(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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(eq(0, 0)) → MARK(true)
ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
ACTIVE(eq(s(X), s(Y))) → EQ(X, Y)
ACTIVE(eq(X, Y)) → MARK(false)
ACTIVE(inf(X)) → MARK(cons(X, inf(s(X))))
ACTIVE(inf(X)) → CONS(X, inf(s(X)))
ACTIVE(inf(X)) → INF(s(X))
ACTIVE(inf(X)) → S(X)
ACTIVE(take(0, X)) → MARK(nil)
ACTIVE(take(s(X), cons(Y, L))) → MARK(cons(Y, take(X, L)))
ACTIVE(take(s(X), cons(Y, L))) → CONS(Y, take(X, L))
ACTIVE(take(s(X), cons(Y, L))) → TAKE(X, L)
ACTIVE(length(nil)) → MARK(0)
ACTIVE(length(cons(X, L))) → MARK(s(length(L)))
ACTIVE(length(cons(X, L))) → S(length(L))
ACTIVE(length(cons(X, L))) → LENGTH(L)
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
MARK(0) → ACTIVE(0)
MARK(true) → ACTIVE(true)
MARK(s(X)) → ACTIVE(s(X))
MARK(false) → ACTIVE(false)
MARK(inf(X)) → ACTIVE(inf(mark(X)))
MARK(inf(X)) → INF(mark(X))
MARK(inf(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → TAKE(mark(X1), mark(X2))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(nil) → ACTIVE(nil)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → LENGTH(mark(X))
MARK(length(X)) → MARK(X)
EQ(mark(X1), X2) → EQ(X1, X2)
EQ(X1, mark(X2)) → EQ(X1, X2)
EQ(active(X1), X2) → EQ(X1, X2)
EQ(X1, active(X2)) → EQ(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
INF(mark(X)) → INF(X)
INF(active(X)) → INF(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)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
The TRS R consists of the following rules:
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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 7 SCCs with 19 less nodes.
(4) Complex Obligation (AND)
(5) 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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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:
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.
(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:
- 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
(9) YES
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)
The TRS R consists of the following rules:
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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:
TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)
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:
- TAKE(X1, mark(X2)) → TAKE(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- TAKE(mark(X1), X2) → TAKE(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- TAKE(active(X1), X2) → TAKE(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- TAKE(X1, active(X2)) → TAKE(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
(14) YES
(15) 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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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:
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.
(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:
- 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
(19) YES
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
INF(active(X)) → INF(X)
INF(mark(X)) → INF(X)
The TRS R consists of the following rules:
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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:
INF(active(X)) → INF(X)
INF(mark(X)) → INF(X)
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:
- INF(active(X)) → INF(X)
The graph contains the following edges 1 > 1
- INF(mark(X)) → INF(X)
The graph contains the following edges 1 > 1
(24) YES
(25) 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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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:
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.
(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:
- 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
(29) YES
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
EQ(X1, mark(X2)) → EQ(X1, X2)
EQ(mark(X1), X2) → EQ(X1, X2)
EQ(active(X1), X2) → EQ(X1, X2)
EQ(X1, active(X2)) → EQ(X1, X2)
The TRS R consists of the following rules:
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(31) 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.
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
EQ(X1, mark(X2)) → EQ(X1, X2)
EQ(mark(X1), X2) → EQ(X1, X2)
EQ(active(X1), X2) → EQ(X1, X2)
EQ(X1, active(X2)) → EQ(X1, X2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(33) 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:
- EQ(X1, mark(X2)) → EQ(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- EQ(mark(X1), X2) → EQ(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- EQ(active(X1), X2) → EQ(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- EQ(X1, active(X2)) → EQ(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
(34) YES
(35) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
ACTIVE(inf(X)) → MARK(cons(X, inf(s(X))))
MARK(s(X)) → ACTIVE(s(X))
ACTIVE(take(s(X), cons(Y, L))) → MARK(cons(Y, take(X, L)))
MARK(inf(X)) → ACTIVE(inf(mark(X)))
ACTIVE(length(cons(X, L))) → MARK(s(length(L)))
MARK(inf(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)
The TRS R consists of the following rules:
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(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,JAR06].
The following pairs can be oriented strictly and are deleted.
