Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
REV(x1)  =  x1
++(x1, x2)  =  ++(x1, x2)
rev(x1)  =  rev(x1)
a  =  a
b  =  b

Lexicographic path order with status [LPO].
Precedence:

rev₁ > ++₂
rev₁ > a
rev₁ > b

Status:

++₂: [1,2]
rev₁: [1]
a: []
b: []

The following usable rules [FROCOS05] were oriented:

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

(4) Obligation:

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.

(5) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(6) TRUE