Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/nonterminSLASHCSRSLASHEx15

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:

and₂(true, X) -> X
and₂(false, Y) -> false
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

and₂(true, X) -> X
and₂(false, Y) -> false
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(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:

and₂(true, X) -> X
and₂(false, Y) -> false
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

and₂(true, x0)
and₂(false, x0)
if₃(true, x0, x1)
if₃(false, x0, x1)
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₂(x1, x2))
from₁(x0)

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

ADD₂(s₁(X), Y) -> ADD₂(X, Y)
FIRST₂(s₁(X), cons₂(Y, Z)) -> FIRST₂(X, Z)
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

and₂(true, X) -> X
and₂(false, Y) -> false
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

and₂(true, x0)
and₂(false, x0)
if₃(true, x0, x1)
if₃(false, x0, x1)
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₂(x1, x2))
from₁(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:

ADD₂(s₁(X), Y) -> ADD₂(X, Y)
FIRST₂(s₁(X), cons₂(Y, Z)) -> FIRST₂(X, Z)
FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

and₂(true, X) -> X
and₂(false, Y) -> false
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

and₂(true, x0)
and₂(false, x0)
if₃(true, x0, x1)
if₃(false, x0, x1)
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₂(x1, x2))
from₁(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ 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:

and₂(true, X) -> X
and₂(false, Y) -> false
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

and₂(true, x0)
and₂(false, x0)
if₃(true, x0, x1)
if₃(false, x0, x1)
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₂(x1, x2))
from₁(x0)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

FIRST₂(s₁(X), cons₂(Y, Z)) -> FIRST₂(X, Z)

The TRS R consists of the following rules:

and₂(true, X) -> X
and₂(false, Y) -> false
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

and₂(true, x0)
and₂(false, x0)
if₃(true, x0, x1)
if₃(false, x0, x1)
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₂(x1, x2))
from₁(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.

FIRST₂(s₁(X), cons₂(Y, Z)) -> FIRST₂(X, Z)

Used argument filtering: FIRST₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

and₂(true, X) -> X
and₂(false, Y) -> false
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

and₂(true, x0)
and₂(false, x0)
if₃(true, x0, x1)
if₃(false, x0, x1)
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₂(x1, x2))
from₁(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

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

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

The TRS R consists of the following rules:

and₂(true, X) -> X
and₂(false, Y) -> false
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

and₂(true, x0)
and₂(false, x0)
if₃(true, x0, x1)
if₃(false, x0, x1)
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₂(x1, x2))
from₁(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.

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

Used argument filtering: ADD₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

and₂(true, X) -> X
and₂(false, Y) -> false
if₃(true, X, Y) -> X
if₃(false, X, Y) -> Y
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
first₂(0, X) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

and₂(true, x0)
and₂(false, x0)
if₃(true, x0, x1)
if₃(false, x0, x1)
add₂(0, x0)
add₂(s₁(x0), x1)
first₂(0, x0)
first₂(s₁(x0), cons₂(x1, x2))
from₁(x0)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.