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

Lexicographic path order with status [LPO].
Quasi-Precedence:
activate1 > from1 > nfrom1 > [0, cons2, negrecip]
pi1 > from1 > nfrom1 > [0, cons2, negrecip]
pi1 > [2ndspos2, rnil, 2ndsneg2] > rcons > [0, cons2, negrecip]
pi1 > [2ndspos2, rnil, 2ndsneg2] > posrecip1 > [0, cons2, negrecip]
square1 > [0, cons2, negrecip]

Status:
nfrom1: [1]
from1: [1]
2ndspos2: [2,1]
0: []
rnil: []
cons2: []
activate1: [1]
rcons: []
posrecip1: [1]
2ndsneg2: [1,2]
negrecip: []
pi1: [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, 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__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.

(8) 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)
from(x1)  =  from
cons(x1, x2)  =  cons
n__from(x1)  =  n__from
2ndspos(x1, x2)  =  2ndspos
0  =  0
rnil  =  rnil
s(x1)  =  s(x1)
cons2(x1, x2)  =  cons2(x1, x2)
activate(x1)  =  activate(x1)
rcons(x1, x2)  =  rcons
posrecip(x1)  =  posrecip
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:
activate1 > [from, nfrom] > ns1
activate1 > [from, nfrom] > cons > [2ndspos, cons22, rcons] > [0, rnil]
activate1 > [from, nfrom] > cons > [2ndspos, cons22, rcons] > posrecip
activate1 > [from, nfrom] > cons > negrecip
activate1 > s1 > ns1
activate1 > s1 > 2ndsneg > [2ndspos, cons22, rcons] > [0, rnil]
activate1 > s1 > 2ndsneg > [2ndspos, cons22, rcons] > posrecip
activate1 > s1 > 2ndsneg > negrecip
pi > [2ndspos, cons22, rcons] > [0, rnil]
pi > [2ndspos, cons22, rcons] > posrecip
square1 > times2 > plus2 > s1 > ns1
square1 > times2 > plus2 > s1 > 2ndsneg > [2ndspos, cons22, rcons] > [0, rnil]
square1 > times2 > plus2 > s1 > 2ndsneg > [2ndspos, cons22, rcons] > posrecip
square1 > times2 > plus2 > s1 > 2ndsneg > negrecip

Status:
ACTIVATE1: [1]
ns1: [1]
from: []
cons: []
nfrom: []
2ndspos: []
0: []
rnil: []
s1: [1]
cons22: [2,1]
activate1: [1]
rcons: []
posrecip: []
2ndsneg: []
negrecip: []
pi: []
plus2: [2,1]
times2: [1,2]
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, 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)
s(x1)  =  s(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  cons
n__from(x1)  =  n__from(x1)
n__s(x1)  =  n__s(x1)
2ndspos(x1, x2)  =  x2
0  =  0
rnil  =  rnil
cons2(x1, x2)  =  cons2
activate(x1)  =  x1
rcons(x1, x2)  =  rcons
posrecip(x1)  =  posrecip(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:
PLUS1 > rnil
posrecip1 > rnil
2ndsneg > [s1, ns1] > rcons > rnil
2ndsneg > cons2 > rcons > rnil
2ndsneg > negrecip > rnil
pi > [from1, cons, nfrom1] > [s1, ns1] > rcons > rnil
pi > [from1, cons, nfrom1] > cons2 > rcons > rnil
pi > [0, times2] > plus2 > [s1, ns1] > rcons > rnil
square1 > [0, times2] > plus2 > [s1, ns1] > rcons > rnil

Status:
PLUS1: [1]
s1: [1]
from1: [1]
cons: []
nfrom1: [1]
ns1: [1]
0: []
rnil: []
cons2: []
rcons: []
posrecip1: [1]
2ndsneg: []
negrecip: []
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, 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)
s(x1)  =  s(x1)
from(x1)  =  from(x1)
cons(x1, x2)  =  cons
n__from(x1)  =  n__from(x1)
n__s(x1)  =  n__s(x1)
2ndspos(x1, x2)  =  x2
0  =  0
rnil  =  rnil
cons2(x1, x2)  =  cons2
activate(x1)  =  x1
rcons(x1, x2)  =  rcons
posrecip(x1)  =  posrecip(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:
TIMES1 > rnil
2ndsneg > [s1, ns1] > rcons > rnil
2ndsneg > cons2 > rcons > rnil
2ndsneg > negrecip > rnil
pi > [from1, cons, nfrom1] > [s1, ns1] > rcons > rnil
pi > [from1, cons, nfrom1] > cons2 > rcons > rnil
pi > [from1, cons, nfrom1] > posrecip1 > rnil
pi > [0, times2] > plus2 > [s1, ns1] > rcons > rnil
square1 > [0, times2] > plus2 > [s1, ns1] > rcons > rnil

Status:
TIMES1: [1]
s1: [1]
from1: [1]
cons: []
nfrom1: [1]
ns1: [1]
0: []
rnil: []
cons2: []
rcons: []
posrecip1: [1]
2ndsneg: []
negrecip: []
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, 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))
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)  =  x2
cons(x1, x2)  =  x2
2NDSNEG(x1, x2)  =  x1
activate(x1)  =  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
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, 0, rnil, 2ndsneg, negrecip, pi] > [2NDSPOS1, s1, ns1] > [from, nfrom]
[2ndspos, 0, rnil, 2ndsneg, negrecip, pi] > rcons > [from, nfrom]
posrecip > [from, nfrom]
square1 > times2 > plus2 > [2NDSPOS1, s1, ns1] > [from, nfrom]

Status:
2NDSPOS1: [1]
s1: [1]
from: []
nfrom: []
ns1: [1]
2ndspos: []
0: []
rnil: []
rcons: []
posrecip: []
2ndsneg: []
negrecip: []
pi: []
plus2: [1,2]
times2: [1,2]
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, 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)))
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.

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

(26) TRUE