(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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(fst(0, Z)) → MARK(nil)
ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
ACTIVE(fst(s(X), cons(Y, Z))) → CONS(Y, fst(X, Z))
ACTIVE(fst(s(X), cons(Y, Z))) → FST(X, Z)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
ACTIVE(add(s(X), Y)) → S(add(X, Y))
ACTIVE(add(s(X), Y)) → ADD(X, Y)
ACTIVE(len(nil)) → MARK(0)
ACTIVE(len(cons(X, Z))) → MARK(s(len(Z)))
ACTIVE(len(cons(X, Z))) → S(len(Z))
ACTIVE(len(cons(X, Z))) → LEN(Z)
MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
MARK(fst(X1, X2)) → FST(mark(X1), mark(X2))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(0) → ACTIVE(0)
MARK(nil) → ACTIVE(nil)
MARK(s(X)) → ACTIVE(s(X))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → ADD(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → ACTIVE(len(mark(X)))
MARK(len(X)) → LEN(mark(X))
MARK(len(X)) → MARK(X)
FST(mark(X1), X2) → FST(X1, X2)
FST(X1, mark(X2)) → FST(X1, X2)
FST(active(X1), X2) → FST(X1, X2)
FST(X1, active(X2)) → FST(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(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)
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)
LEN(mark(X)) → LEN(X)
LEN(active(X)) → LEN(X)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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 18 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LEN(active(X)) → LEN(X)
LEN(mark(X)) → LEN(X)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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:
LEN(active(X)) → LEN(X)
LEN(mark(X)) → LEN(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:
- LEN(active(X)) → LEN(X)
The graph contains the following edges 1 > 1
- LEN(mark(X)) → LEN(X)
The graph contains the following edges 1 > 1
(9) TRUE
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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:
ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(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:
- ADD(X1, mark(X2)) → ADD(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- ADD(mark(X1), X2) → ADD(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- ADD(active(X1), X2) → ADD(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- ADD(X1, active(X2)) → ADD(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
(14) TRUE
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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:
FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(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:
- FROM(active(X)) → FROM(X)
The graph contains the following edges 1 > 1
- FROM(mark(X)) → FROM(X)
The graph contains the following edges 1 > 1
(19) TRUE
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- CONS(X1, mark(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- CONS(mark(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- CONS(active(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- CONS(X1, active(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
(24) TRUE
(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(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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) TRUE
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FST(X1, mark(X2)) → FST(X1, X2)
FST(mark(X1), X2) → FST(X1, X2)
FST(active(X1), X2) → FST(X1, X2)
FST(X1, active(X2)) → FST(X1, X2)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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:
FST(X1, mark(X2)) → FST(X1, X2)
FST(mark(X1), X2) → FST(X1, X2)
FST(active(X1), X2) → FST(X1, X2)
FST(X1, active(X2)) → FST(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:
- FST(X1, mark(X2)) → FST(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- FST(mark(X1), X2) → FST(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- FST(active(X1), X2) → FST(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- FST(X1, active(X2)) → FST(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
(34) TRUE
(35) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(s(X)) → ACTIVE(s(X))
ACTIVE(add(0, X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(len(cons(X, Z))) → MARK(s(len(Z)))
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → ACTIVE(len(mark(X)))
MARK(len(X)) → MARK(X)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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(s(X)) → ACTIVE(s(X))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(fst(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(add(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(X)
fst(X1, mark(X2)) → fst(X1, X2)
fst(mark(X1), X2) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
len(active(X)) → len(X)
len(mark(X)) → len(X)
from(active(X)) → from(X)
from(mark(X)) → from(X)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
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)
(37) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(len(cons(X, Z))) → MARK(s(len(Z)))
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → ACTIVE(len(mark(X)))
MARK(len(X)) → MARK(X)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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].
The following pairs can be oriented strictly and are deleted.
ACTIVE(len(cons(X, Z))) → MARK(s(len(Z)))
MARK(len(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(ACTIVE(x1)) = x1
POL(MARK(x1)) = x1
POL(active(x1)) = x1
POL(add(x1, x2)) = x1 + x2
POL(cons(x1, x2)) = x1
POL(from(x1)) = x1
POL(fst(x1, x2)) = x1 + x2
POL(len(x1)) = 1 + x1
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = 0
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(X)
fst(X1, mark(X2)) → fst(X1, X2)
fst(mark(X1), X2) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
mark(nil) → active(nil)
mark(0) → active(0)
active(len(nil)) → mark(0)
active(add(0, X)) → mark(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(from(X)) → active(from(mark(X)))
mark(len(X)) → active(len(mark(X)))
active(len(cons(X, Z))) → mark(s(len(Z)))
active(fst(0, Z)) → mark(nil)
len(active(X)) → len(X)
len(mark(X)) → len(X)
from(active(X)) → from(X)
from(mark(X)) → from(X)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
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)
(39) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → ACTIVE(len(mark(X)))
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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(len(X)) → ACTIVE(len(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(ACTIVE(x1)) = x1
POL(MARK(x1)) = 1
POL(active(x1)) = 0
POL(add(x1, x2)) = 1
POL(cons(x1, x2)) = 0
POL(from(x1)) = 1
POL(fst(x1, x2)) = 1
POL(len(x1)) = 0
POL(mark(x1)) = 0
POL(nil) = 0
POL(s(x1)) = 0
The following usable rules [FROCOS05] were oriented:
fst(X1, mark(X2)) → fst(X1, X2)
fst(mark(X1), X2) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
len(active(X)) → len(X)
len(mark(X)) → len(X)
from(active(X)) → from(X)
from(mark(X)) → from(X)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
(41) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(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.
