Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳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

(0) Obligation:

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

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) 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, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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.


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

Lexicographic Path Order [LPO].
Precedence:
from > [cons, 2ndspos, rnil, 2ndsneg]
0 > [cons, 2ndspos, rnil, 2ndsneg]
negrecip1 > [cons, 2ndspos, rnil, 2ndsneg]
square1 > times2 > plus2 > [s1, posrecip1] > [cons, 2ndspos, rnil, 2ndsneg]


The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

(9) Obligation:

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

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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:

TIMES(s(X), Y) → TIMES(X, Y)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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.


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

Lexicographic Path Order [LPO].
Precedence:
posrecip1 > [2ndspos, rnil, 2ndsneg]
negrecip1 > [2ndspos, rnil, 2ndsneg]
square1 > times2 > 0 > [2ndspos, rnil, 2ndsneg]
square1 > times2 > plus2 > [s1, from1] > [2ndspos, rnil, 2ndsneg]


The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

(14) Obligation:

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

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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:

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

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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.


2NDSNEG(s(N), cons(X, cons(Y, Z))) → 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, cons(Y, Z))) → 2NDSNEG(N, Z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
2NDSNEG(x1, x2)  =  2NDSNEG(x1)
s(x1)  =  s(x1)
cons(x1, x2)  =  x2
2NDSPOS(x1, x2)  =  x1
from(x1)  =  from
2ndspos(x1, x2)  =  2ndspos
0  =  0
rnil  =  rnil
rcons(x1, x2)  =  x2
posrecip(x1)  =  posrecip
2ndsneg(x1, x2)  =  2ndsneg
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 [LPO].
Precedence:
from > [s1, posrecip] > [2NDSNEG1, 0]
[2ndspos, rnil, 2ndsneg, pi] > [s1, posrecip] > [2NDSNEG1, 0]
negrecip1 > [2NDSNEG1, 0]
square1 > times2 > plus2 > [s1, posrecip] > [2NDSNEG1, 0]


The following usable rules [FROCOS05] were oriented:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

(19) Obligation:

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

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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:

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
2ndspos(0, Z) → rnil
2ndspos(s(N), cons(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, cons(Y, Z))) → rcons(negrecip(Y), 2ndspos(N, 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)

The set Q consists of the following terms:

from(x0)
2ndspos(0, x0)
2ndspos(s(x0), cons(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)

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