(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)))
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(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)
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)

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)
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(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)
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)

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)))
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(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)
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)

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 9 SCCs with 28 less nodes.

(4) Complex Obligation (AND)

(5) 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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(active(X1), X2) → FIRST(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(FIRST(x1, x2)) = x1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FIRST(X1, mark(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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(X1, active(X2)) → FIRST(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(FIRST(x1, x2)) = x2   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

Q DP problem:
P is empty.
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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) 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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DBL(active(X)) → DBL(X)
DBL(mark(X)) → DBL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(DBL(x1)) = x1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE

(17) 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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ADD(mark(X1), X2) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ADD(x1, x2)) = x1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ADD(X1, mark(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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ADD(x1, x2)) = x2   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

Q DP problem:
P is empty.
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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(22) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(23) TRUE

(24) 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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(active(X)) → S(X)
S(mark(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(S(x1)) = x1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(26) Obligation:

Q DP problem:
P is empty.
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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(27) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(28) TRUE

(29) 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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SQR(active(X)) → SQR(X)
SQR(mark(X)) → SQR(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(SQR(x1)) = x1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(31) Obligation:

Q DP problem:
P is empty.
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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(32) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(33) TRUE

(34) 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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


RECIP(active(X)) → RECIP(X)
RECIP(mark(X)) → RECIP(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(RECIP(x1)) = x1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(36) Obligation:

Q DP problem:
P is empty.
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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(37) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(38) TRUE

(39) 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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(CONS(x1, x2)) = x1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(41) Obligation:

Q DP problem:
The TRS P consists of the following rules:

CONS(X1, mark(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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(CONS(x1, x2)) = x2   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

Q DP problem:
P is empty.
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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(44) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(45) TRUE

(46) 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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TERMS(active(X)) → TERMS(X)
TERMS(mark(X)) → TERMS(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(TERMS(x1)) = x1   
POL(active(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

Q DP problem:
P is empty.
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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(49) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(50) TRUE

(51) 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(X))
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
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)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(recip(X)) → ACTIVE(recip(mark(X)))
MARK(s(X)) → ACTIVE(s(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(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:

terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
recip(active(X)) → recip(X)
recip(mark(X)) → recip(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
dbl(active(X)) → dbl(X)
dbl(mark(X)) → dbl(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)
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)

(53) 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(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)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(add(0, X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(add(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1   
POL(dbl(x1)) = x1   
POL(first(x1, x2)) = x1 + x2   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(recip(x1)) = x1   
POL(s(x1)) = 0   
POL(sqr(x1)) = x1   
POL(terms(x1)) = x1   

The following usable rules [FROCOS05] were oriented:

mark(terms(X)) → active(terms(mark(X)))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(sqr(X)) → active(sqr(mark(X)))
active(add(0, X)) → mark(X)
mark(s(X)) → active(s(X))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
recip(active(X)) → recip(X)
recip(mark(X)) → recip(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
dbl(active(X)) → dbl(X)
dbl(mark(X)) → dbl(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)
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)
active(sqr(0)) → mark(0)
active(dbl(0)) → mark(0)
active(first(0, X)) → mark(nil)

(55) 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)))
MARK(sqr(X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
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)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(terms(X)) → MARK(X)
MARK(recip(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(add(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1   
POL(dbl(x1)) = x1   
POL(first(x1, x2)) = x1 + x2   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(recip(x1)) = 1 + x1   
POL(s(x1)) = 0   
POL(sqr(x1)) = x1   
POL(terms(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

mark(terms(X)) → active(terms(mark(X)))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(sqr(X)) → active(sqr(mark(X)))
active(add(0, X)) → mark(X)
mark(s(X)) → active(s(X))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
recip(active(X)) → recip(X)
recip(mark(X)) → recip(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
dbl(active(X)) → dbl(X)
dbl(mark(X)) → dbl(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)
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)
active(sqr(0)) → mark(0)
active(dbl(0)) → mark(0)
active(first(0, X)) → mark(nil)

