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(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

f2(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)

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(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)

The set Q consists of the following terms:

f2(x0, g1(x0))
f2(x0, h1(x1))


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

F2(x, h1(y)) -> F2(h1(x), y)

The TRS R consists of the following rules:

f2(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)

The set Q consists of the following terms:

f2(x0, g1(x0))
f2(x0, h1(x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ QDPAfsSolverProof

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

F2(x, h1(y)) -> F2(h1(x), y)

The TRS R consists of the following rules:

f2(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)

The set Q consists of the following terms:

f2(x0, g1(x0))
f2(x0, h1(x1))

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.

F2(x, h1(y)) -> F2(h1(x), y)
Used argument filtering: F2(x1, x2)  =  x2
h1(x1)  =  h1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPAfsSolverProof
QDP
              ↳ PisEmptyProof

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

f2(x, g1(x)) -> x
f2(x, h1(y)) -> f2(h1(x), y)

The set Q consists of the following terms:

f2(x0, g1(x0))
f2(x0, h1(x1))

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