(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, 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:
REV(cons(x, l)) → REV1(x, l)
REV(cons(x, l)) → REV2(x, l)
REV1(x, cons(y, l)) → REV1(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
REV2(x, cons(y, l)) → REV2(y, l)
The TRS R consists of the following rules:
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, 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:
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, 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(
x1,
x2) =
REV1(
x1,
x2)
Tags:
REV1 has tags [1,1]
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Recursive path order with status [RPO].
Quasi-Precedence:
trivial
Status:
cons2: multiset
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(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, 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, rev2(y, l)))
REV2(x, cons(y, l)) → REV2(y, l)
The TRS R consists of the following rules:
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, 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)) → 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(
x1) =
REV(
x1)
REV2(
x1,
x2) =
REV2(
x2)
Tags:
REV has tags [1]
REV2 has tags [2,1]
Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
cons(
x1,
x2) =
cons(
x2)
rev2(
x1,
x2) =
x2
nil =
nil
rev(
x1) =
x1
rev1(
x1,
x2) =
x2
0 =
0
s(
x1) =
x1
Recursive path order with status [RPO].
Quasi-Precedence:
[nil, 0]
Status:
cons1: [1]
nil: multiset
0: multiset
The following usable rules [FROCOS05] were oriented:
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))
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, rev2(y, l)))
The TRS R consists of the following rules:
rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, 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