ACTIVE(take(s(X), cons(Y, L))) → MARK(cons(Y, take(X, L)))
ACTIVE(length(cons(X, L))) → MARK(s(length(L)))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(length(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(
x1) =
x1
eq(
x1,
x2) =
eq
MARK(
x1) =
x1
inf(
x1) =
x1
cons(
x1,
x2) =
cons
s(
x1) =
s
take(
x1,
x2) =
take(
x1,
x2)
mark(
x1) =
x1
length(
x1) =
length(
x1)
active(
x1) =
x1
0 =
0
true =
true
false =
false
nil =
nil
Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
s=5
true=5
eq=7
length_1=4
take_2=3
0=8
cons=3
false=5
nil=8
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
eq(X1, mark(X2)) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
inf(active(X)) → inf(X)
inf(mark(X)) → inf(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)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
mark(eq(X1, X2)) → active(eq(X1, X2))
active(inf(X)) → mark(cons(X, inf(s(X))))
mark(s(X)) → active(s(X))
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
mark(inf(X)) → active(inf(mark(X)))
active(length(cons(X, L))) → mark(s(length(L)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(length(X)) → active(length(mark(X)))
mark(0) → active(0)
mark(true) → active(true)
mark(false) → active(false)
mark(nil) → active(nil)
length(active(X)) → length(X)
length(mark(X)) → length(X)
active(eq(0, 0)) → mark(true)
active(eq(X, Y)) → mark(false)
active(take(0, X)) → mark(nil)
active(length(nil)) → mark(0)
(37) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
ACTIVE(inf(X)) → MARK(cons(X, inf(s(X))))
MARK(s(X)) → ACTIVE(s(X))
MARK(inf(X)) → ACTIVE(inf(mark(X)))
MARK(inf(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(length(X)) → ACTIVE(length(mark(X)))
The TRS R consists of the following rules:
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(38) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
ACTIVE(inf(X)) → MARK(cons(X, inf(s(X))))
MARK(inf(X)) → ACTIVE(inf(mark(X)))
MARK(inf(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( eq(x1, x2) ) = x1 + x2 |
POL( mark(x1) ) = 2x1 + 1 |
POL( active(x1) ) = x1 + 1 |
POL( cons(x1, x2) ) = max{0, -2} |
POL( take(x1, x2) ) = x2 + 1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
eq(X1, mark(X2)) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
inf(active(X)) → inf(X)
inf(mark(X)) → inf(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)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
mark(eq(X1, X2)) → active(eq(X1, X2))
active(inf(X)) → mark(cons(X, inf(s(X))))
mark(s(X)) → active(s(X))
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
mark(inf(X)) → active(inf(mark(X)))
active(length(cons(X, L))) → mark(s(length(L)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(length(X)) → active(length(mark(X)))
mark(0) → active(0)
mark(true) → active(true)
mark(false) → active(false)
mark(nil) → active(nil)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
active(eq(0, 0)) → mark(true)
active(eq(X, Y)) → mark(false)
active(take(0, X)) → mark(nil)
active(length(nil)) → mark(0)
(39) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
MARK(s(X)) → ACTIVE(s(X))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(length(X)) → ACTIVE(length(mark(X)))
The TRS R consists of the following rules:
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(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,JAR06].
The following pairs can be oriented strictly and are deleted.
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( ACTIVE(x1) ) = max{0, x1 - 2} |
POL( MARK(x1) ) = max{0, x1 - 1} |
POL( take(x1, x2) ) = 2x2 + 1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
eq(X1, mark(X2)) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
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)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
mark(eq(X1, X2)) → active(eq(X1, X2))
active(inf(X)) → mark(cons(X, inf(s(X))))
mark(s(X)) → active(s(X))
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
mark(inf(X)) → active(inf(mark(X)))
active(length(cons(X, L))) → mark(s(length(L)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(length(X)) → active(length(mark(X)))
mark(0) → active(0)
mark(true) → active(true)
mark(false) → active(false)
mark(nil) → active(nil)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
active(eq(0, 0)) → mark(true)
active(eq(X, Y)) → mark(false)
active(take(0, X)) → mark(nil)
active(length(nil)) → mark(0)
inf(active(X)) → inf(X)
inf(mark(X)) → inf(X)
(41) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
MARK(s(X)) → ACTIVE(s(X))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(length(X)) → ACTIVE(length(mark(X)))
The TRS R consists of the following rules:
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(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,JAR06].
The following pairs can be oriented strictly and are deleted.
ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(
x1) =
x1
eq(
x1,
x2) =
x2
s(
x1) =
s(
x1)
MARK(
x1) =
x1
take(
x1,
x2) =
take
length(
x1) =
length
mark(
x1) =
x1
active(
x1) =
x1
Knuth-Bendix order [KBO] with precedence:
trivial
and weight map:
s_1=1
length=1
take=2
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
eq(X1, mark(X2)) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(active(X)) → length(X)
length(mark(X)) → length(X)
(43) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
MARK(s(X)) → ACTIVE(s(X))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(length(X)) → ACTIVE(length(mark(X)))
The TRS R consists of the following rules:
active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(44) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 4 less nodes.
(45) TRUE