Termination w.r.t. Q proof of /tmp/tmpBudQqk/Ex1_2_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 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 PisEmptyProof (⇔)

    ↳14 TRUE

  ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), 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:

FROM(X) → CONS(X, n__from(n__s(X)))
2NDSPOS(s(N), cons(X, n__cons(Y, Z))) → ACTIVATE(Y)
2NDSPOS(s(N), cons(X, n__cons(Y, Z))) → 2NDSNEG(N, activate(Z))
2NDSPOS(s(N), cons(X, n__cons(Y, Z))) → ACTIVATE(Z)
2NDSNEG(s(N), cons(X, n__cons(Y, Z))) → ACTIVATE(Y)
2NDSNEG(s(N), cons(X, n__cons(Y, Z))) → 2NDSPOS(N, activate(Z))
2NDSNEG(s(N), cons(X, n__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)
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)

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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), 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 4 SCCs with 13 less nodes.

(4) Complex Obligation (AND)

(5) 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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), 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.

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) = 2ndspos
0 = 0
rnil = rnil
n__cons(x1, x2) = x2
rcons(x1, x2) = x2
posrecip(x1) = posrecip
activate(x1) = x1
2ndsneg(x1, x2) = 2ndsneg
negrecip(x1) = negrecip
pi(x1) = pi
plus(x1, x2) = plus(x1, x2)
times(x1, x2) = times(x1, x2)
square(x1) = square(x1)

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

square₁ > times₂ > [from, n_{_from}, 2ndspos, 0, 2ndsneg, negrecip, pi] > [s₁, n_{_s₁}] > posrecip
square₁ > times₂ > [from, n_{_from}, 2ndspos, 0, 2ndsneg, negrecip, pi] > rnil
square₁ > times₂ > plus₂ > [s₁, n_{_s₁}] > posrecip

Status:

PLUS₂: [2,1]
s₁: [1]
from: []
n_{_from}: []
n_{_s₁}: [1]
2ndspos: []
0: []
rnil: []
posrecip: []
2ndsneg: []
negrecip: []
pi: []
plus₂: [1,2]
times₂: [2,1]
square₁: [1]

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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

(7) 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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), 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:

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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), 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.

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) = 2ndspos
0 = 0
rnil = rnil
n__cons(x1, x2) = x2
rcons(x1, x2) = x2
posrecip(x1) = posrecip
activate(x1) = x1
2ndsneg(x1, x2) = 2ndsneg
negrecip(x1) = negrecip
pi(x1) = pi
plus(x1, x2) = plus(x1, x2)
times(x1, x2) = times(x1, x2)
square(x1) = square(x1)

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

square₁ > times₂ > [from, n_{_from}, 2ndspos, 0, 2ndsneg, negrecip, pi] > [s₁, n_{_s₁}] > posrecip
square₁ > times₂ > [from, n_{_from}, 2ndspos, 0, 2ndsneg, negrecip, pi] > rnil
square₁ > times₂ > plus₂ > [s₁, n_{_s₁}] > posrecip

Status:

TIMES₂: [2,1]
s₁: [1]
from: []
n_{_from}: []
n_{_s₁}: [1]
2ndspos: []
0: []
rnil: []
posrecip: []
2ndsneg: []
negrecip: []
pi: []
plus₂: [1,2]
times₂: [2,1]
square₁: [1]

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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

(12) 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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), 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:

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)

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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), 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.

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

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

[2ndspos, pi] > posrecip > [n_{_s₁}, rnil]
[2ndspos, pi] > [2ndsneg, negrecip] > [n_{_s₁}, rnil]
activate₁ > s₁ > [n_{_s₁}, rnil]
square₁ > [0, times₂] > plus₂ > s₁ > [n_{_s₁}, rnil]

Status:

n_{_s₁}: [1]
2ndspos: []
0: []
rnil: []
s₁: [1]
posrecip: []
activate₁: [1]
2ndsneg: []
negrecip: []
pi: []
plus₂: [1,2]
times₂: [2,1]
square₁: [1]

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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

(17) Obligation:

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

ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)

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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
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.

ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)

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

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

ACTIVATE₁ > [0, rnil, rcons₁, posrecip, negrecip]
[n_{_from₁}, from₁] > [n_{_cons₁}, cons₁] > 2ndsneg > [0, rnil, rcons₁, posrecip, negrecip]
[2ndspos, pi] > 2ndsneg > [0, rnil, rcons₁, posrecip, negrecip]
[times₁, square₁] > [0, rnil, rcons₁, posrecip, negrecip]

Status:

ACTIVATE₁: [1]
n_{_from₁}: [1]
n_{_cons₁}: [1]
from₁: [1]
cons₁: [1]
2ndspos: []
0: []
rnil: []
rcons₁: [1]
posrecip: []
2ndsneg: []
negrecip: []
pi: []
times₁: [1]
square₁: [1]

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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
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), cons(X, n__cons(Y, Z))) → 2NDSNEG(N, activate(Z))
2NDSNEG(s(N), cons(X, n__cons(Y, Z))) → 2NDSPOS(N, 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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
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.

2NDSNEG(s(N), cons(X, n__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) = x1
s(x1) = s(x1)
cons(x1, x2) = cons
n__cons(x1, x2) = n__cons
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) = rcons
posrecip(x1) = posrecip
2ndsneg(x1, x2) = 2ndsneg(x2)
negrecip(x1) = negrecip(x1)
pi(x1) = pi
plus(x1, x2) = plus(x1, x2)
times(x1, x2) = times(x1, x2)
square(x1) = square(x1)

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

[2ndspos, rcons, 2ndsneg₁, pi] > [cons, n_{_cons}, activate₁, from] > n_{_from} > [s₁, 2NDSNEG₁, n_{_s₁}]
[2ndspos, rcons, 2ndsneg₁, pi] > 0 > [s₁, 2NDSNEG₁, n_{_s₁}]
[2ndspos, rcons, 2ndsneg₁, pi] > rnil > [s₁, 2NDSNEG₁, n_{_s₁}]
[2ndspos, rcons, 2ndsneg₁, pi] > posrecip > [s₁, 2NDSNEG₁, n_{_s₁}]
[2ndspos, rcons, 2ndsneg₁, pi] > negrecip₁ > [s₁, 2NDSNEG₁, n_{_s₁}]
square₁ > times₂ > 0 > [s₁, 2NDSNEG₁, n_{_s₁}]
square₁ > times₂ > plus₂ > [s₁, 2NDSNEG₁, n_{_s₁}]

Status:

s₁: [1]
cons: []
n_{_cons}: []
2NDSNEG₁: [1]
activate₁: [1]
from: []
n_{_from}: []
n_{_s₁}: [1]
2ndspos: []
0: []
rnil: []
rcons: []
posrecip: []
2ndsneg₁: [1]
negrecip₁: [1]
pi: []
plus₂: [2,1]
times₂: [2,1]
square₁: [1]

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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

(24) Obligation:

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

2NDSPOS(s(N), cons(X, n__cons(Y, Z))) → 2NDSNEG(N, 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, n__cons(Y, Z))) → rcons(posrecip(activate(Y)), 2ndsneg(N, activate(Z)))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, n__cons(Y, Z))) → rcons(negrecip(activate(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
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 1 less node.

(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) PisEmptyProof (EQUIVALENT transformation)

(9) TRUE

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) PisEmptyProof (EQUIVALENT transformation)

(14) TRUE

(15) Obligation:

(16) QDPOrderProof (EQUIVALENT transformation)

(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