms:

rev12(0, nil)
rev12(s1(x0), nil)
rev12(x0, cons2(x1, x2))
rev1(nil)
rev1(cons2(x0, x1))
rev22(x0, nil)
rev22(x0, cons2(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
                ↳ QDPAfsSolverProof

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

REV22(X, cons2(Y, L)) -> REV1(cons2(X, rev1(rev22(Y, L))))
REV22(X, cons2(Y, L)) -> REV22(Y, L)
REV1(cons2(X, L)) -> REV22(X, L)
REV22(X, cons2(Y, L)) -> REV1(rev22(Y, L))

The TRS R consists of the following rules:

rev12(0, nil) -> 0
rev12(s1(X), nil) -> s1(X)
rev12(X, cons2(Y, L)) -> rev12(Y, L)
rev1(nil) -> nil
rev1(cons2(X, L)) -> cons2(rev12(X, L), rev22(X, L))
rev22(X, nil) -> nil
rev22(X, cons2(Y, L)) -> rev1(cons2(X, rev1(rev22(Y, L))))

The set Q consists of the following terms:

rev12(0, nil)
rev12(s1(x0), nil)
rev12(x0, cons2(x1, x2))
rev1(nil)
rev1(cons2(x0, x1))
rev22(x0, nil)
rev22(x0, cons2(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.

REV22(X, cons2(Y, L)) -> REV22(Y, L)
REV1(cons2(X, L)) -> REV22(X, L)
REV22(X, cons2(Y, L)) -> REV1(rev22(Y, L))
Used argument filtering: REV22(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
REV1(x1)  =  x1
rev1(x1)  =  x1
rev22(x1, x2)  =  x2
nil  =  nil
rev12(x1, x2)  =  rev1
0  =  0
s1(x1)  =  s
Used ordering: Quasi Precedence: [rev1, 0, s]


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

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

REV22(X, cons2(Y, L)) -> REV1(cons2(X, rev1(rev22(Y, L))))

The TRS R consists of the following rules:

rev12(0, nil) -> 0
rev12(s1(X), nil) -> s1(X)
rev12(X, cons2(Y, L)) -> rev12(Y, L)
rev1(nil) -> nil
rev1(cons2(X, L)) -> cons2(rev12(X, L), rev22(X, L))
rev22(X, nil) -> nil
rev22(X, cons2(Y, L)) -> rev1(cons2(X, rev1(rev22(Y, L))))

The set Q consists of the following terms:

rev12(0, nil)
rev12(s1(x0), nil)
rev12(x0, cons2(x1, x2))
rev1(nil)
rev1(cons2(x0, x1))
rev22(x0, nil)
rev22(x0, cons2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.