Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 UsableRulesProof (⇔)
    ↳12 QDP
    ↳13 QDPSizeChangeProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 UsableRulesProof (⇔)
    ↳17 QDP
    ↳18 QDPSizeChangeProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 UsableRulesProof (⇔)
    ↳22 QDP
    ↳23 QDPSizeChangeProof (⇔)
    ↳24 TRUE
  ↳25 QDP
    ↳26 UsableRulesProof (⇔)
    ↳27 QDP
    ↳28 QDPSizeChangeProof (⇔)
    ↳29 TRUE
  ↳30 QDP
    ↳31 UsableRulesProof (⇔)
    ↳32 QDP
    ↳33 QDPSizeChangeProof (⇔)
    ↳34 TRUE
  ↳35 QDP
    ↳36 UsableRulesProof (⇔)
    ↳37 QDP
    ↳38 QDPSizeChangeProof (⇔)
    ↳39 TRUE
  ↳40 QDP
    ↳41 UsableRulesProof (⇔)
    ↳42 QDP
    ↳43 QDPSizeChangeProof (⇔)
    ↳44 TRUE
  ↳45 QDP
    ↳46 QDPOrderProof (⇔)
    ↳47 QDP
    ↳48 QDPOrderProof (⇔)
    ↳49 QDP
    ↳50 QDPOrderProof (⇔)
    ↳51 QDP
    ↳52 QDPOrderProof (⇔)
    ↳53 QDP
    ↳54 QDPOrderProof (⇔)
    ↳55 QDP
    ↳56 QDPOrderProof (⇔)
    ↳57 QDP
    ↳58 QDPOrderProof (⇔)
    ↳59 QDP
    ↳60 QDPOrderProof (⇔)
    ↳61 QDP
    ↳62 QDPOrderProof (⇔)
    ↳63 QDP
    ↳64 QDPOrderProof (⇔)
    ↳65 QDP
    ↳66 QDPOrderProof (⇔)
    ↳67 QDP
    ↳68 DependencyGraphProof (⇔)
    ↳69 QDP
    ↳70 UsableRulesProof (⇔)
    ↳71 QDP
    ↳72 QDPSizeChangeProof (⇔)
    ↳73 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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(active(X1), X2) → FIRST(X1, X2)
FIRST(X1, active(X2)) → FIRST(X1, X2)

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

(8) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • FIRST(X1, mark(X2)) → FIRST(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • FIRST(mark(X1), X2) → FIRST(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • FIRST(active(X1), X2) → FIRST(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • FIRST(X1, active(X2)) → FIRST(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(9) TRUE

(10) 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.

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(12) Obligation:

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

DBL(active(X)) → DBL(X)
DBL(mark(X)) → DBL(X)

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • DBL(active(X)) → DBL(X)
    The graph contains the following edges 1 > 1

  • DBL(mark(X)) → DBL(X)
    The graph contains the following edges 1 > 1

(14) TRUE

(15) 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.

(16) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

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

ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)

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

(18) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • ADD(X1, mark(X2)) → ADD(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • ADD(mark(X1), X2) → ADD(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • ADD(active(X1), X2) → ADD(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • ADD(X1, active(X2)) → ADD(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(19) TRUE

(20) 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.

(21) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(22) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

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

(23) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • S(active(X)) → S(X)
    The graph contains the following edges 1 > 1

  • S(mark(X)) → S(X)
    The graph contains the following edges 1 > 1

(24) TRUE

(25) 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.

(26) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(27) Obligation:

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

SQR(active(X)) → SQR(X)
SQR(mark(X)) → SQR(X)

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

(28) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • SQR(active(X)) → SQR(X)
    The graph contains the following edges 1 > 1

  • SQR(mark(X)) → SQR(X)
    The graph contains the following edges 1 > 1

(29) TRUE

(30) 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.

(31) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(32) Obligation:

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

RECIP(active(X)) → RECIP(X)
RECIP(mark(X)) → RECIP(X)

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

(33) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RECIP(active(X)) → RECIP(X)
    The graph contains the following edges 1 > 1

  • RECIP(mark(X)) → RECIP(X)
    The graph contains the following edges 1 > 1

(34) TRUE

(35) 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.

(36) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(37) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

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

(38) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • CONS(X1, mark(X2)) → CONS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • CONS(mark(X1), X2) → CONS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS(active(X1), X2) → CONS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS(X1, active(X2)) → CONS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(39) TRUE

(40) 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.

(41) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(42) Obligation:

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

TERMS(active(X)) → TERMS(X)
TERMS(mark(X)) → TERMS(X)

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

(43) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • TERMS(active(X)) → TERMS(X)
    The graph contains the following edges 1 > 1

  • TERMS(mark(X)) → TERMS(X)
    The graph contains the following edges 1 > 1

(44) TRUE

(45) 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.

