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 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 PisEmptyProof (⇔)
    ↳24 TRUE
  ↳25 QDP
    ↳26 QDPOrderProof (⇔)
    ↳27 QDP
    ↳28 DependencyGraphProof (⇔)
    ↳29 QDP
    ↳30 QDPOrderProof (⇔)
    ↳31 QDP
    ↳32 PisEmptyProof (⇔)
    ↳33 TRUE

(0) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → 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:

TERMS(N) → SQR(N)
SQR(s(X)) → S(add(sqr(X), dbl(X)))
SQR(s(X)) → ADD(sqr(X), dbl(X))
SQR(s(X)) → SQR(X)
SQR(s(X)) → DBL(X)
DBL(s(X)) → S(s(dbl(X)))
DBL(s(X)) → S(dbl(X))
DBL(s(X)) → DBL(X)
ADD(s(X), Y) → S(add(X, Y))
ADD(s(X), Y) → ADD(X, Y)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
HALF(s(s(X))) → S(half(X))
HALF(s(s(X))) → HALF(X)
ACTIVATE(n__terms(X)) → TERMS(activate(X))
ACTIVATE(n__terms(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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 5 SCCs with 10 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

HALF(s(s(X))) → HALF(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


HALF(s(s(X))) → HALF(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HALF(x1)  =  HALF(x1)
s(x1)  =  s(x1)
terms(x1)  =  x1
cons(x1, x2)  =  cons
recip(x1)  =  recip(x1)
sqr(x1)  =  sqr(x1)
n__terms(x1)  =  x1
n__s(x1)  =  n__s(x1)
0  =  0
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  first
nil  =  nil
n__first(x1, x2)  =  n__first
activate(x1)  =  activate(x1)
half(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
HALF1 > [cons, 0, nil]
recip1 > [cons, 0, nil]
sqr1 > add2 > [s1, ns1] > [cons, 0, nil]
sqr1 > dbl1 > [s1, ns1] > [cons, 0, nil]
activate1 > [s1, ns1] > [cons, 0, nil]
activate1 > [first, nfirst] > [cons, 0, nil]

Status:
HALF1: [1]
s1: [1]
cons: []
recip1: [1]
sqr1: [1]
ns1: [1]
0: []
add2: [1,2]
dbl1: [1]
first: []
nil: []
nfirst: []
activate1: [1]


The following usable rules [FROCOS05] were oriented:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

(7) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ADD(s(X), Y) → ADD(X, Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2)  =  x1
s(x1)  =  s(x1)
terms(x1)  =  x1
cons(x1, x2)  =  cons
recip(x1)  =  recip
sqr(x1)  =  sqr(x1)
n__terms(x1)  =  x1
n__s(x1)  =  n__s(x1)
0  =  0
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  x2
nil  =  nil
n__first(x1, x2)  =  x2
activate(x1)  =  x1
half(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
recip > [cons, 0, nil]
sqr1 > add2 > [s1, ns1] > [cons, 0, nil]
sqr1 > dbl1 > [s1, ns1] > [cons, 0, nil]

Status:
s1: [1]
cons: []
recip: []
sqr1: [1]
ns1: [1]
0: []
add2: [1,2]
dbl1: [1]
nil: []


The following usable rules [FROCOS05] were oriented:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

(12) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

DBL(s(X)) → DBL(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DBL(s(X)) → DBL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DBL(x1)  =  DBL(x1)
s(x1)  =  s(x1)
terms(x1)  =  terms(x1)
cons(x1, x2)  =  x1
recip(x1)  =  recip
sqr(x1)  =  sqr(x1)
n__terms(x1)  =  n__terms(x1)
n__s(x1)  =  n__s(x1)
0  =  0
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  x2
nil  =  nil
n__first(x1, x2)  =  x2
activate(x1)  =  x1
half(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
[terms1, sqr1, nterms1, dbl1] > add2 > [DBL1, s1, recip, ns1, nil]
0 > [DBL1, s1, recip, ns1, nil]

