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:

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, x)

Q is empty.

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

F1(h2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> G2(f1(x), f1(y))
F1(h2(x, y)) -> G2(h2(y, f1(x)), h2(x, f1(y)))
F1(g2(x, y)) -> F1(x)
F1(h2(x, y)) -> F1(x)

The TRS R consists of the following rules:

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, 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:

F1(h2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> G2(f1(x), f1(y))
F1(h2(x, y)) -> G2(h2(y, f1(x)), h2(x, f1(y)))
F1(g2(x, y)) -> F1(x)
F1(h2(x, y)) -> F1(x)

The TRS R consists of the following rules:

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, 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:

F1(h2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(x)
F1(h2(x, y)) -> F1(x)

The TRS R consists of the following rules:

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, 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.

F1(h2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(y)
F1(g2(x, y)) -> F1(x)
F1(h2(x, y)) -> F1(x)
Used argument filtering: F1(x1)  =  x1
h2(x1, x2)  =  h2(x1, x2)
g2(x1, x2)  =  g2(x1, x2)
Used ordering: Quasi Precedence: trivial 


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

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

f1(a) -> b
f1(c) -> d
f1(g2(x, y)) -> g2(f1(x), f1(y))
f1(h2(x, y)) -> g2(h2(y, f1(x)), h2(x, f1(y)))
g2(x, x) -> h2(e, 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.