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:

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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

G1(g1(x)) -> H1(g1(x))
G1(g1(x)) -> G1(h1(g1(x)))
H1(h1(x)) -> H1(f2(h1(x), x))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

G1(g1(x)) -> H1(g1(x))
G1(g1(x)) -> G1(h1(g1(x)))
H1(h1(x)) -> H1(f2(h1(x), x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPAfsSolverProof

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

G1(g1(x)) -> G1(h1(g1(x)))

The TRS R consists of the following rules:

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

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.

G1(g1(x)) -> G1(h1(g1(x)))
Used argument filtering: G1(x1)  =  x1
g1(x1)  =  g
h1(x1)  =  h
Used ordering: Quasi Precedence: g > h


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

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

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

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.