(46) 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)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
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)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)

(47) 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.

(48) 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) = 1   
POL(recip(x1)) = x1   
POL(s(x1)) = 0   
POL(sqr(x1)) = x1   
POL(terms(x1)) = x1   

The following usable rules [FROCOS05] were oriented:

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

(49) 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.

(50) 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))))
MARK(dbl(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)) = 1 + 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(nil) → active(nil)
terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(add(0, X)) → mark(X)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
mark(sqr(X)) → active(sqr(mark(X)))
mark(add(X1, X2)) → active(add(mark(X1), mark(X2)))
mark(dbl(X)) → active(dbl(mark(X)))
active(dbl(s(X))) → mark(s(s(dbl(X))))
mark(s(X)) → active(s(X))
mark(recip(X)) → active(recip(mark(X)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
mark(terms(X)) → active(terms(mark(X)))
mark(0) → active(0)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
active(sqr(0)) → mark(0)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
active(dbl(0)) → mark(0)
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)
active(first(0, X)) → mark(nil)
s(active(X)) → s(X)
s(mark(X)) → s(X)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)
recip(mark(X)) → recip(X)
recip(active(X)) → recip(X)

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

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = 1   
POL(active(x1)) = 0   
POL(add(x1, x2)) = 1   
POL(cons(x1, x2)) = 0   
POL(dbl(x1)) = 0   
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)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
dbl(mark(X)) → dbl(X)
dbl(active(X)) → dbl(X)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)

(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)
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(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.


MARK(terms(X)) → MARK(X)
ACTIVE(sqr(s(X))) → MARK(s(add(sqr(X), dbl(X))))
MARK(sqr(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)) = 1 + x1   
POL(terms(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

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

(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(cons(X1, X2)) → MARK(X1)
MARK(recip(X)) → MARK(X)
MARK(sqr(X)) → ACTIVE(sqr(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(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.


ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
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)) = 1 + x1   
POL(first(x1, x2)) = x1 + x2   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(recip(x1)) = x1   
POL(s(x1)) = 1   
POL(sqr(x1)) = 1 + x1   
POL(terms(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

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

(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))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(recip(X)) → MARK(X)
MARK(sqr(X)) → ACTIVE(sqr(mark(X)))
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
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.


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

The following usable rules [FROCOS05] were oriented:

terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(mark(X1), X2) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
sqr(active(X)) → sqr(X)
sqr(mark(X)) → sqr(X)

(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))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(recip(X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → 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.


MARK(add(X1, X2)) → ACTIVE(add(mark(X1), mark(X2)))
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)) = 0   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(add(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = x1   
POL(dbl(x1)) = 0   
POL(first(x1, x2)) = x1 + x2   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(recip(x1)) = x1   
POL(s(x1)) = 0   
POL(sqr(x1)) = 0   
POL(terms(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

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

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

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

The following usable rules [FROCOS05] were oriented:

terms(active(X)) → terms(X)
terms(mark(X)) → terms(X)

(63) 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(cons(X1, X2)) → MARK(X1)
MARK(recip(X)) → MARK(X)
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.

(64) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(first(X1, X2)) → MARK(X1)
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)) = 0   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(add(x1, x2)) = 0   
POL(cons(x1, x2)) = x1   
POL(dbl(x1)) = 0   
POL(first(x1, x2)) = 1 + x1 + x2   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(recip(x1)) = x1   
POL(s(x1)) = 0   
POL(sqr(x1)) = 0   
POL(terms(x1)) = 0   

The following usable rules [FROCOS05] were oriented:

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

(65) 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(cons(X1, X2)) → MARK(X1)
MARK(recip(X)) → MARK(X)

The TRS R consists of the following rules:

active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
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) 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) = 0   
POL(ACTIVE(x1)) = 1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(add(x1, x2)) = 0   
POL(cons(x1, x2)) = x1   
POL(dbl(x1)) = 0   
POL(first(x1, x2)) = 0   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(recip(x1)) = x1   
POL(s(x1)) = 0   
POL(sqr(x1)) = 0   
POL(terms(x1)) = 1   

The following usable rules [FROCOS05] were oriented:

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

(67) Obligation:

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

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

The TRS R consists of the following rules:

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

(68) DependencyGraphProof (EQUIVALENT transformation)

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

(69) Obligation:

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

MARK(recip(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

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.

(70) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(71) Obligation:

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

MARK(recip(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

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

(72) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • MARK(recip(X)) → MARK(X)
    The graph contains the following edges 1 > 1

  • MARK(cons(X1, X2)) → MARK(X1)
    The graph contains the following edges 1 > 1

(73) TRUE