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

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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

F2(a, x) -> G1(x)
F2(a, x) -> F2(g1(x), x)
G1(h1(x)) -> G1(x)
H1(g1(x)) -> H1(a)

The TRS R consists of the following rules:

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

F2(a, x) -> G1(x)
F2(a, x) -> F2(g1(x), x)
G1(h1(x)) -> G1(x)
H1(g1(x)) -> H1(a)

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

f2(a, x) -> f2(g1(x), x)
h1(g1(x)) -> h1(a)
g1(h1(x)) -> g1(x)
h1(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(h1(x)) -> G1(x)
Used argument filtering: G1(x1)  =  x1
h1(x1)  =  h1(x1)
Used ordering: Quasi Precedence: trivial


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

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

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

F2(a, x) -> F2(g1(x), x)
Used argument filtering: F2(x1, x2)  =  x1
a  =  a
g1(x1)  =  g
Used ordering: Quasi Precedence: a > g


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

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

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