(57) 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)))
MARK(sqr(X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
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)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(add(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = x1   
POL(dbl(x1)) = x1   
POL(first(x1, x2)) = x1 + x2   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(recip(x1)) = 0   
POL(s(x1)) = 0   
POL(sqr(x1)) = x1   
POL(terms(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

mark(terms(X)) → active(terms(mark(X)))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(sqr(X)) → active(sqr(mark(X)))
active(add(0, X)) → mark(X)
mark(s(X)) → active(s(X))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
recip(active(X)) → recip(X)
recip(mark(X)) → recip(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
dbl(active(X)) → dbl(X)
dbl(mark(X)) → dbl(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)
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)
active(sqr(0)) → mark(0)
active(dbl(0)) → mark(0)
active(first(0, X)) → mark(nil)

(59) 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)))
MARK(sqr(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(dbl(X)) → ACTIVE(dbl(mark(X)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(60) 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: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(add(x1, x2)) = x2   
POL(cons(x1, x2)) = x1   
POL(dbl(x1)) = x1   
POL(first(x1, x2)) = x1 + x2   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(recip(x1)) = 0   
POL(s(x1)) = 0   
POL(sqr(x1)) = x1   
POL(terms(x1)) = 1   

The following usable rules [FROCOS05] were oriented:

mark(terms(X)) → active(terms(mark(X)))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(sqr(X)) → active(sqr(mark(X)))
active(add(0, X)) → mark(X)
mark(s(X)) → active(s(X))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
recip(active(X)) → recip(X)
recip(mark(X)) → recip(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
dbl(active(X)) → dbl(X)
dbl(mark(X)) → dbl(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)
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)
active(sqr(0)) → mark(0)
active(dbl(0)) → mark(0)
active(first(0, X)) → mark(nil)

(61) 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)))
MARK(sqr(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(dbl(X)) → ACTIVE(dbl(mark(X)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(dbl(s(X))) → MARK(s(s(dbl(X))))
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(add(x1, x2)) = x2   
POL(cons(x1, x2)) = x1   
POL(dbl(x1)) = 1 + x1   
POL(first(x1, x2)) = 1 + x1 + x2   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(recip(x1)) = 0   
POL(s(x1)) = 0   
POL(sqr(x1)) = x1   
POL(terms(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

mark(terms(X)) → active(terms(mark(X)))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(recip(X)) → active(recip(mark(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(sqr(X)) → active(sqr(mark(X)))
active(add(0, X)) → mark(X)
mark(s(X)) → active(s(X))
active(add(s(X), Y)) → mark(s(add(X, Y)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
mark(dbl(X)) → active(dbl(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
dbl(active(X)) → dbl(X)
dbl(mark(X)) → dbl(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)
s(active(X)) → s(X)
s(mark(X)) → s(X)
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)
cons(X1, mark(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
active(sqr(0)) → mark(0)
active(dbl(0)) → mark(0)
active(first(0, X)) → mark(nil)
recip(active(X)) → recip(X)
recip(mark(X)) → recip(X)

(63) 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)
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
MARK(sqr(X)) → MARK(X)
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)))

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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(64) 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)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
MARK(sqr(X)) → MARK(X)
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)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = 0   
POL(MARK(x1)) = x1   
POL(active(x1)) = 0   
POL(add(x1, x2)) = 1   
POL(cons(x1, x2)) = 1 + x1   
POL(dbl(x1)) = 1   
POL(first(x1, x2)) = 1   
POL(mark(x1)) = 0   
POL(nil) = 0   
POL(recip(x1)) = 0   
POL(s(x1)) = 0   
POL(sqr(x1)) = 1 + x1   
POL(terms(x1)) = 1   

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)

(65) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(sqr(s(X))) → MARK(s(add(sqr(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)))
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(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)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(66) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(67) TRUE