Termination w.r.t. Q proof of /tmp/tmpdX3f6e/ExAppendixB_AEL03

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 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 DependencyGraphProof (⇔)

    ↳26 TRUE

(0) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
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:

2NDSPOS(s(N), cons(X, Z)) → 2NDSPOS(s(N), cons2(X, activate(Z)))
2NDSPOS(s(N), cons(X, Z)) → ACTIVATE(Z)
2NDSPOS(s(N), cons2(X, cons(Y, Z))) → 2NDSNEG(N, activate(Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) → ACTIVATE(Z)
2NDSNEG(s(N), cons(X, Z)) → 2NDSNEG(s(N), cons2(X, activate(Z)))
2NDSNEG(s(N), cons(X, Z)) → ACTIVATE(Z)
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → 2NDSPOS(N, activate(Z))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → ACTIVATE(Z)
PI(X) → 2NDSPOS(X, from(0))
PI(X) → FROM(0)
PLUS(s(X), Y) → S(plus(X, Y))
PLUS(s(X), Y) → PLUS(X, Y)
TIMES(s(X), Y) → PLUS(Y, times(X, Y))
TIMES(s(X), Y) → TIMES(X, Y)
SQUARE(X) → TIMES(X, X)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
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 4 SCCs with 11 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
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.

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)
n__from(x1) = x1
from(x1) = x1
cons(x1, x2) = x1
2ndspos(x1, x2) = 2ndspos(x1, x2)
0 = 0
rnil = rnil
s(x1) = s(x1)
cons2(x1, x2) = x1
activate(x1) = activate(x1)
rcons(x1, x2) = rcons
posrecip(x1) = posrecip(x1)
2ndsneg(x1, x2) = 2ndsneg(x1)
negrecip(x1) = x1
pi(x1) = pi(x1)
plus(x1, x2) = plus(x1, x2)
times(x1, x2) = times(x1, x2)
square(x1) = square(x1)

Recursive Path Order [RPO].
Precedence:

ACTIVATE₁ > n_{_s₁}
pi₁ > 2ndspos₂ > activate₁ > s₁ > rcons > n_{_s₁}
pi₁ > 2ndspos₂ > activate₁ > s₁ > posrecip₁ > n_{_s₁}
pi₁ > 2ndspos₂ > 2ndsneg₁ > rnil > n_{_s₁}
pi₁ > 2ndspos₂ > 2ndsneg₁ > rcons > n_{_s₁}
pi₁ > 0 > n_{_s₁}
square₁ > times₂ > 0 > n_{_s₁}
square₁ > times₂ > plus₂ > s₁ > rcons > n_{_s₁}
square₁ > times₂ > plus₂ > s₁ > posrecip₁ > n_{_s₁}

The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

(7) Obligation:

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

ACTIVATE(n__from(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → 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.

ACTIVATE(n__from(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__from(x1) = n__from(x1)
from(x1) = from(x1)
cons(x1, x2) = cons(x1, x2)
n__s(x1) = x1
2ndspos(x1, x2) = x2
0 = 0
rnil = rnil
s(x1) = x1
cons2(x1, x2) = cons2
activate(x1) = activate(x1)
rcons(x1, x2) = x1
posrecip(x1) = posrecip
2ndsneg(x1, x2) = 2ndsneg(x1)
negrecip(x1) = negrecip
pi(x1) = pi(x1)
plus(x1, x2) = x2
times(x1, x2) = x1
square(x1) = x1

Recursive Path Order [RPO].
Precedence:

ACTIVATE₁ > rnil
activate₁ > from₁ > n_{_from₁} > rnil
activate₁ > from₁ > cons₂ > cons2 > posrecip > rnil
2ndsneg₁ > cons2 > posrecip > rnil
2ndsneg₁ > negrecip > rnil
pi₁ > from₁ > n_{_from₁} > rnil
pi₁ > from₁ > cons₂ > cons2 > posrecip > rnil
pi₁ > 0 > rnil

The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

(9) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → 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:

PLUS(s(X), Y) → PLUS(X, Y)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → 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.

PLUS(s(X), Y) → PLUS(X, Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = PLUS(x1, x2)
s(x1) = s(x1)
from(x1) = from
cons(x1, x2) = x2
n__from(x1) = n__from
n__s(x1) = n__s(x1)
2ndspos(x1, x2) = x1
0 = 0
rnil = rnil
cons2(x1, x2) = x2
activate(x1) = activate(x1)
rcons(x1, x2) = rcons(x1)
posrecip(x1) = posrecip
2ndsneg(x1, x2) = x1
negrecip(x1) = negrecip
pi(x1) = pi(x1)
plus(x1, x2) = plus(x1, x2)
times(x1, x2) = times(x1, x2)
square(x1) = square(x1)

Recursive Path Order [RPO].
Precedence:

activate₁ > s₁ > PLUS₂ > negrecip
activate₁ > s₁ > n_{_s₁} > negrecip
activate₁ > s₁ > rcons₁ > negrecip
activate₁ > s₁ > posrecip > negrecip
activate₁ > from > n_{_from} > negrecip
activate₁ > from > n_{_s₁} > negrecip
pi₁ > from > n_{_from} > negrecip
pi₁ > from > n_{_s₁} > negrecip
pi₁ > 0 > rnil > negrecip
square₁ > times₂ > plus₂ > s₁ > PLUS₂ > negrecip
square₁ > times₂ > plus₂ > s₁ > n_{_s₁} > negrecip
square₁ > times₂ > plus₂ > s₁ > rcons₁ > negrecip
square₁ > times₂ > plus₂ > s₁ > posrecip > negrecip

