(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(active(x1)) = x1
POL(and(x1, x2)) = x1 + x2
POL(cons(x1, x2)) = x1 + x2
POL(length(x1)) = 1 + x1
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
POL(tt) = 1
POL(zeros) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(active(x1)) = x1
POL(and(x1, x2)) = 2 + 2·x1 + x2
POL(cons(x1, x2)) = 2·x1 + 2·x2
POL(length(x1)) = 2·x1
POL(mark(x1)) = 2·x1
POL(nil) = 2
POL(s(x1)) = x1
POL(tt) = 0
POL(zeros) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(nil) → active(nil)
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
Q is empty.
(5) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(active(x1)) = x1
POL(and(x1, x2)) = 2·x1 + x2
POL(cons(x1, x2)) = 2·x1 + 2·x2
POL(length(x1)) = x1
POL(mark(x1)) = 2·x1
POL(s(x1)) = x1
POL(tt) = 1
POL(zeros) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
mark(tt) → active(tt)
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
Q is empty.
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
ACTIVE(length(cons(N, L))) → S(length(L))
ACTIVE(length(cons(N, L))) → LENGTH(L)
MARK(zeros) → ACTIVE(zeros)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(0) → ACTIVE(0)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → LENGTH(mark(X))
MARK(length(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(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)
AND(mark(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 7 less nodes.
(10) Complex Obligation (AND)
(11) 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(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) 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.
(13) 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.
(14) 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
(15) TRUE
(16) 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(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) 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.
(18) 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.
(19) 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
(20) TRUE
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AND(X1, mark(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(22) 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.
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
AND(X1, mark(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(24) 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:
- AND(X1, mark(X2)) → AND(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- AND(mark(X1), X2) → AND(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- AND(active(X1), X2) → AND(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- AND(X1, active(X2)) → AND(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
(25) TRUE
(26) 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(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(27) 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.
(28) 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.
(29) 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
(30) TRUE
(31) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(32) 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.
(33) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(34) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
MARK(length(X)) → MARK(X)
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(ACTIVE(x1)) = x1
POL(MARK(x1)) = x1
POL(active(x1)) = x1
POL(cons(x1, x2)) = x1 + x2
POL(length(x1)) = 1 + x1
POL(mark(x1)) = x1
POL(s(x1)) = x1
POL(zeros) = 0
(35) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(36) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(zeros) → ACTIVE(zeros)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
(37) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(38) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(39) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(40) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
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: Matrix interpretation [MATRO]:
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
(41) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(42) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
(43) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(44) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
MARK(
length(
X)) →
ACTIVE(
length(
mark(
X))) at position [0] we obtained the following new rules [LPAR04]:
MARK(length(x0)) → ACTIVE(length(x0))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(0)) → ACTIVE(length(active(0)))
(45) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(0)) → ACTIVE(length(active(0)))
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(46) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
MARK(
length(
0)) →
ACTIVE(
length(
active(
0))) at position [0] we obtained the following new rules [LPAR04]:
MARK(length(0)) → ACTIVE(length(0))
(47) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(0)) → ACTIVE(length(0))
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(48) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(49) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(50) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(ACTIVE(x1)) = 1 + x1
POL(MARK(x1)) = 2·x1
POL(active(x1)) = x1
POL(cons(x1, x2)) = x1 + 2·x2
POL(length(x1)) = 1 + x1
POL(mark(x1)) = x1
POL(s(x1)) = x1
POL(zeros) = 0
(51) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(52) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
(53) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(54) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(MARK(x1)) = | 1A | + | 2A | · | x1 |
POL(length(x1)) = | -I | + | 0A | · | x1 |
POL(ACTIVE(x1)) = | 1A | + | 0A | · | x1 |
POL(cons(x1, x2)) = | 1A | + | -I | · | x1 | + | 2A | · | x2 |
POL(active(x1)) = | 2A | + | 0A | · | x1 |
POL(mark(x1)) = | 2A | + | 0A | · | x1 |
The following usable rules [FROCOS05] were oriented:
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
(55) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(56) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(length(x0)) → ACTIVE(length(x0))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(MARK(x1)) = | -I | + | 1A | · | x1 |
POL(length(x1)) = | -I | + | 0A | · | x1 |
POL(ACTIVE(x1)) = | -I | + | 0A | · | x1 |
POL(cons(x1, x2)) = | -I | + | 0A | · | x1 | + | 1A | · | x2 |
POL(active(x1)) = | 1A | + | 0A | · | x1 |
POL(mark(x1)) = | 1A | + | 0A | · | x1 |
The following usable rules [FROCOS05] were oriented:
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
(57) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
The TRS R consists of the following rules:
mark(s(X)) → active(s(mark(X)))
mark(length(X)) → active(length(mark(X)))
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(58) NonTerminationProof (EQUIVALENT transformation)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
s =
MARK(
length(
zeros)) evaluates to t =
MARK(
length(
zeros))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceMARK(length(zeros)) →
ACTIVE(
length(
active(
zeros)))
with rule
MARK(
length(
zeros)) →
ACTIVE(
length(
active(
zeros))) at position [] and matcher [ ]
ACTIVE(length(active(zeros))) →
ACTIVE(
length(
mark(
cons(
0,
zeros))))
with rule
active(
zeros) →
mark(
cons(
0,
zeros)) at position [0,0] and matcher [ ]
ACTIVE(length(mark(cons(0, zeros)))) →
ACTIVE(
length(
cons(
0,
zeros)))
with rule
length(
mark(
X)) →
length(
X) at position [0] and matcher [
X /
cons(
0,
zeros)]
ACTIVE(length(cons(0, zeros))) →
MARK(
s(
length(
zeros)))
with rule
ACTIVE(
length(
cons(
N,
L))) →
MARK(
s(
length(
L))) at position [] and matcher [
N /
0,
L /
zeros]
MARK(s(length(zeros))) →
MARK(
length(
zeros))
with rule
MARK(
s(
X)) →
MARK(
X)
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
(59) FALSE