(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, 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:

A__TERMS(N) → A__SQR(mark(N))
A__TERMS(N) → MARK(N)
A__SQR(s(X)) → A__ADD(a__sqr(mark(X)), a__dbl(mark(X)))
A__SQR(s(X)) → A__SQR(mark(X))
A__SQR(s(X)) → MARK(X)
A__SQR(s(X)) → A__DBL(mark(X))
A__DBL(s(X)) → A__DBL(mark(X))
A__DBL(s(X)) → MARK(X)
A__ADD(0, X) → MARK(X)
A__ADD(s(X), Y) → A__ADD(mark(X), mark(Y))
A__ADD(s(X), Y) → MARK(X)
A__ADD(s(X), Y) → MARK(Y)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(terms(X)) → A__TERMS(mark(X))
MARK(terms(X)) → MARK(X)
MARK(sqr(X)) → A__SQR(mark(X))
MARK(sqr(X)) → MARK(X)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → A__DBL(mark(X))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(recip(X)) → MARK(X)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A__TERMS(N) → MARK(N)
A__SQR(s(X)) → A__ADD(a__sqr(mark(X)), a__dbl(mark(X)))
A__SQR(s(X)) → A__SQR(mark(X))
A__SQR(s(X)) → MARK(X)
A__SQR(s(X)) → A__DBL(mark(X))
A__DBL(s(X)) → A__DBL(mark(X))
A__DBL(s(X)) → MARK(X)
A__ADD(0, X) → MARK(X)
A__ADD(s(X), Y) → A__ADD(mark(X), mark(Y))
A__ADD(s(X), Y) → MARK(X)
A__ADD(s(X), Y) → MARK(Y)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(terms(X)) → A__TERMS(mark(X))
MARK(terms(X)) → MARK(X)
MARK(sqr(X)) → A__SQR(mark(X))
MARK(sqr(X)) → MARK(X)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(dbl(X)) → A__DBL(mark(X))
MARK(dbl(X)) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__TERMS(x1)  =  A__TERMS(x1)
A__SQR(x1)  =  A__SQR(x1)
mark(x1)  =  x1
MARK(x1)  =  MARK(x1)
s(x1)  =  s(x1)
A__ADD(x1, x2)  =  A__ADD(x1, x2)
a__sqr(x1)  =  a__sqr(x1)
a__dbl(x1)  =  a__dbl(x1)
A__DBL(x1)  =  A__DBL(x1)
0  =  0
A__FIRST(x1, x2)  =  A__FIRST(x1, x2)
cons(x1, x2)  =  x1
terms(x1)  =  terms(x1)
sqr(x1)  =  sqr(x1)
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  first(x1, x2)
recip(x1)  =  x1
a__terms(x1)  =  a__terms(x1)
a__add(x1, x2)  =  a__add(x1, x2)
a__first(x1, x2)  =  a__first(x1, x2)
nil  =  nil

Recursive path order with status [RPO].
Quasi-Precedence:
[ATERMS1, ASQR1, asqr1, terms1, sqr1, aterms1] > [MARK1, AADD2, ADBL1, AFIRST2]
[ATERMS1, ASQR1, asqr1, terms1, sqr1, aterms1] > [adbl1, dbl1] > s1 > [first2, afirst2]
[ATERMS1, ASQR1, asqr1, terms1, sqr1, aterms1] > [adbl1, dbl1] > 0 > nil
[ATERMS1, ASQR1, asqr1, terms1, sqr1, aterms1] > [add2, aadd2] > s1 > [first2, afirst2]

Status:
ATERMS1: [1]
ASQR1: [1]
MARK1: multiset
s1: multiset
AADD2: multiset
asqr1: [1]
adbl1: multiset
ADBL1: multiset
0: multiset
AFIRST2: multiset
terms1: [1]
sqr1: [1]
add2: multiset
dbl1: multiset
first2: multiset
aterms1: [1]
aadd2: multiset
afirst2: multiset
nil: multiset


The following usable rules [FROCOS05] were oriented:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

(4) Obligation:

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

A__TERMS(N) → A__SQR(mark(N))
MARK(cons(X1, X2)) → MARK(X1)
MARK(recip(X)) → MARK(X)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) 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:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive path order with status [RPO].
Quasi-Precedence:
[aterms, terms] > [asqr, mark1, aadd2, adbl1, afirst1, first1, add2, dbl1] > recip1
[aterms, terms] > [asqr, mark1, aadd2, adbl1, afirst1, first1, add2, dbl1] > s
[aterms, terms] > [asqr, mark1, aadd2, adbl1, afirst1, first1, add2, dbl1] > 0 > nil
[aterms, terms] > [asqr, mark1, aadd2, adbl1, afirst1, first1, add2, dbl1] > sqr

Status:
MARK1: [1]
recip1: [1]
aterms: []
asqr: []
mark1: [1]
terms: []
s: multiset
0: multiset
aadd2: [2,1]
adbl1: [1]
afirst1: [1]
nil: multiset
first1: [1]
sqr: multiset
add2: [2,1]
dbl1: [1]


The following usable rules [FROCOS05] were oriented:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

(8) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
cons(x1, x2)  =  cons(x1)
a__terms(x1)  =  a__terms
recip(x1)  =  recip
a__sqr(x1)  =  x1
mark(x1)  =  mark(x1)
terms(x1)  =  terms
s(x1)  =  s
0  =  0
a__add(x1, x2)  =  a__add(x2)
a__dbl(x1)  =  x1
a__first(x1, x2)  =  a__first(x1, x2)
nil  =  nil
first(x1, x2)  =  first(x1, x2)
sqr(x1)  =  x1
add(x1, x2)  =  add(x2)
dbl(x1)  =  x1

Recursive path order with status [RPO].
Quasi-Precedence:
[aterms, terms] > [mark1, s, aadd1, afirst2, first2, add1] > cons1
[aterms, terms] > [mark1, s, aadd1, afirst2, first2, add1] > recip
[0, nil] > [mark1, s, aadd1, afirst2, first2, add1] > cons1
[0, nil] > [mark1, s, aadd1, afirst2, first2, add1] > recip

Status:
cons1: multiset
aterms: []
recip: multiset
mark1: multiset
terms: []
s: multiset
0: multiset
aadd1: multiset
afirst2: multiset
nil: multiset
first2: multiset
add1: multiset


The following usable rules [FROCOS05] were oriented:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

(10) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(dbl(X)) → a__dbl(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(recip(X)) → recip(mark(X))
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__terms(X) → terms(X)
a__sqr(X) → sqr(X)
a__add(X1, X2) → add(X1, X2)
a__dbl(X) → dbl(X)
a__first(X1, X2) → first(X1, X2)

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) TRUE