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))


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)) -> 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 [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, 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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


REV12(X, cons2(Y, L)) -> REV12(Y, L)
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)
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[REV12, cons2]


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:

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
                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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 remaining pairs can at least by weakly be oriented.

REV22(X, cons2(Y, L)) -> REV1(cons2(X, rev1(rev22(Y, L))))
Used ordering: Combined order from the following AFS and order.
REV22(x1, x2)  =  REV21(x2)
cons2(x1, x2)  =  cons1(x2)
REV1(x1)  =  REV1(x1)
rev1(x1)  =  x1
rev22(x1, x2)  =  x2
rev12(x1, x2)  =  x1
0  =  0
nil  =  nil
s1(x1)  =  s

Lexicographic Path Order [19].
Precedence:
[REV21, REV1] > cons1
[0, nil] > s


The following usable rules [14] were oriented:

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



↳ 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, 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 [13,14,18] contains 0 SCCs with 1 less node.