Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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.


↳ QTRS
  ↳ Overlay + Local Confluence

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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(x1, cons(x2, x3)))
pi(x0)
plus(0, x0)
plus(s(x0), x1)
times(0, x0)
times(s(x0), x1)
square(x0)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

PLUS(s(X), Y) → PLUS(X, Y)
PI(X) → 2NDSPOS(X, from(0))
TIMES(s(X), Y) → PLUS(Y, times(X, Y))
2NDSNEG(s(N), cons(X, Z)) → 2NDSNEG(s(N), cons2(X, Z))
TIMES(s(X), Y) → TIMES(X, Y)
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → 2NDSPOS(N, Z)
SQUARE(X) → TIMES(X, X)
PI(X) → FROM(0)
2NDSPOS(s(N), cons2(X, cons(Y, Z))) → 2NDSNEG(N, Z)
2NDSPOS(s(N), cons(X, Z)) → 2NDSPOS(s(N), cons2(X, Z))
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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(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.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

PLUS(s(X), Y) → PLUS(X, Y)
PI(X) → 2NDSPOS(X, from(0))
TIMES(s(X), Y) → PLUS(Y, times(X, Y))
2NDSNEG(s(N), cons(X, Z)) → 2NDSNEG(s(N), cons2(X, Z))
TIMES(s(X), Y) → TIMES(X, Y)
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → 2NDSPOS(N, Z)
SQUARE(X) → TIMES(X, X)
PI(X) → FROM(0)
2NDSPOS(s(N), cons2(X, cons(Y, Z))) → 2NDSNEG(N, Z)
2NDSPOS(s(N), cons(X, Z)) → 2NDSPOS(s(N), cons2(X, Z))
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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

PI(X) → 2NDSPOS(X, from(0))
PLUS(s(X), Y) → PLUS(X, Y)
2NDSNEG(s(N), cons(X, Z)) → 2NDSNEG(s(N), cons2(X, Z))
TIMES(s(X), Y) → PLUS(Y, times(X, Y))
TIMES(s(X), Y) → TIMES(X, Y)
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → 2NDSPOS(N, Z)
PI(X) → FROM(0)
SQUARE(X) → TIMES(X, X)
2NDSPOS(s(N), cons(X, Z)) → 2NDSPOS(s(N), cons2(X, Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) → 2NDSNEG(N, Z)
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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(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.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 4 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(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.
We use the reduction pair processor [13].


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.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(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.
We use the reduction pair processor [13].


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.
none
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

2NDSNEG(s(N), cons(X, Z)) → 2NDSNEG(s(N), cons2(X, Z))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, Z)) → 2NDSPOS(s(N), cons2(X, Z))
2NDSPOS(s(N), cons2(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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


2NDSNEG(s(N), cons(X, Z)) → 2NDSNEG(s(N), cons2(X, Z))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) → 2NDSPOS(N, Z)
2NDSPOS(s(N), cons(X, Z)) → 2NDSPOS(s(N), cons2(X, Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) → 2NDSNEG(N, Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
2NDSNEG(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)
cons2(x1, x2)  =  x2
2NDSPOS(x1, x2)  =  x2

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(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.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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, Z)) → 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) → rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) → rnil
2ndsneg(s(N), cons(X, Z)) → 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(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, x2))
2ndspos(s(x0), cons2(x1, cons(x2, x3)))
2ndsneg(0, x0)
2ndsneg(s(x0), cons(x1, x2))
2ndsneg(s(x0), cons2(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.