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

(0) Obligation:

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

from(X) → cons(X, n__from(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(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(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) → 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(X)
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(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 3 SCCs with 11 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(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(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)  =  x1
s(x1)  =  s(x1)
from(x1)  =  x1
cons(x1, x2)  =  cons
n__from(x1)  =  x1
2ndspos(x1, x2)  =  x1
0  =  0
rnil  =  rnil
n__cons(x1, x2)  =  n__cons
rcons(x1, x2)  =  x1
posrecip(x1)  =  posrecip
activate(x1)  =  activate(x1)
2ndsneg(x1, x2)  =  2ndsneg(x2)
negrecip(x1)  =  negrecip
pi(x1)  =  pi(x1)
plus(x1, x2)  =  plus(x1, x2)
times(x1, x2)  =  times(x1, x2)
square(x1)  =  square(x1)

Lexicographic Path Order [LPO].
Precedence:
activate1 > [cons, 0, rnil, ncons, posrecip, 2ndsneg1, negrecip]
pi1 > [cons, 0, rnil, ncons, posrecip, 2ndsneg1, negrecip]
square1 > times2 > plus2 > s1 > [cons, 0, rnil, ncons, posrecip, 2ndsneg1, negrecip]


The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(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(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(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(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(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)  =  x1
cons(x1, x2)  =  x1
n__from(x1)  =  x1
2ndspos(x1, x2)  =  2ndspos
0  =  0
rnil  =  rnil
n__cons(x1, x2)  =  x1
rcons(x1, x2)  =  x2
posrecip(x1)  =  x1
activate(x1)  =  x1
2ndsneg(x1, x2)  =  2ndsneg
negrecip(x1)  =  negrecip
pi(x1)  =  pi(x1)
plus(x1, x2)  =  plus(x1, x2)
times(x1, x2)  =  times(x1, x2)
square(x1)  =  square(x1)

Lexicographic Path Order [LPO].
Precedence:
0 > [2ndspos, rnil, 2ndsneg, negrecip]
pi1 > [2ndspos, rnil, 2ndsneg, negrecip]
square1 > times2 > plus2 > [TIMES2, s1] > [2ndspos, rnil, 2ndsneg, negrecip]


The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(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(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(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:

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(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(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.


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

Lexicographic Path Order [LPO].
Precedence:
[2NDSPOS1, 2NDSNEG1] > activate1 > [cons, ncons, from, nfrom, 2ndspos, 0, rnil, posrecip, 2ndsneg]
pi > [cons, ncons, from, nfrom, 2ndspos, 0, rnil, posrecip, 2ndsneg]
square1 > times2 > plus2 > s1 > activate1 > [cons, ncons, from, nfrom, 2ndspos, 0, rnil, posrecip, 2ndsneg]


The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, n__from(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

(17) Obligation:

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

from(X) → cons(X, n__from(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)
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__cons(X1, X2)) → cons(X1, 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