Status:
DBL1: [1]
s1: [1]
terms1: [1]
recip: []
sqr1: [1]
nterms1: [1]
ns1: [1]
0: []
add2: [1,2]
dbl1: [1]
nil: []


The following usable rules [FROCOS05] were oriented:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

(17) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

SQR(s(X)) → SQR(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SQR(s(X)) → SQR(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SQR(x1)  =  SQR(x1)
s(x1)  =  s(x1)
terms(x1)  =  terms(x1)
cons(x1, x2)  =  x1
recip(x1)  =  recip
sqr(x1)  =  sqr(x1)
n__terms(x1)  =  n__terms(x1)
n__s(x1)  =  n__s(x1)
0  =  0
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  x2
nil  =  nil
n__first(x1, x2)  =  x2
activate(x1)  =  x1
half(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
[terms1, sqr1, nterms1, dbl1] > add2 > [SQR1, s1, recip, ns1, nil]
0 > [SQR1, s1, recip, ns1, nil]

Status:
SQR1: [1]
s1: [1]
terms1: [1]
recip: []
sqr1: [1]
nterms1: [1]
ns1: [1]
0: []
add2: [1,2]
dbl1: [1]
nil: []


The following usable rules [FROCOS05] were oriented:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

(22) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE

(25) Obligation:

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

ACTIVATE(n__terms(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__terms(X)) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE(x1)
n__terms(x1)  =  n__terms(x1)
n__s(x1)  =  x1
n__first(x1, x2)  =  n__first(x1, x2)
FIRST(x1, x2)  =  FIRST(x2)
activate(x1)  =  x1
s(x1)  =  x1
cons(x1, x2)  =  x2
terms(x1)  =  terms(x1)
recip(x1)  =  recip
sqr(x1)  =  sqr(x1)
0  =  0
add(x1, x2)  =  add(x2)
dbl(x1)  =  x1
first(x1, x2)  =  first(x1, x2)
nil  =  nil
half(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
[ACTIVATE1, nfirst2, FIRST1, first2] > nil
[nterms1, terms1, sqr1] > recip
[nterms1, terms1, sqr1] > add1
0 > nil

Status:
ACTIVATE1: [1]
nterms1: [1]
nfirst2: [1,2]
FIRST1: [1]
terms1: [1]
recip: []
sqr1: [1]
0: []
add1: [1]
first2: [1,2]
nil: []


The following usable rules [FROCOS05] were oriented:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

(27) Obligation:

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

ACTIVATE(n__s(X)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(28) DependencyGraphProof (EQUIVALENT transformation)

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

(29) Obligation:

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

ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__s(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE(x1)
n__s(x1)  =  n__s(x1)
terms(x1)  =  terms
cons(x1, x2)  =  cons
recip(x1)  =  recip(x1)
sqr(x1)  =  sqr(x1)
n__terms(x1)  =  n__terms
0  =  0
s(x1)  =  s(x1)
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  first
nil  =  nil
n__first(x1, x2)  =  n__first
activate(x1)  =  x1
half(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
[terms, nterms] > sqr1 > add2 > [ns1, s1] > ACTIVATE1 > [cons, first, nil, nfirst]
[terms, nterms] > sqr1 > add2 > [ns1, s1] > 0 > [cons, first, nil, nfirst]
[terms, nterms] > sqr1 > dbl1 > [ns1, s1] > ACTIVATE1 > [cons, first, nil, nfirst]
[terms, nterms] > sqr1 > dbl1 > [ns1, s1] > 0 > [cons, first, nil, nfirst]
recip1 > [cons, first, nil, nfirst]

Status:
ACTIVATE1: [1]
ns1: [1]
terms: []
cons: []
recip1: [1]
sqr1: [1]
nterms: []
0: []
s1: [1]
add2: [2,1]
dbl1: [1]
first: []
nil: []
nfirst: []


The following usable rules [FROCOS05] were oriented:

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

(31) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(n__s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
s(X) → n__s(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(X) → X

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE