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(a) -> a
rev1(b) -> b
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(++2(x, x)) -> rev1(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

rev1(a) -> a
rev1(b) -> b
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(++2(x, x)) -> rev1(x)

Q is empty.

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

REV1(++2(x, y)) -> REV1(x)
REV1(++2(x, y)) -> REV1(y)
REV1(++2(x, x)) -> REV1(x)

The TRS R consists of the following rules:

rev1(a) -> a
rev1(b) -> b
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(++2(x, x)) -> rev1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof

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

REV1(++2(x, y)) -> REV1(x)
REV1(++2(x, y)) -> REV1(y)
REV1(++2(x, x)) -> REV1(x)

The TRS R consists of the following rules:

rev1(a) -> a
rev1(b) -> b
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(++2(x, x)) -> rev1(x)

Q is empty.
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.

REV1(++2(x, y)) -> REV1(x)
REV1(++2(x, y)) -> REV1(y)
REV1(++2(x, x)) -> REV1(x)
Used argument filtering: REV1(x1)  =  x1
++2(x1, x2)  =  ++2(x1, x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

rev1(a) -> a
rev1(b) -> b
rev1(++2(x, y)) -> ++2(rev1(y), rev1(x))
rev1(++2(x, x)) -> rev1(x)

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.