Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/nonterminSLASHCSRSLASHExAppendixB

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:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

The set Q consists of the following terms:

from₁(x0)
2ndspos₂(0, x0)
2ndspos₂(s₁(x0), cons₂(x1, x2))
2ndspos₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
2ndsneg₂(0, x0)
2ndsneg₂(s₁(x0), cons₂(x1, x2))
2ndsneg₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
pi₁(x0)
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
square₁(x0)

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

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)
PI₁(X) -> 2NDSPOS₂(X, from₁(0))
TIMES₂(s₁(X), Y) -> PLUS₂(Y, times₂(X, Y))
2NDSNEG₂(s₁(N), cons₂(X, Z)) -> 2NDSNEG₂(s₁(N), cons2₂(X, Z))
TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)
2NDSNEG₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSPOS₂(N, Z)
SQUARE₁(X) -> TIMES₂(X, X)
PI₁(X) -> FROM₁(0)
2NDSPOS₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSNEG₂(N, Z)
2NDSPOS₂(s₁(N), cons₂(X, Z)) -> 2NDSPOS₂(s₁(N), cons2₂(X, Z))
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

The set Q consists of the following terms:

from₁(x0)
2ndspos₂(0, x0)
2ndspos₂(s₁(x0), cons₂(x1, x2))
2ndspos₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
2ndsneg₂(0, x0)
2ndsneg₂(s₁(x0), cons₂(x1, x2))
2ndsneg₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
pi₁(x0)
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
square₁(x0)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)
PI₁(X) -> 2NDSPOS₂(X, from₁(0))
TIMES₂(s₁(X), Y) -> PLUS₂(Y, times₂(X, Y))
2NDSNEG₂(s₁(N), cons₂(X, Z)) -> 2NDSNEG₂(s₁(N), cons2₂(X, Z))
TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)
2NDSNEG₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSPOS₂(N, Z)
SQUARE₁(X) -> TIMES₂(X, X)
PI₁(X) -> FROM₁(0)
2NDSPOS₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSNEG₂(N, Z)
2NDSPOS₂(s₁(N), cons₂(X, Z)) -> 2NDSPOS₂(s₁(N), cons2₂(X, Z))
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

The set Q consists of the following terms:

from₁(x0)
2ndspos₂(0, x0)
2ndspos₂(s₁(x0), cons₂(x1, x2))
2ndspos₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
2ndsneg₂(0, x0)
2ndsneg₂(s₁(x0), cons₂(x1, x2))
2ndsneg₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
pi₁(x0)
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
square₁(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 4 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

The set Q consists of the following terms:

from₁(x0)
2ndspos₂(0, x0)
2ndspos₂(s₁(x0), cons₂(x1, x2))
2ndspos₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
2ndsneg₂(0, x0)
2ndsneg₂(s₁(x0), cons₂(x1, x2))
2ndsneg₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
pi₁(x0)
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
square₁(x0)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PLUS₂(s₁(X), Y) -> PLUS₂(X, Y)

Used argument filtering: PLUS₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

The set Q consists of the following terms:

from₁(x0)
2ndspos₂(0, x0)
2ndspos₂(s₁(x0), cons₂(x1, x2))
2ndspos₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
2ndsneg₂(0, x0)
2ndsneg₂(s₁(x0), cons₂(x1, x2))
2ndsneg₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
pi₁(x0)
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
square₁(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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

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

TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

The set Q consists of the following terms:

from₁(x0)
2ndspos₂(0, x0)
2ndspos₂(s₁(x0), cons₂(x1, x2))
2ndspos₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
2ndsneg₂(0, x0)
2ndsneg₂(s₁(x0), cons₂(x1, x2))
2ndsneg₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
pi₁(x0)
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
square₁(x0)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TIMES₂(s₁(X), Y) -> TIMES₂(X, Y)

Used argument filtering: TIMES₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

The set Q consists of the following terms:

from₁(x0)
2ndspos₂(0, x0)
2ndspos₂(s₁(x0), cons₂(x1, x2))
2ndspos₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
2ndsneg₂(0, x0)
2ndsneg₂(s₁(x0), cons₂(x1, x2))
2ndsneg₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
pi₁(x0)
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
square₁(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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

2NDSNEG₂(s₁(N), cons₂(X, Z)) -> 2NDSNEG₂(s₁(N), cons2₂(X, Z))
2NDSNEG₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSPOS₂(N, Z)
2NDSPOS₂(s₁(N), cons₂(X, Z)) -> 2NDSPOS₂(s₁(N), cons2₂(X, Z))
2NDSPOS₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSNEG₂(N, Z)

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

The set Q consists of the following terms:

from₁(x0)
2ndspos₂(0, x0)
2ndspos₂(s₁(x0), cons₂(x1, x2))
2ndspos₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
2ndsneg₂(0, x0)
2ndsneg₂(s₁(x0), cons₂(x1, x2))
2ndsneg₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
pi₁(x0)
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
square₁(x0)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

2NDSNEG₂(s₁(N), cons₂(X, Z)) -> 2NDSNEG₂(s₁(N), cons2₂(X, Z))
2NDSNEG₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSPOS₂(N, Z)
2NDSPOS₂(s₁(N), cons₂(X, Z)) -> 2NDSPOS₂(s₁(N), cons2₂(X, Z))
2NDSPOS₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> 2NDSNEG₂(N, Z)

Used argument filtering: 2NDSNEG₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
cons2₂(x1, x2) = x2
2NDSPOS₂(x1, x2) = x2
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

The set Q consists of the following terms:

from₁(x0)
2ndspos₂(0, x0)
2ndspos₂(s₁(x0), cons₂(x1, x2))
2ndspos₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
2ndsneg₂(0, x0)
2ndsneg₂(s₁(x0), cons₂(x1, x2))
2ndsneg₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
pi₁(x0)
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
square₁(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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

from₁(X) -> cons₂(X, from₁(s₁(X)))
2ndspos₂(0, Z) -> rnil
2ndspos₂(s₁(N), cons₂(X, Z)) -> 2ndspos₂(s₁(N), cons2₂(X, Z))
2ndspos₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(posrecip₁(Y), 2ndsneg₂(N, Z))
2ndsneg₂(0, Z) -> rnil
2ndsneg₂(s₁(N), cons₂(X, Z)) -> 2ndsneg₂(s₁(N), cons2₂(X, Z))
2ndsneg₂(s₁(N), cons2₂(X, cons₂(Y, Z))) -> rcons₂(negrecip₁(Y), 2ndspos₂(N, Z))
pi₁(X) -> 2ndspos₂(X, from₁(0))
plus₂(0, Y) -> Y
plus₂(s₁(X), Y) -> s₁(plus₂(X, Y))
times₂(0, Y) -> 0
times₂(s₁(X), Y) -> plus₂(Y, times₂(X, Y))
square₁(X) -> times₂(X, X)

The set Q consists of the following terms:

from₁(x0)
2ndspos₂(0, x0)
2ndspos₂(s₁(x0), cons₂(x1, x2))
2ndspos₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
2ndsneg₂(0, x0)
2ndsneg₂(s₁(x0), cons₂(x1, x2))
2ndsneg₂(s₁(x0), cons2₂(x1, cons₂(x2, x3)))
pi₁(x0)
plus₂(0, x0)
plus₂(s₁(x0), x1)
times₂(0, x0)
times₂(s₁(x0), x1)
square₁(x0)

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