Termination w.r.t. Q proof of TRS/TRCSREx6_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:

first(0, X) → nil
first(s(X), cons(Y)) → cons(Y)
from(X) → cons(X)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

first(0, X) → nil
first(s(X), cons(Y)) → cons(Y)
from(X) → cons(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:

first(0, X) → nil
first(s(X), cons(Y)) → cons(Y)
from(X) → cons(X)

The set Q consists of the following terms:

first(0, x0)
first(s(x0), cons(x1))
from(x0)

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

first(0, X) → nil
first(s(X), cons(Y)) → cons(Y)
from(X) → cons(X)

The set Q consists of the following terms:

first(0, x0)
first(s(x0), cons(x1))
from(x0)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ PisEmptyProof

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

first(0, X) → nil
first(s(X), cons(Y)) → cons(Y)
from(X) → cons(X)

The set Q consists of the following terms:

first(0, x0)
first(s(x0), cons(x1))
from(x0)

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