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) -> x
g1(a) -> h3(a, b, a)
i1(x) -> f2(x, x)
h3(x, x, y) -> g1(x)

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

f2(x, y) -> x
g1(a) -> h3(a, b, a)
i1(x) -> f2(x, x)
h3(x, x, y) -> g1(x)

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

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

f2(x, y) -> x
g1(a) -> h3(a, b, a)
i1(x) -> f2(x, x)
h3(x, x, y) -> g1(x)

The set Q consists of the following terms:

f2(x0, x1)
g1(a)
i1(x0)
h3(x0, x0, x1)


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

G1(a) -> H3(a, b, a)
H3(x, x, y) -> G1(x)
I1(x) -> F2(x, x)

The TRS R consists of the following rules:

f2(x, y) -> x
g1(a) -> h3(a, b, a)
i1(x) -> f2(x, x)
h3(x, x, y) -> g1(x)

The set Q consists of the following terms:

f2(x0, x1)
g1(a)
i1(x0)
h3(x0, x0, x1)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

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

G1(a) -> H3(a, b, a)
H3(x, x, y) -> G1(x)
I1(x) -> F2(x, x)

The TRS R consists of the following rules:

f2(x, y) -> x
g1(a) -> h3(a, b, a)
i1(x) -> f2(x, x)
h3(x, x, y) -> g1(x)

The set Q consists of the following terms:

f2(x0, x1)
g1(a)
i1(x0)
h3(x0, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 3 less nodes.