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)


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:

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.