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(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

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)) → REV1(x, y)
REV(++(x, y)) → REV2(x, y)
REV1(x, ++(y, z)) → REV1(y, z)
REV2(x, ++(y, z)) → REV(++(x, rev(rev2(y, z))))
REV2(x, ++(y, z)) → REV(rev2(y, z))
REV2(x, ++(y, z)) → REV2(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

REV1(x, ++(y, z)) → REV1(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

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

(6) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
++(x1, x2)  =  ++(x2)

From the DPs we obtained the following set of size-change graphs:

• REV1(x, ++(y, z)) → REV1(y, z) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) Obligation:

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

REV(++(x, y)) → REV2(x, y)
REV2(x, ++(y, z)) → REV(++(x, rev(rev2(y, z))))
REV2(x, ++(y, z)) → REV(rev2(y, z))
REV2(x, ++(y, z)) → REV2(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Combined order from the following AFS and order.
rev2(x1, x2)  =  x2
nil  =  nil
++(x1, x2)  =  ++(x2)
rev(x1)  =  x1
rev1(x1, x2)  =  rev1(x2)

Homeomorphic Embedding Order

AFS:
rev2(x1, x2)  =  x2
nil  =  nil
++(x1, x2)  =  ++(x2)
rev(x1)  =  x1
rev1(x1, x2)  =  rev1(x2)

From the DPs we obtained the following set of size-change graphs:

• REV2(x, ++(y, z)) → REV2(y, z) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 2 > 2

• REV(++(x, y)) → REV2(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 2

• REV2(x, ++(y, z)) → REV(++(x, rev(rev2(y, z)))) (allowed arguments on rhs = {1})
  The graph contains the following edges 2 >= 1

• REV2(x, ++(y, z)) → REV(rev2(y, z)) (allowed arguments on rhs = {1})
  The graph contains the following edges 2 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05].

rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(y, z))))
rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))

(10) TRUE