(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) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
REV1(
x0,
x1,
x2) =
REV1(
x2)
Tags:
REV1 has argument tags [1,3,0] and root tag 0
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
REV1(
x1,
x2) =
REV1(
x1,
x2)
++(
x1,
x2) =
++(
x2)
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
none
(7) 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))))
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)) → 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.
(11) 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(++(x, rev(rev2(y, z))))
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: SCNP Order with the following components:
Level mapping:
Top level AFS:
REV(
x0,
x1) =
REV(
x1)
REV2(
x0,
x1,
x2) =
REV2(
x2)
Tags:
REV has argument tags [0,7] and root tag 0
REV2 has argument tags [0,5,7] and root tag 1
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
REV(
x1) =
x1
++(
x1,
x2) =
++(
x2)
REV2(
x1,
x2) =
REV2(
x1)
rev(
x1) =
x1
rev2(
x1,
x2) =
x2
nil =
nil
rev1(
x1,
x2) =
rev1
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
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))
(12) 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))))
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