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, y) -> g2(x, y)
g2(h1(x), y) -> h1(f2(x, y))
g2(h1(x), y) -> h1(g2(x, y))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f2(x, y) -> g2(x, y)
g2(h1(x), y) -> h1(f2(x, y))
g2(h1(x), y) -> h1(g2(x, y))

Q is empty.

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

G2(h1(x), y) -> G2(x, y)
F2(x, y) -> G2(x, y)
G2(h1(x), y) -> F2(x, y)

The TRS R consists of the following rules:

f2(x, y) -> g2(x, y)
g2(h1(x), y) -> h1(f2(x, y))
g2(h1(x), y) -> h1(g2(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:

G2(h1(x), y) -> G2(x, y)
F2(x, y) -> G2(x, y)
G2(h1(x), y) -> F2(x, y)

The TRS R consists of the following rules:

f2(x, y) -> g2(x, y)
g2(h1(x), y) -> h1(f2(x, y))
g2(h1(x), y) -> h1(g2(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.

G2(h1(x), y) -> G2(x, y)
G2(h1(x), y) -> F2(x, y)
Used argument filtering: G2(x1, x2)  =  x1
h1(x1)  =  h1(x1)
F2(x1, x2)  =  x1
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
QDP
          ↳ DependencyGraphProof

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

F2(x, y) -> G2(x, y)

The TRS R consists of the following rules:

f2(x, y) -> g2(x, y)
g2(h1(x), y) -> h1(f2(x, y))
g2(h1(x), y) -> h1(g2(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.