The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

(14) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

TIMES(s(X), Y) → TIMES(X, Y)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TIMES(s(X), Y) → TIMES(X, Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2) = TIMES(x1, x2)
s(x1) = s(x1)
from(x1) = from
cons(x1, x2) = x2
n__from(x1) = n__from
n__s(x1) = n__s(x1)
2ndspos(x1, x2) = x1
0 = 0
rnil = rnil
cons2(x1, x2) = x2
activate(x1) = activate(x1)
rcons(x1, x2) = rcons(x1)
posrecip(x1) = posrecip
2ndsneg(x1, x2) = x1
negrecip(x1) = negrecip
pi(x1) = pi(x1)
plus(x1, x2) = plus(x1, x2)
times(x1, x2) = times(x1, x2)
square(x1) = square(x1)

Recursive Path Order [RPO].
Precedence:

activate₁ > s₁ > TIMES₂ > negrecip
activate₁ > s₁ > n_{_s₁} > negrecip
activate₁ > s₁ > rcons₁ > negrecip
activate₁ > s₁ > posrecip > negrecip
activate₁ > from > n_{_from} > negrecip
activate₁ > from > n_{_s₁} > negrecip
pi₁ > from > n_{_from} > negrecip
pi₁ > from > n_{_s₁} > negrecip
pi₁ > 0 > rnil > negrecip
square₁ > times₂ > plus₂ > s₁ > TIMES₂ > negrecip
square₁ > times₂ > plus₂ > s₁ > n_{_s₁} > negrecip
square₁ > times₂ > plus₂ > s₁ > rcons₁ > negrecip
square₁ > times₂ > plus₂ > s₁ > posrecip > negrecip

The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

(19) Obligation:

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

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

2NDSPOS(s(N), cons2(X, cons(Y, Z))) → 2NDSNEG(N, activate(Z))
2NDSNEG(s(N), cons(X, Z)) → 2NDSNEG(s(N), cons2(X, activate(Z)))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → 2NDSPOS(N, activate(Z))
2NDSPOS(s(N), cons(X, Z)) → 2NDSPOS(s(N), cons2(X, activate(Z)))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

2NDSPOS(s(N), cons2(X, cons(Y, Z))) → 2NDSNEG(N, activate(Z))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → 2NDSPOS(N, activate(Z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
2NDSPOS(x1, x2) = 2NDSPOS(x1)
s(x1) = s(x1)
cons2(x1, x2) = x1
cons(x1, x2) = cons(x2)
2NDSNEG(x1, x2) = 2NDSNEG(x1)
activate(x1) = activate(x1)
from(x1) = from
n__from(x1) = n__from
n__s(x1) = n__s(x1)
2ndspos(x1, x2) = 2ndspos
0 = 0
rnil = rnil
rcons(x1, x2) = x1
posrecip(x1) = posrecip
2ndsneg(x1, x2) = x1
negrecip(x1) = negrecip
pi(x1) = pi(x1)
plus(x1, x2) = plus(x1, x2)
times(x1, x2) = times(x1, x2)
square(x1) = square(x1)

Recursive Path Order [RPO].
Precedence:

activate₁ > s₁ > 2NDSPOS₁ > 2NDSNEG₁ > n_{_s₁}
activate₁ > s₁ > 2ndspos > rnil > n_{_s₁}
activate₁ > s₁ > 2ndspos > posrecip > n_{_s₁}
activate₁ > s₁ > negrecip > n_{_s₁}
activate₁ > from > cons₁ > 2NDSNEG₁ > n_{_s₁}
activate₁ > from > cons₁ > negrecip > n_{_s₁}
activate₁ > from > n_{_from} > n_{_s₁}
pi₁ > from > cons₁ > 2NDSNEG₁ > n_{_s₁}
pi₁ > from > cons₁ > negrecip > n_{_s₁}
pi₁ > from > n_{_from} > n_{_s₁}
pi₁ > 2ndspos > rnil > n_{_s₁}
pi₁ > 2ndspos > posrecip > n_{_s₁}
pi₁ > 0 > rnil > n_{_s₁}
square₁ > times₂ > 0 > rnil > n_{_s₁}
square₁ > times₂ > plus₂ > s₁ > 2NDSPOS₁ > 2NDSNEG₁ > n_{_s₁}
square₁ > times₂ > plus₂ > s₁ > 2ndspos > rnil > n_{_s₁}
square₁ > times₂ > plus₂ > s₁ > 2ndspos > posrecip > n_{_s₁}
square₁ > times₂ > plus₂ > s₁ > negrecip > n_{_s₁}

The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

(24) Obligation:

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

2NDSNEG(s(N), cons(X, Z)) → 2NDSNEG(s(N), cons2(X, activate(Z)))
2NDSPOS(s(N), cons(X, Z)) → 2NDSPOS(s(N), cons2(X, activate(Z)))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, Z)) → 2ndspos(s(N), cons2(X, activate(Z)))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, activate(Z)))
2ndsneg(s(N), cons2(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, activate(Z)))
pi(X) → 2ndspos(X, from(0))
plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
times(0, Y) → 0
times(s(X), Y) → plus(Y, times(X, Y))
square(X) → times(X, X)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) QDPOrderProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPOrderProof (EQUIVALENT transformation)

(9) Obligation:

(10) PisEmptyProof (EQUIVALENT transformation)

(11) TRUE

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) PisEmptyProof (EQUIVALENT transformation)

(16) TRUE

(17) Obligation:

(18) QDPOrderProof (EQUIVALENT transformation)

(19) Obligation:

(20) PisEmptyProof (EQUIVALENT transformation)

(21) TRUE

(22) Obligation:

(23) QDPOrderProof (EQUIVALENT transformation)

(24) Obligation:

(25) DependencyGraphProof (EQUIVALENT transformation)

(26) TRUE