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(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

Q is empty.

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

F1(f1(x)) -> F1(g2(f1(x), x))
F1(f1(x)) -> H2(f1(x), f1(x))
H2(x, x) -> G2(x, 0)
F1(f1(x)) -> G2(f1(x), x)
F1(f1(x)) -> F1(h2(f1(x), f1(x)))

The TRS R consists of the following rules:

f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

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(f1(x)) -> F1(g2(f1(x), x))
F1(f1(x)) -> H2(f1(x), f1(x))
H2(x, x) -> G2(x, 0)
F1(f1(x)) -> G2(f1(x), x)
F1(f1(x)) -> F1(h2(f1(x), f1(x)))

The TRS R consists of the following rules:

f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

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(f1(x)) -> F1(g2(f1(x), x))
F1(f1(x)) -> F1(h2(f1(x), f1(x)))
Used argument filtering: F1(x1)  =  x1
f1(x1)  =  f1(x1)
g2(x1, x2)  =  x2
h2(x1, x2)  =  h
0  =  0
Used ordering: Quasi Precedence: f_1 > h > 0


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

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

f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

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.