(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
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:
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(length(nil)) → MARK(0)
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
ACTIVE(length(cons(X, Y))) → S(length1(Y))
ACTIVE(length(cons(X, Y))) → LENGTH1(Y)
ACTIVE(length1(X)) → MARK(length(X))
ACTIVE(length1(X)) → LENGTH(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(nil) → ACTIVE(nil)
MARK(0) → ACTIVE(0)
MARK(length1(X)) → ACTIVE(length1(X))
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
LENGTH1(mark(X)) → LENGTH1(X)
LENGTH1(active(X)) → LENGTH1(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
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 6 SCCs with 12 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH1(active(X)) → LENGTH1(X)
LENGTH1(mark(X)) → LENGTH1(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH1(active(X)) → LENGTH1(X)
LENGTH1(mark(X)) → LENGTH1(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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:
- LENGTH1(active(X)) → LENGTH1(X)
The graph contains the following edges 1 > 1
- LENGTH1(mark(X)) → LENGTH1(X)
The graph contains the following edges 1 > 1
(9) YES
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(active(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(active(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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:
- LENGTH(active(X)) → LENGTH(X)
The graph contains the following edges 1 > 1
- LENGTH(mark(X)) → LENGTH(X)
The graph contains the following edges 1 > 1
(14) YES
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(active(X)) → S(X)
S(mark(X)) → S(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
S(active(X)) → S(X)
S(mark(X)) → S(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) 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:
- S(active(X)) → S(X)
The graph contains the following edges 1 > 1
- S(mark(X)) → S(X)
The graph contains the following edges 1 > 1
(19) YES
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) 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:
- CONS(X1, mark(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- CONS(mark(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- CONS(active(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- CONS(X1, active(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
(24) YES
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(26) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(28) 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:
- FROM(active(X)) → FROM(X)
The graph contains the following edges 1 > 1
- FROM(mark(X)) → FROM(X)
The graph contains the following edges 1 > 1
(29) YES
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(length1(X)) → MARK(length(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(31) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( ACTIVE(x1) ) = x1 + 2 |
POL( cons(x1, x2) ) = x1 + 2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
mark(from(X)) → active(from(mark(X)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(length(cons(X, Y))) → mark(s(length1(Y)))
mark(s(X)) → active(s(mark(X)))
active(length1(X)) → mark(length(X))
mark(length(X)) → active(length(X))
mark(length1(X)) → active(length1(X))
mark(nil) → active(nil)
mark(0) → active(0)
from(active(X)) → from(X)
from(mark(X)) → from(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
length1(active(X)) → length1(X)
length1(mark(X)) → length1(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
active(length(nil)) → mark(0)
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(length1(X)) → MARK(length(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(33) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → ACTIVE(s(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(
x1) =
MARK
from(
x1) =
from
ACTIVE(
x1) =
x1
mark(
x1) =
mark
cons(
x1,
x2) =
cons
s(
x1) =
s
length(
x1) =
length
length1(
x1) =
length1
active(
x1) =
active
nil =
nil
0 =
0
Recursive path order with status [RPO].
Quasi-Precedence:
[mark, nil, 0] > [MARK, from, length, length1] > active > cons
[mark, nil, 0] > [MARK, from, length, length1] > active > s
Status:
MARK: multiset
from: multiset
mark: multiset
cons: []
s: []
length: multiset
length1: multiset
active: multiset
nil: multiset
0: multiset
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
from(active(X)) → from(X)
from(mark(X)) → from(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
length1(active(X)) → length1(X)
length1(mark(X)) → length1(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
ACTIVE(length1(X)) → MARK(length(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(35) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
MARK(from(X)) → ACTIVE(from(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( ACTIVE(x1) ) = x1 + 2 |
POL( MARK(x1) ) = 2x1 + 2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
mark(from(X)) → active(from(mark(X)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(length(cons(X, Y))) → mark(s(length1(Y)))
mark(s(X)) → active(s(mark(X)))
active(length1(X)) → mark(length(X))
mark(length(X)) → active(length(X))
mark(length1(X)) → active(length1(X))
mark(nil) → active(nil)
mark(0) → active(0)
from(active(X)) → from(X)
from(mark(X)) → from(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
length1(active(X)) → length1(X)
length1(mark(X)) → length1(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
active(length(nil)) → mark(0)
(36) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
ACTIVE(length1(X)) → MARK(length(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(37) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( cons(x1, x2) ) = max{0, -2} |
POL( from(x1) ) = 2x1 + 2 |
POL( ACTIVE(x1) ) = x1 + 1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
length1(active(X)) → length1(X)
length1(mark(X)) → length1(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
ACTIVE(length1(X)) → MARK(length(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(X))
mark(nil) → active(nil)
mark(0) → active(0)
mark(length1(X)) → active(length1(X))
from(mark(X)) → from(X)
from(active(X)) → from(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(mark(X)) → length(X)
length(active(X)) → length(X)
length1(mark(X)) → length1(X)
length1(active(X)) → length1(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(39) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(40) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
ACTIVE(length1(X)) → MARK(length(X))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))
The TRS R consists of the following rules:
length1(active(X)) → length1(X)
length1(mark(X)) → length1(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(41) UsableRulesReductionPairsProof (EQUIVALENT transformation)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
ACTIVE(length(cons(X, Y))) → MARK(s(length1(Y)))
MARK(s(X)) → MARK(X)
The following rules are removed from R:
length1(active(X)) → length1(X)
length1(mark(X)) → length1(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
Used ordering: POLO with Polynomial interpretation [POLO]:
POL(ACTIVE(x1)) = 2·x1
POL(MARK(x1)) = 2·x1
POL(active(x1)) = x1
POL(cons(x1, x2)) = 1 + x1 + 2·x2
POL(length(x1)) = 2·x1
POL(length1(x1)) = 2·x1
POL(mark(x1)) = x1
POL(s(x1)) = 1 + x1
(42) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(length1(X)) → MARK(length(X))
MARK(length(X)) → ACTIVE(length(X))
MARK(length1(X)) → ACTIVE(length1(X))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(43) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.
(44) TRUE