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:

rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)

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:

REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
REV(++(x, x)) → REV(x)

The TRS R consists of the following rules:

rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
REV(++(x, x)) → REV(x)

The TRS R consists of the following rules:

rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(x)

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.


REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
REV(++(x, x)) → REV(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
REV(x1)  =  REV(x1)
++(x1, x2)  =  ++(x1, x2)

Recursive Path Order [2].
Precedence:
[REV1, ++2]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ PisEmptyProof

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

rev(a) → a
rev(b) → b
rev(++(x, y)) → ++(rev(y), rev(x))
rev(++(x, x)) → rev(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.