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:

f2(s1(X), Y) -> h1(s1(f2(h1(Y), X)))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

f2(s1(X), Y) -> h1(s1(f2(h1(Y), 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:

f2(s1(X), Y) -> h1(s1(f2(h1(Y), X)))

The set Q consists of the following terms:

f2(s1(x0), x1)


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

F2(s1(X), Y) -> F2(h1(Y), X)

The TRS R consists of the following rules:

f2(s1(X), Y) -> h1(s1(f2(h1(Y), X)))

The set Q consists of the following terms:

f2(s1(x0), x1)

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:

F2(s1(X), Y) -> F2(h1(Y), X)

The TRS R consists of the following rules:

f2(s1(X), Y) -> h1(s1(f2(h1(Y), X)))

The set Q consists of the following terms:

f2(s1(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.