(0) Obligation:

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

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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)) → ++1(rev(y), rev(x))
REV(++(x, y)) → REV(y)
REV(++(x, y)) → REV(x)
++1(.(x, y), z) → ++1(y, z)
++1(x, ++(y, z)) → ++1(++(x, y), z)
++1(x, ++(y, z)) → ++1(x, y)

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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:

++1(x, ++(y, z)) → ++1(++(x, y), z)
++1(.(x, y), z) → ++1(y, z)
++1(x, ++(y, z)) → ++1(x, y)

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


++1(x, ++(y, z)) → ++1(++(x, y), z)
++1(.(x, y), z) → ++1(y, z)
++1(x, ++(y, z)) → ++1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:
++^12 > ++2 > .2
rev1 > ++2 > .2
rev1 > nil > .2
make1 > nil > .2

Status:
++^12: [2,1]
++2: [2,1]
.2: multiset
rev1: [1]
nil: multiset
make1: multiset

The following usable rules [FROCOS05] were oriented:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

(7) Obligation:

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

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

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

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Precedence:
REV1 > .2
rev1 > ++2 > .2
rev1 > nil > .2
make1 > nil > .2

Status:
REV1: [1]
++2: [2,1]
rev1: [1]
nil: multiset
.2: multiset
make1: multiset

The following usable rules [FROCOS05] were oriented:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

(12) Obligation:

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

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE