(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(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(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
ACTIVE(terms(N)) → CONS(recip(sqr(N)), terms(s(N)))
ACTIVE(terms(N)) → RECIP(sqr(N))
ACTIVE(terms(N)) → SQR(N)
ACTIVE(terms(N)) → TERMS(s(N))
ACTIVE(terms(N)) → S(N)
ACTIVE(sqr(0)) → MARK(0)
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
ACTIVE(sqr(s(X))) → S(add(sqr(X), dbl(X)))
ACTIVE(sqr(s(X))) → ADD(sqr(X), dbl(X))
ACTIVE(sqr(s(X))) → SQR(X)
ACTIVE(sqr(s(X))) → DBL(X)
ACTIVE(dbl(0)) → MARK(0)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
ACTIVE(dbl(s(X))) → S(s(dbl(X)))
ACTIVE(dbl(s(X))) → S(dbl(X))
ACTIVE(dbl(s(X))) → DBL(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(first(0, X)) → MARK(nil)
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
ACTIVE(first(s(X), cons(Y, Z))) → CONS(Y, first(X, Z))
ACTIVE(first(s(X), cons(Y, Z))) → FIRST(X, Z)
ACTIVE(half(0)) → MARK(0)
ACTIVE(half(s(0))) → MARK(0)
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
ACTIVE(half(s(s(X)))) → S(half(X))
ACTIVE(half(s(s(X)))) → HALF(X)
ACTIVE(half(dbl(X))) → MARK(X)
MARK(terms(X)) → ACTIVE(terms(mark(X)))
MARK(terms(X)) → TERMS(mark(X))
MARK(terms(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(recip(X)) → ACTIVE(recip(mark(X)))
MARK(recip(X)) → RECIP(mark(X))
MARK(recip(X)) → MARK(X)
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
MARK(sqr(X)) → SQR(mark(X))
MARK(sqr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(0) → ACTIVE(0)
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(dbl(X)) → ACTIVE(dbl(mark(X)))
MARK(dbl(X)) → DBL(mark(X))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(nil) → ACTIVE(nil)
MARK(half(X)) → ACTIVE(half(mark(X)))
MARK(half(X)) → HALF(mark(X))
MARK(half(X)) → MARK(X)
TERMS(mark(X)) → TERMS(X)
TERMS(active(X)) → TERMS(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)
RECIP(mark(X)) → RECIP(X)
RECIP(active(X)) → RECIP(X)
SQR(mark(X)) → SQR(X)
SQR(active(X)) → SQR(X)
S(mark(X)) → S(X)
S(active(X)) → S(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)
DBL(mark(X)) → DBL(X)
DBL(active(X)) → DBL(X)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(active(X1), X2) → FIRST(X1, X2)
FIRST(X1, active(X2)) → FIRST(X1, X2)
HALF(mark(X)) → HALF(X)
HALF(active(X)) → HALF(X)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(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 10 SCCs with 34 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HALF(active(X)) → HALF(X)
HALF(mark(X)) → HALF(X)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(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:
HALF(active(X)) → HALF(X)
HALF(mark(X)) → HALF(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:
- HALF(active(X)) → HALF(X)
The graph contains the following edges 1 > 1
- HALF(mark(X)) → HALF(X)
The graph contains the following edges 1 > 1
(9) TRUE
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(active(X1), X2) → FIRST(X1, X2)
FIRST(X1, active(X2)) → FIRST(X1, X2)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(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:
FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(active(X1), X2) → FIRST(X1, X2)
FIRST(X1, active(X2)) → FIRST(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:
- FIRST(X1, mark(X2)) → FIRST(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- FIRST(mark(X1), X2) → FIRST(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- FIRST(active(X1), X2) → FIRST(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- FIRST(X1, active(X2)) → FIRST(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:
DBL(active(X)) → DBL(X)
DBL(mark(X)) → DBL(X)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(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:
DBL(active(X)) → DBL(X)
DBL(mark(X)) → DBL(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:
- DBL(active(X)) → DBL(X)
The graph contains the following edges 1 > 1
- DBL(mark(X)) → DBL(X)
The graph contains the following edges 1 > 1
(19) TRUE
(20) 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(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(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:
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.
(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:
- 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
(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(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(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:
SQR(active(X)) → SQR(X)
SQR(mark(X)) → SQR(X)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(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:
SQR(active(X)) → SQR(X)
SQR(mark(X)) → SQR(X)
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:
- SQR(active(X)) → SQR(X)
The graph contains the following edges 1 > 1
- SQR(mark(X)) → SQR(X)
The graph contains the following edges 1 > 1
(34) TRUE
(35) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RECIP(active(X)) → RECIP(X)
RECIP(mark(X)) → RECIP(X)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(36) 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.
(37) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RECIP(active(X)) → RECIP(X)
RECIP(mark(X)) → RECIP(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(38) 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:
- RECIP(active(X)) → RECIP(X)
The graph contains the following edges 1 > 1
- RECIP(mark(X)) → RECIP(X)
The graph contains the following edges 1 > 1
(39) TRUE
(40) 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(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(41) 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.
(42) 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.
(43) 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
(44) TRUE
(45) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TERMS(active(X)) → TERMS(X)
TERMS(mark(X)) → TERMS(X)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(46) 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.
(47) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TERMS(active(X)) → TERMS(X)
TERMS(mark(X)) → TERMS(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(48) 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:
- TERMS(active(X)) → TERMS(X)
The graph contains the following edges 1 > 1
- TERMS(mark(X)) → TERMS(X)
The graph contains the following edges 1 > 1
(49) TRUE
(50) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
MARK(terms(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(recip(X)) → ACTIVE(recip(mark(X)))
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(recip(X)) → MARK(X)
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
ACTIVE(half(dbl(X))) → MARK(X)
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(half(X)) → ACTIVE(half(mark(X)))
MARK(half(X)) → MARK(X)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(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.
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(recip(X)) → ACTIVE(recip(mark(X)))
MARK(s(X)) → ACTIVE(s(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(dbl(x1)) = 1
POL(first(x1, x2)) = 1
POL(half(x1)) = 1
POL(mark(x1)) = 0
POL(nil) = 0
POL(recip(x1)) = 0
POL(s(x1)) = 0
POL(sqr(x1)) = 1
POL(terms(x1)) = 1
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
(52) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
MARK(terms(X)) → MARK(X)
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(recip(X)) → MARK(X)
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
ACTIVE(half(dbl(X))) → MARK(X)
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(half(X)) → ACTIVE(half(mark(X)))
MARK(half(X)) → MARK(X)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(53) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(first(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(MARK(x1)) = | -I | + | 0A | · | x1 |
POL(terms(x1)) = | 0A | + | 0A | · | x1 |
POL(ACTIVE(x1)) = | -I | + | 0A | · | x1 |
POL(mark(x1)) = | 0A | + | 0A | · | x1 |
POL(cons(x1, x2)) = | -I | + | 0A | · | x1 | + | -I | · | x2 |
POL(recip(x1)) = | 0A | + | 0A | · | x1 |
POL(sqr(x1)) = | 0A | + | 0A | · | x1 |
POL(add(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(dbl(x1)) = | 0A | + | 0A | · | x1 |
POL(first(x1, x2)) = | 1A | + | 1A | · | x1 | + | 0A | · | x2 |
POL(half(x1)) = | 0A | + | 0A | · | x1 |
POL(active(x1)) = | -I | + | 0A | · | x1 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
mark(nil) → active(nil)
mark(0) → active(0)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
active(sqr(0)) → mark(0)
active(dbl(0)) → mark(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
(54) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
MARK(terms(X)) → MARK(X)
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(recip(X)) → MARK(X)
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
ACTIVE(half(dbl(X))) → MARK(X)
MARK(first(X1, X2)) → MARK(X2)
MARK(half(X)) → ACTIVE(half(mark(X)))
