Termination w.r.t. Q proof of /tmp/tmpvpodmR/Ex26_Luc03b

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

  ↳6 QDP

  ↳7 QDP

  ↳8 QDP

  ↳9 QDP

  ↳10 QDP

  ↳11 QDP

  ↳12 QDP

  ↳13 QDP

    ↳14 QDPOrderProof (⇔)

    ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

(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) 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.

(7) 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.

(8) 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.

(9) 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.

(10) 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.

(11) 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.

(12) 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.

(13) 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.

(14) 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(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
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)) → 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: Combined order from the following AFS and order.
MARK(x1) = x1
terms(x1) = x1
ACTIVE(x1) = x1
mark(x1) = x1
cons(x1, x2) = x1
recip(x1) = x1
sqr(x1) = x1
s(x1) = s
add(x1, x2) = add(x1, x2)
dbl(x1) = dbl(x1)
0 = 0
first(x1, x2) = first(x1, x2)
active(x1) = x1
nil = nil

Lexicographic path order with status [LPO].
Precedence:

s > add₂
s > dbl₁ > 0
s > first₂ > nil

Status:

s: []
add₂: [2,1]
dbl₁: [1]
0: []
first₂: [2,1]
nil: []

The following usable rules [FROCOS05] were oriented:

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)

(15) 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)))
MARK(recip(X)) → MARK(X)
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
MARK(sqr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(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.

(16) 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))))
MARK(terms(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = x1
terms(x1) = terms(x1)
ACTIVE(x1) = x1
mark(x1) = x1
cons(x1, x2) = cons(x1)
recip(x1) = x1
sqr(x1) = x1
s(x1) = s
add(x1, x2) = add(x1, x2)
dbl(x1) = dbl(x1)
first(x1, x2) = first(x1, x2)
active(x1) = x1
0 = 0
nil = nil

Lexicographic path order with status [LPO].
Precedence:

terms₁ > cons₁
s > add₂
s > dbl₁ > 0
s > first₂ > cons₁
s > first₂ > nil

Status:

terms₁: [1]
cons₁: [1]
s: []
add₂: [2,1]
dbl₁: [1]
first₂: [2,1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

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)

(17) Obligation:

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

MARK(terms(X)) → ACTIVE(terms(mark(X)))
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)))
MARK(recip(X)) → MARK(X)
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
MARK(sqr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(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.

(18) 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)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(recip(X)) → ACTIVE(recip(mark(X)))
MARK(recip(X)) → MARK(X)
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
MARK(sqr(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(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: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
terms(x1) = terms(x1)
ACTIVE(x1) = x1
mark(x1) = x1
cons(x1, x2) = x2
sqr(x1) = sqr(x1)
s(x1) = x1
add(x1, x2) = add(x2)
dbl(x1) = dbl(x1)
recip(x1) = recip(x1)
first(x1, x2) = x1
active(x1) = x1
0 = 0
nil = nil

Lexicographic path order with status [LPO].
Precedence:

sqr₁ > MARK₁ > terms₁
sqr₁ > add₁ > terms₁
sqr₁ > dbl₁ > terms₁
sqr₁ > 0 > nil > terms₁
recip₁ > MARK₁ > terms₁

Status:

MARK₁: [1]
terms₁: [1]
sqr₁: [1]
add₁: [1]
dbl₁: [1]
recip₁: [1]
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

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)

(19) 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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) Obligation:

(7) Obligation:

(8) Obligation:

(9) Obligation:

(10) Obligation:

(11) Obligation:

(12) Obligation:

(13) Obligation:

(14) QDPOrderProof (EQUIVALENT transformation)

(15) Obligation:

(16) QDPOrderProof (EQUIVALENT transformation)

(17) Obligation:

(18) QDPOrderProof (EQUIVALENT transformation)

(19) Obligation:

(20) PisEmptyProof (EQUIVALENT transformation)

(21) TRUE