Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx6_Luc98

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

first₂(0, X) -> nil
first₂(s₁(X), cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

first₂(0, X) -> nil
first₂(s₁(X), cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(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:

first₂(0, X) -> nil
first₂(s₁(X), cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)

The set Q consists of the following terms:

first₂(0, x0)
first₂(s₁(x0), cons₁(x1))
from₁(x0)

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

first₂(0, X) -> nil
first₂(s₁(X), cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)

The set Q consists of the following terms:

first₂(0, x0)
first₂(s₁(x0), cons₁(x1))
from₁(x0)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ PisEmptyProof

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

first₂(0, X) -> nil
first₂(s₁(X), cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)

The set Q consists of the following terms:

first₂(0, x0)
first₂(s₁(x0), cons₁(x1))
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.