MARK(half(X)) → MARK(X)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(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(terms(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(MARK(x1)) = | 0A | + | 0A | · | x1 |
POL(terms(x1)) = | 1A | + | 1A | · | x1 |
POL(ACTIVE(x1)) = | 0A | + | 0A | · | x1 |
POL(mark(x1)) = | -I | + | 0A | · | x1 |
POL(cons(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(recip(x1)) = | 0A | + | 0A | · | x1 |
POL(sqr(x1)) = | 1A | + | 0A | · | x1 |
POL(add(x1, x2)) = | -I | + | 0A | · | x1 | + | 0A | · | x2 |
POL(dbl(x1)) = | 0A | + | 0A | · | x1 |
POL(first(x1, x2)) = | -I | + | 0A | · | x1 | + | 0A | · | x2 |
POL(half(x1)) = | -I | + | 0A | · | x1 |
POL(active(x1)) = | -I | + | 0A | · | x1 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
mark(nil) → active(nil)
mark(0) → active(0)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
active(sqr(0)) → mark(0)
active(dbl(0)) → mark(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
(56) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(recip(X)) → MARK(X)
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
ACTIVE(half(dbl(X))) → MARK(X)
MARK(first(X1, X2)) → MARK(X2)
MARK(half(X)) → ACTIVE(half(mark(X)))
MARK(half(X)) → MARK(X)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(57) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(recip(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(MARK(x1)) = | -I | + | 0A | · | x1 |
POL(terms(x1)) = | -I | + | 4A | · | x1 |
POL(ACTIVE(x1)) = | -I | + | 0A | · | x1 |
POL(mark(x1)) = | -I | + | 0A | · | x1 |
POL(cons(x1, x2)) = | -I | + | 0A | · | x1 | + | -I | · | x2 |
POL(recip(x1)) = | -I | + | 4A | · | x1 |
POL(sqr(x1)) = | -I | + | 0A | · | x1 |
POL(add(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(dbl(x1)) = | 0A | + | 0A | · | x1 |
POL(first(x1, x2)) = | -I | + | 0A | · | x1 | + | 0A | · | x2 |
POL(half(x1)) = | 0A | + | 0A | · | x1 |
POL(active(x1)) = | -I | + | 0A | · | x1 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
mark(nil) → active(nil)
mark(0) → active(0)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
active(sqr(0)) → mark(0)
active(dbl(0)) → mark(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
(58) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
ACTIVE(half(dbl(X))) → MARK(X)
MARK(first(X1, X2)) → MARK(X2)
MARK(half(X)) → ACTIVE(half(mark(X)))
MARK(half(X)) → MARK(X)
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(59) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
ACTIVE(half(dbl(X))) → MARK(X)
MARK(half(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(MARK(x1)) = | -I | + | 0A | · | x1 |
POL(terms(x1)) = | 0A | + | 0A | · | x1 |
POL(ACTIVE(x1)) = | -I | + | 0A | · | x1 |
POL(mark(x1)) = | 0A | + | 0A | · | x1 |
POL(cons(x1, x2)) = | -I | + | 0A | · | x1 | + | 0A | · | x2 |
POL(recip(x1)) = | 0A | + | 0A | · | x1 |
POL(sqr(x1)) = | 0A | + | 0A | · | x1 |
POL(add(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(dbl(x1)) = | 0A | + | 0A | · | x1 |
POL(first(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(half(x1)) = | 1A | + | 1A | · | x1 |
POL(active(x1)) = | 0A | + | 0A | · | x1 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(sqr(0)) → mark(0)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
active(dbl(0)) → mark(0)
mark(nil) → active(nil)
mark(0) → active(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
(60) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
MARK(sqr(X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → MARK(X2)
MARK(half(X)) → ACTIVE(half(mark(X)))
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(61) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(sqr(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(MARK(x1)) = | -I | + | 0A | · | x1 |
POL(terms(x1)) = | 0A | + | -I | · | x1 |
POL(ACTIVE(x1)) = | -I | + | 0A | · | x1 |
POL(mark(x1)) = | -I | + | 0A | · | x1 |
POL(cons(x1, x2)) = | 0A | + | 0A | · | x1 | + | -I | · | x2 |
POL(recip(x1)) = | 0A | + | -I | · | x1 |
POL(sqr(x1)) = | -I | + | 1A | · | x1 |
POL(add(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(dbl(x1)) = | 1A | + | 0A | · | x1 |
POL(first(x1, x2)) = | -I | + | 0A | · | x1 | + | 0A | · | x2 |
POL(half(x1)) = | 0A | + | 0A | · | x1 |
POL(active(x1)) = | -I | + | 0A | · | x1 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(sqr(0)) → mark(0)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
active(dbl(0)) → mark(0)
mark(nil) → active(nil)
mark(0) → active(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
(62) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → MARK(X2)
MARK(half(X)) → ACTIVE(half(mark(X)))
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(63) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(first(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(MARK(x1)) = | -I | + | 0A | · | x1 |
