(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))
a__half(0) → 0
a__half(s(0)) → 0
a__half(s(s(X))) → s(a__half(mark(X)))
a__half(dbl(X)) → mark(X)
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(half(X)) → a__half(mark(X))
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)
a__half(X) → half(X)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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)
A__HALF(s(s(X))) → A__HALF(mark(X))
A__HALF(s(s(X))) → MARK(X)
A__HALF(dbl(X)) → MARK(X)
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(half(X)) → A__HALF(mark(X))
MARK(half(X)) → MARK(X)
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))
a__half(0) → 0
a__half(s(0)) → 0
a__half(s(s(X))) → s(a__half(mark(X)))
a__half(dbl(X)) → mark(X)
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(half(X)) → a__half(mark(X))
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)
a__half(X) → half(X)

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) → MARK(X)
A__ADD(s(X), Y) → MARK(Y)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
A__HALF(s(s(X))) → A__HALF(mark(X))
A__HALF(s(s(X))) → MARK(X)
A__HALF(dbl(X)) → MARK(X)
MARK(terms(X)) → 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
A__TERMS(x0, x1)  =  A__TERMS(x0)
A__SQR(x0, x1)  =  A__SQR(x0)
MARK(x0, x1)  =  MARK(x1)
A__ADD(x0, x1, x2)  =  A__ADD(x0, x2)
A__DBL(x0, x1)  =  A__DBL(x1)
A__FIRST(x0, x1, x2)  =  A__FIRST(x0)
A__HALF(x0, x1)  =  A__HALF(x1)

Tags:
A__TERMS has argument tags [0,0] and root tag 0
A__SQR has argument tags [0,0] and root tag 0
MARK has argument tags [0,0] and root tag 0
A__ADD has argument tags [3,2,0] and root tag 6
A__DBL has argument tags [2,1] and root tag 5
A__FIRST has argument tags [0,12,8] and root tag 7
A__HALF has argument tags [8,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
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
s(x1)  =  s(x1)
A__ADD(x1, x2)  =  x1
a__sqr(x1)  =  a__sqr(x1)
a__dbl(x1)  =  a__dbl(x1)
A__DBL(x1)  =  A__DBL
0  =  0
A__FIRST(x1, x2)  =  A__FIRST(x2)
cons(x1, x2)  =  x1
A__HALF(x1)  =  A__HALF
dbl(x1)  =  dbl(x1)
terms(x1)  =  terms(x1)
sqr(x1)  =  sqr(x1)
add(x1, x2)  =  add(x1, x2)
first(x1, x2)  =  first(x1, x2)
half(x1)  =  x1
recip(x1)  =  x1
a__terms(x1)  =  a__terms(x1)
a__add(x1, x2)  =  a__add(x1, x2)
a__half(x1)  =  x1
a__first(x1, x2)  =  a__first(x1, x2)
nil  =  nil

Lexicographic path order with status [LPO].
Quasi-Precedence:
[MARK, first2, afirst2] > [ATERMS1, ASQR1, asqr1, adbl1, dbl1, terms1, sqr1, aterms1] > [add2, aadd2] > [s1, ADBL] > 0 > nil
[MARK, first2, afirst2] > [ATERMS1, ASQR1, asqr1, adbl1, dbl1, terms1, sqr1, aterms1] > [add2, aadd2] > [s1, ADBL] > AHALF
[MARK, first2, afirst2] > AFIRST1

Status:
ATERMS1: [1]
ASQR1: [1]
MARK: []
s1: [1]
asqr1: [1]
adbl1: [1]
ADBL: []
0: []
AFIRST1: [1]
AHALF: []
dbl1: [1]
terms1: [1]
sqr1: [1]
add2: [2,1]
first2: [2,1]
aterms1: [1]
aadd2: [2,1]
afirst2: [2,1]
nil: []


The following usable rules [FROCOS05] were oriented:

mark(terms(X)) → a__terms(mark(X))
mark(sqr(X)) → a__sqr(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
a__add(0, X) → mark(X)
mark(half(X)) → a__half(mark(X))
a__half(dbl(X)) → mark(X)
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__sqr(0) → 0
a__sqr(s(X)) → s(a__add(a__sqr(mark(X)), a__dbl(mark(X))))
a__sqr(X) → sqr(X)
a__dbl(0) → 0
a__dbl(s(X)) → s(s(a__dbl(mark(X))))
a__dbl(X) → dbl(X)
a__terms(N) → cons(recip(a__sqr(mark(N))), terms(s(N)))
a__add(s(X), Y) → s(a__add(mark(X), mark(Y)))
a__half(s(s(X))) → s(a__half(mark(X)))
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__terms(X) → terms(X)
a__add(X1, X2) → add(X1, X2)
a__half(0) → 0
a__half(s(0)) → 0
a__half(X) → half(X)
a__first(0, X) → nil
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))
A__ADD(s(X), Y) → A__ADD(mark(X), mark(Y))
MARK(terms(X)) → A__TERMS(mark(X))
MARK(sqr(X)) → A__SQR(mark(X))
MARK(half(X)) → A__HALF(mark(X))
MARK(half(X)) → MARK(X)
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))
a__half(0) → 0
a__half(s(0)) → 0
a__half(s(s(X))) → s(a__half(mark(X)))
a__half(dbl(X)) → mark(X)
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(half(X)) → a__half(mark(X))
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)
a__half(X) → half(X)

