(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))
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:
REV1(X, cons(Y, L)) → REV1(Y, L)
REV(cons(X, L)) → REV1(X, L)
REV(cons(X, L)) → REV2(X, L)
REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))
REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV2(X, cons(Y, L)) → REV2(Y, L)
The TRS R consists of the following rules:
rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))
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, cons(Y, L)) → REV1(Y, L)
The TRS R consists of the following rules:
rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))
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, cons(Y, L)) → REV1(Y, L)
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)
cons(
x1,
x2) =
cons(
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:
rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))
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(cons(X, L)) → REV2(X, L)
REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))
REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV2(X, cons(Y, L)) → REV2(Y, L)
The TRS R consists of the following rules:
rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))
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(cons(X, L)) → REV2(X, L)
REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV2(X, cons(Y, L)) → REV2(Y, L)
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(
x0)
REV2(
x0,
x1,
x2) =
REV2(
x0)
Tags:
REV has argument tags [4,7] and root tag 0
REV2 has argument tags [4,7,0] 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.
REV(
x1) =
x1
cons(
x1,
x2) =
cons(
x2)
REV2(
x1,
x2) =
x2
rev(
x1) =
x1
rev2(
x1,
x2) =
x2
nil =
nil
rev1(
x1,
x2) =
rev1(
x2)
0 =
0
s(
x1) =
x1
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented:
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))
The TRS R consists of the following rules:
rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(14) TRUE