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) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) 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))))

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) 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))))

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) 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))))

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

REV1(x, ++(y, z)) → REV1(y, z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
REV1(x1, x2)  =  x2
++(x1, x2)  =  ++(x2)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
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))))

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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))))

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
REV(x1)  =  x1
++(x1, x2)  =  ++(x2)
REV2(x1, x2)  =  x2
rev(x1)  =  x1
rev2(x1, x2)  =  x2
nil  =  nil

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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

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

The set Q consists of the following terms:

rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(16) TRUE