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(ACTIVE(x1)) = x1
POL(MARK(x1)) = x1
POL(active(x1)) = x1
POL(add(x1, x2)) = 1 + x1 + x2
POL(cons(x1, x2)) = x1
POL(from(x1)) = x1
POL(fst(x1, x2)) = x1 + x2
POL(len(x1)) = 0
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = 0
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(X)
fst(X1, mark(X2)) → fst(X1, X2)
fst(mark(X1), X2) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
mark(nil) → active(nil)
mark(0) → active(0)
active(len(nil)) → mark(0)
active(add(0, X)) → mark(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(from(X)) → active(from(mark(X)))
mark(len(X)) → active(len(mark(X)))
active(len(cons(X, Z))) → mark(s(len(Z)))
active(fst(0, Z)) → mark(nil)
len(active(X)) → len(X)
len(mark(X)) → len(X)
from(active(X)) → from(X)
from(mark(X)) → from(X)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
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)
(43) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(44) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(ACTIVE(x1)) = 1 + x1
POL(MARK(x1)) = 1 + x1
POL(active(x1)) = x1
POL(add(x1, x2)) = x2
POL(cons(x1, x2)) = x1
POL(from(x1)) = 1 + x1
POL(fst(x1, x2)) = x1 + x2
POL(len(x1)) = 0
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(X)
fst(X1, mark(X2)) → fst(X1, X2)
fst(mark(X1), X2) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
mark(nil) → active(nil)
mark(0) → active(0)
active(len(nil)) → mark(0)
active(add(0, X)) → mark(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(from(X)) → active(from(mark(X)))
mark(len(X)) → active(len(mark(X)))
active(len(cons(X, Z))) → mark(s(len(Z)))
active(fst(0, Z)) → mark(nil)
len(active(X)) → len(X)
len(mark(X)) → len(X)
from(active(X)) → from(X)
from(mark(X)) → from(X)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
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)
(45) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(46) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(ACTIVE(x1)) = x1
POL(MARK(x1)) = 1
POL(active(x1)) = 0
POL(add(x1, x2)) = 0
POL(cons(x1, x2)) = 0
POL(from(x1)) = 0
POL(fst(x1, x2)) = 1
POL(len(x1)) = 0
POL(mark(x1)) = 0
POL(nil) = 0
POL(s(x1)) = 0
The following usable rules [FROCOS05] were oriented:
fst(X1, mark(X2)) → fst(X1, X2)
fst(mark(X1), X2) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
from(active(X)) → from(X)
from(mark(X)) → from(X)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
(47) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(48) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(fst(X1, X2)) → ACTIVE(fst(mark(X1), mark(X2)))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0) = 0
POL(ACTIVE(x1)) = x1
POL(MARK(x1)) = 1 + x1
POL(active(x1)) = x1
POL(add(x1, x2)) = 1 + x2
POL(cons(x1, x2)) = x1
POL(from(x1)) = x1
POL(fst(x1, x2)) = 1 + x1 + x2
POL(len(x1)) = 1
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(add(0, X)) → mark(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(from(X)) → active(from(mark(X)))
mark(len(X)) → active(len(mark(X)))
active(len(cons(X, Z))) → mark(s(len(Z)))
fst(X1, mark(X2)) → fst(X1, X2)
fst(mark(X1), X2) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
active(fst(0, Z)) → mark(nil)
len(active(X)) → len(X)
len(mark(X)) → len(X)
from(active(X)) → from(X)
from(mark(X)) → from(X)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
mark(nil) → active(nil)
mark(0) → active(0)
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)
active(len(nil)) → mark(0)
(49) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(fst(s(X), cons(Y, Z))) → MARK(cons(Y, fst(X, Z)))
MARK(cons(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(50) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(51) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons(X1, X2)) → MARK(X1)
The TRS R consists of the following rules:
active(fst(0, Z)) → mark(nil)
active(fst(s(X), cons(Y, Z))) → mark(cons(Y, fst(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(len(nil)) → mark(0)
active(len(cons(X, Z))) → mark(s(len(Z)))
mark(fst(X1, X2)) → active(fst(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(X))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(len(X)) → active(len(mark(X)))
fst(mark(X1), X2) → fst(X1, X2)
fst(X1, mark(X2)) → fst(X1, X2)
fst(active(X1), X2) → fst(X1, X2)
fst(X1, active(X2)) → fst(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)
from(mark(X)) → from(X)
from(active(X)) → from(X)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
len(mark(X)) → len(X)
len(active(X)) → len(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(52) 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.
(53) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(cons(X1, X2)) → MARK(X1)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(54) 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(cons(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1
(55) TRUE