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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

F2(g1(x), y) -> F2(x, y)

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof

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

F2(g1(x), y) -> F2(x, y)

The TRS R consists of the following rules:

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

Q is empty.
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(g1(x), y) -> F2(x, y)
Used argument filtering: F2(x1, x2)  =  x1
g1(x1)  =  g1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ PisEmptyProof

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

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

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