POL(terms(x1)) = | 0A | + | 0A | · | x1 |
POL(ACTIVE(x1)) = | 0A | + | 0A | · | x1 |
POL(mark(x1)) = | 0A | + | 0A | · | x1 |
POL(cons(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(recip(x1)) = | 0A | + | -I | · | x1 |
POL(sqr(x1)) = | 0A | + | 0A | · | x1 |
POL(add(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(dbl(x1)) = | 0A | + | 0A | · | x1 |
POL(first(x1, x2)) = | 5A | + | -I | · | x1 | + | 1A | · | x2 |
POL(half(x1)) = | 1A | + | 1A | · | x1 |
POL(active(x1)) = | -I | + | 0A | · | x1 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(sqr(0)) → mark(0)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
active(dbl(0)) → mark(0)
mark(nil) → active(nil)
mark(0) → active(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
(64) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(half(X)) → ACTIVE(half(mark(X)))
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(65) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(dbl(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(MARK(x1)) = | -I | + | 0A | · | x1 |
POL(terms(x1)) = | 0A | + | 0A | · | x1 |
POL(ACTIVE(x1)) = | -I | + | 0A | · | x1 |
POL(mark(x1)) = | -I | + | 0A | · | x1 |
POL(cons(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(recip(x1)) = | 0A | + | -I | · | x1 |
POL(sqr(x1)) = | 1A | + | 1A | · | x1 |
POL(add(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(dbl(x1)) = | 0A | + | 1A | · | x1 |
POL(first(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(half(x1)) = | -I | + | 0A | · | x1 |
POL(active(x1)) = | -I | + | 0A | · | x1 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(sqr(0)) → mark(0)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
active(dbl(0)) → mark(0)
mark(nil) → active(nil)
mark(0) → active(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
(66) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(half(X)) → ACTIVE(half(mark(X)))
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(67) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(add(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(first(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(sqr(0)) → mark(0)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
active(dbl(0)) → mark(0)
mark(nil) → active(nil)
mark(0) → active(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
(68) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(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(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(half(X)) → ACTIVE(half(mark(X)))
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(69) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(first(X1, X2)) → ACTIVE(first(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)) = 1
POL(cons(x1, x2)) = 0
POL(dbl(x1)) = 1
POL(first(x1, x2)) = 0
POL(half(x1)) = 1
POL(mark(x1)) = 0
POL(nil) = 0
POL(recip(x1)) = 0
POL(s(x1)) = 0
POL(sqr(x1)) = 1
POL(terms(x1)) = 1
The following usable rules [FROCOS05] were oriented:
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)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
(70) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(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(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(half(X)) → ACTIVE(half(mark(X)))
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(71) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
ACTIVE(terms(N)) → MARK(cons(recip(sqr(N)), terms(s(N))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(add(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(first(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(sqr(0)) → mark(0)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
active(dbl(0)) → mark(0)
mark(nil) → active(nil)
mark(0) → active(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
(72) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(terms(X)) → ACTIVE(terms(mark(X)))
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(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(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(half(X)) → ACTIVE(half(mark(X)))
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(73) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(terms(X)) → ACTIVE(terms(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(dbl(x1)) = 1
POL(first(x1, x2)) = 0
POL(half(x1)) = 1
POL(mark(x1)) = 0
POL(nil) = 0
POL(recip(x1)) = 0
POL(s(x1)) = 0
POL(sqr(x1)) = 1
POL(terms(x1)) = 0
The following usable rules [FROCOS05] were oriented:
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)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
half(active(X)) → half(X)
half(mark(X)) → half(X)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
