(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) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REV1(x, ++(y, z)) → REV1(y, z)
R is empty.
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) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
rev(nil)
rev(++(x0, x1))
rev1(x0, nil)
rev1(x0, ++(x1, x2))
rev2(x0, nil)
rev2(x0, ++(x1, x2))
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REV1(x, ++(y, z)) → REV1(y, z)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- REV1(x, ++(y, z)) → REV1(y, z)
The graph contains the following edges 2 > 1, 2 > 2
(13) YES
(14) 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.
(15) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
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: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( ++(x1, x2) ) = 2x2 + 1 |
| POL( rev1(x1, x2) ) = 2x1 + 2x2 |
| POL( REV2(x1, x2) ) = x2 + 1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] 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))
(16) 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.
(17) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(18) TRUE