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 2 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

A__ADD(s(X), Y) → A__ADD(mark(X), mark(Y))

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))
a__half(0) → 0
a__half(s(0)) → 0
a__half(s(s(X))) → s(a__half(mark(X)))
a__half(dbl(X)) → mark(X)
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(half(X)) → a__half(mark(X))
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)
a__half(X) → half(X)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A__ADD(s(X), Y) → A__ADD(mark(X), mark(Y))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
A__ADD(x0, x1, x2)  =  A__ADD(x0, x2)

Tags:
A__ADD has argument tags [1,1,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
A__ADD(x1, x2)  =  A__ADD(x1, x2)
s(x1)  =  s(x1)
mark(x1)  =  x1
terms(x1)  =  x1
a__terms(x1)  =  x1
sqr(x1)  =  sqr(x1)
a__sqr(x1)  =  a__sqr(x1)
add(x1, x2)  =  add(x1, x2)
a__add(x1, x2)  =  a__add(x1, x2)
0  =  0
half(x1)  =  x1
a__half(x1)  =  x1
dbl(x1)  =  dbl(x1)
a__dbl(x1)  =  a__dbl(x1)
first(x1, x2)  =  x1
a__first(x1, x2)  =  x1
cons(x1, x2)  =  cons
recip(x1)  =  x1
nil  =  nil

Lexicographic path order with status [LPO].
Quasi-Precedence:
[sqr1, asqr1] > [add2, aadd2] > [AADD2, s1] > cons
[sqr1, asqr1] > 0 > nil > cons
[sqr1, asqr1] > [dbl1, adbl1] > [AADD2, s1] > cons

Status:
AADD2: [2,1]
s1: [1]
sqr1: [1]
asqr1: [1]
add2: [1,2]
aadd2: [1,2]
0: []
dbl1: [1]
adbl1: [1]
cons: []
nil: []


The following usable rules [FROCOS05] were oriented:

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

(9) 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))
a__half(0) → 0
a__half(s(0)) → 0
a__half(s(s(X))) → s(a__half(mark(X)))
a__half(dbl(X)) → mark(X)
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(half(X)) → a__half(mark(X))
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)
a__half(X) → half(X)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)
MARK(half(X)) → MARK(X)
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))
a__half(0) → 0
a__half(s(0)) → 0
a__half(s(s(X))) → s(a__half(mark(X)))
a__half(dbl(X)) → mark(X)
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(half(X)) → a__half(mark(X))
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)
a__half(X) → half(X)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x1)

Tags:
MARK has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
MARK(x1)  =  MARK
cons(x1, x2)  =  cons(x1)
half(x1)  =  x1
recip(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
MARK: []
cons1: [1]


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

MARK(half(X)) → MARK(X)
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))
a__half(0) → 0
a__half(s(0)) → 0
a__half(s(s(X))) → s(a__half(mark(X)))
a__half(dbl(X)) → mark(X)
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(half(X)) → a__half(mark(X))
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)
a__half(X) → half(X)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(half(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x1)

Tags:
MARK has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
MARK(x1)  =  MARK
half(x1)  =  half(x1)
recip(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
MARK: []
half1: [1]


The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

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))
a__half(0) → 0
a__half(s(0)) → 0
a__half(s(s(X))) → s(a__half(mark(X)))
a__half(dbl(X)) → mark(X)
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(half(X)) → a__half(mark(X))
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)
a__half(X) → half(X)

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

(17) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x1)

Tags:
MARK has argument tags [1,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
MARK(x1)  =  MARK
recip(x1)  =  recip(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
MARK: []
recip1: [1]


The following usable rules [FROCOS05] were oriented: none

(18) 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))
a__half(0) → 0
a__half(s(0)) → 0
a__half(s(s(X))) → s(a__half(mark(X)))
a__half(dbl(X)) → mark(X)
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(half(X)) → a__half(mark(X))
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)
a__half(X) → half(X)

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE