(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(active(x1)) = x1
POL(cons(x1, x2)) = x1 + x2
POL(head(x1)) = 1 + x1
POL(incr(x1)) = x1
POL(mark(x1)) = x1
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x1)) = x1
POL(tail(x1)) = 1 + x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
(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:
ACTIVE(nats) → MARK(cons(0, incr(nats)))
ACTIVE(nats) → CONS(0, incr(nats))
ACTIVE(nats) → INCR(nats)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
ACTIVE(pairs) → CONS(0, incr(odds))
ACTIVE(pairs) → INCR(odds)
ACTIVE(odds) → MARK(incr(pairs))
ACTIVE(odds) → INCR(pairs)
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
ACTIVE(incr(cons(X, XS))) → CONS(s(X), incr(XS))
ACTIVE(incr(cons(X, XS))) → S(X)
ACTIVE(incr(cons(X, XS))) → INCR(XS)
MARK(nats) → ACTIVE(nats)
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(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
MARK(odds) → ACTIVE(odds)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(head(X)) → HEAD(mark(X))
MARK(head(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → TAIL(mark(X))
MARK(tail(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)
INCR(mark(X)) → INCR(X)
INCR(active(X)) → INCR(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
HEAD(mark(X)) → HEAD(X)
HEAD(active(X)) → HEAD(X)
TAIL(mark(X)) → TAIL(X)
TAIL(active(X)) → TAIL(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 14 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) 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:
- TAIL(active(X)) → TAIL(X)
The graph contains the following edges 1 > 1
- TAIL(mark(X)) → TAIL(X)
The graph contains the following edges 1 > 1
(11) TRUE
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HEAD(active(X)) → HEAD(X)
HEAD(mark(X)) → HEAD(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HEAD(active(X)) → HEAD(X)
HEAD(mark(X)) → HEAD(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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:
- HEAD(active(X)) → HEAD(X)
The graph contains the following edges 1 > 1
- HEAD(mark(X)) → HEAD(X)
The graph contains the following edges 1 > 1
(16) TRUE
(17) 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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) 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.
(19) 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.
(20) 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
(21) TRUE
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INCR(active(X)) → INCR(X)
INCR(mark(X)) → INCR(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) 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.
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INCR(active(X)) → INCR(X)
INCR(mark(X)) → INCR(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) 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:
- INCR(active(X)) → INCR(X)
The graph contains the following edges 1 > 1
- INCR(mark(X)) → INCR(X)
The graph contains the following edges 1 > 1
(26) TRUE
(27) 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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(28) 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.
(29) 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.
(30) 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
(31) TRUE
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(head(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → MARK(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(33) 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(head(X)) → MARK(X)
MARK(tail(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(head(x1)) = 1 + x1
POL(incr(x1)) = x1
POL(mark(x1)) = x1
POL(nats) = 0
POL(odds) = 0
POL(pairs) = 0
POL(s(x1)) = x1
POL(tail(x1)) = 1 + x1
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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].
The following pairs can be oriented strictly and are deleted.
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
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)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
(36) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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].
The following pairs can be oriented strictly and are deleted.
MARK(tail(X)) → ACTIVE(tail(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)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(odds) → active(odds)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(nats) → active(nats)
mark(pairs) → active(pairs)
mark(tail(X)) → active(tail(mark(X)))
active(odds) → mark(incr(pairs))
mark(0) → active(0)
(38) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(39) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(head(X)) → ACTIVE(head(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:
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
(40) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(41) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
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:
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
(42) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(43) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(nats) → ACTIVE(nats)
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)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(odds) → active(odds)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(nats) → active(nats)
mark(pairs) → active(pairs)
mark(tail(X)) → active(tail(mark(X)))
active(odds) → mark(incr(pairs))
mark(0) → active(0)
(44) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(45) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(46) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons(X1, X2)) → MARK(X1)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(47) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
ACTIVE(odds) → MARK(incr(pairs))
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)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(odds) → active(odds)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(nats) → active(nats)
mark(pairs) → active(pairs)
mark(tail(X)) → active(tail(mark(X)))
active(odds) → mark(incr(pairs))
mark(0) → active(0)
(48) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons(X1, X2)) → MARK(X1)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(odds) → ACTIVE(odds)
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(49) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(50) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(51) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
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)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(odds) → active(odds)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(nats) → active(nats)
mark(pairs) → active(pairs)
mark(tail(X)) → active(tail(mark(X)))
active(odds) → mark(incr(pairs))
mark(0) → active(0)
(52) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(53) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(54) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(55) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → 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)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
head(active(X)) → head(X)
head(mark(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(head(X)) → active(head(mark(X)))
mark(odds) → active(odds)
mark(incr(X)) → active(incr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(nats) → active(nats)
mark(pairs) → active(pairs)
mark(tail(X)) → active(tail(mark(X)))
active(odds) → mark(incr(pairs))
mark(0) → active(0)
(56) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(57) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(58) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(59) 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.
(60) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(61) 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:
- MARK(s(X)) → MARK(X)
The graph contains the following edges 1 > 1
- MARK(cons(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1
(62) TRUE