(74) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(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(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(half(X)) → ACTIVE(half(mark(X)))
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(75) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(add(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(cons(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(first(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(sqr(0)) → mark(0)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
active(dbl(0)) → mark(0)
mark(nil) → active(nil)
mark(0) → active(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
(76) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(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(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(half(X)) → ACTIVE(half(mark(X)))
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(77) 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)
MARK(add(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(ACTIVE(x1)) = | -I | + | 0A | · | x1 |
POL(sqr(x1)) = | 2A | + | 1A | · | x1 |
POL(MARK(x1)) = | -I | + | 0A | · | x1 |
POL(add(x1, x2)) = | 2A | + | 0A | · | x1 | + | 1A | · | x2 |
POL(dbl(x1)) = | -I | + | 0A | · | x1 |
POL(mark(x1)) = | -I | + | 0A | · | x1 |
POL(half(x1)) = | -I | + | 0A | · | x1 |
POL(active(x1)) = | -I | + | 0A | · | x1 |
POL(recip(x1)) = | -I | + | 0A | · | x1 |
POL(cons(x1, x2)) = | 0A | + | 0A | · | x1 | + | 0A | · | x2 |
POL(terms(x1)) = | 3A | + | 1A | · | x1 |
POL(first(x1, x2)) = | 0A | + | -I | · | x1 | + | 0A | · | x2 |
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(sqr(0)) → mark(0)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
active(dbl(0)) → mark(0)
mark(nil) → active(nil)
mark(0) → active(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
(78) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → MARK(X1)
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
MARK(half(X)) → ACTIVE(half(mark(X)))
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(79) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(s(X)) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
ACTIVE(half(s(s(X)))) → MARK(s(half(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(
x1) =
x1
sqr(
x1) =
sqr(
x1)
s(
x1) =
s(
x1)
MARK(
x1) =
x1
add(
x1,
x2) =
add(
x1,
x2)
dbl(
x1) =
dbl(
x1)
mark(
x1) =
x1
half(
x1) =
half(
x1)
active(
x1) =
x1
recip(
x1) =
recip
cons(
x1,
x2) =
x2
0 =
0
terms(
x1) =
terms
nil =
nil
first(
x1,
x2) =
x1
Recursive path order with status [RPO].
Quasi-Precedence:
sqr1 > add2 > s1
sqr1 > dbl1 > s1
sqr1 > dbl1 > [0, nil]
half1 > s1
half1 > [0, nil]
terms > s1
terms > recip
Status:
add2: [1,2]
half1: multiset
sqr1: multiset
dbl1: [1]
s1: multiset
recip: multiset
0: multiset
nil: multiset
terms: multiset
The following usable rules [FROCOS05] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(sqr(0)) → mark(0)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
active(dbl(0)) → mark(0)
mark(nil) → active(nil)
mark(0) → active(0)
half(active(X)) → half(X)
half(mark(X)) → half(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
active(add(0, X)) → mark(X)
mark(half(X)) → active(half(mark(X)))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(half(s(s(X)))) → mark(s(half(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(s(X)) → active(s(mark(X)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
active(first(0, X)) → mark(nil)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
active(half(s(0))) → mark(0)
active(half(0)) → mark(0)
(80) Obligation:
Q DP problem:
The TRS P consists of the following rules:
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(dbl(X)) → ACTIVE(dbl(mark(X)))
MARK(half(X)) → ACTIVE(half(mark(X)))
The TRS R consists of the following rules:
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
mark(terms(X)) → active(terms(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(recip(X)) → active(recip(mark(X)))
mark(sqr(X)) → active(sqr(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(0) → active(0)
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(half(X)) → active(half(mark(X)))
terms(mark(X)) → terms(X)
terms(active(X)) → terms(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)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)
sqr(mark(X)) → sqr(X)
sqr(active(X)) → sqr(X)
s(mark(X)) → s(X)
s(active(X)) → s(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)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
half(mark(X)) → half(X)
half(active(X)) → half(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(81) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 4 less nodes.
(82) TRUE