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:

f3(g1(x), y, y) -> g1(f3(x, x, y))

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

f3(g1(x), y, y) -> g1(f3(x, 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:

f3(g1(x), y, y) -> g1(f3(x, x, y))

The set Q consists of the following terms:

f3(g1(x0), x1, x1)


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

F3(g1(x), y, y) -> F3(x, x, y)

The TRS R consists of the following rules:

f3(g1(x), y, y) -> g1(f3(x, x, y))

The set Q consists of the following terms:

f3(g1(x0), x1, 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:

F3(g1(x), y, y) -> F3(x, x, y)

The TRS R consists of the following rules:

f3(g1(x), y, y) -> g1(f3(x, x, y))

The set Q consists of the following terms:

f3(g1(x0), x1, 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.

F3(g1(x), y, y) -> F3(x, x, y)
Used argument filtering: F3(x1, x2, x3)  =  x1
g1(x1)  =  g1(x1)
Used ordering: Quasi 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:

f3(g1(x), y, y) -> g1(f3(x, x, y))

The set Q consists of the following terms:

f3(g1(x0), x1, x1)

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