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) -> g13(x, x, y)
f2(x, y) -> g13(y, x, x)
f2(x, y) -> g23(x, y, y)
f2(x, y) -> g23(y, y, x)
g13(x, x, y) -> h2(x, y)
g13(y, x, x) -> h2(x, y)
g23(x, y, y) -> h2(x, y)
g23(y, y, x) -> h2(x, y)
h2(x, x) -> x

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f2(x, y) -> g13(x, x, y)
f2(x, y) -> g13(y, x, x)
f2(x, y) -> g23(x, y, y)
f2(x, y) -> g23(y, y, x)
g13(x, x, y) -> h2(x, y)
g13(y, x, x) -> h2(x, y)
g23(x, y, y) -> h2(x, y)
g23(y, y, x) -> h2(x, y)
h2(x, x) -> x

Q is empty.

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

G23(x, y, y) -> H2(x, y)
F2(x, y) -> G13(y, x, x)
F2(x, y) -> G13(x, x, y)
G13(x, x, y) -> H2(x, y)
G13(y, x, x) -> H2(x, y)
F2(x, y) -> G23(y, y, x)
F2(x, y) -> G23(x, y, y)
G23(y, y, x) -> H2(x, y)

The TRS R consists of the following rules:

f2(x, y) -> g13(x, x, y)
f2(x, y) -> g13(y, x, x)
f2(x, y) -> g23(x, y, y)
f2(x, y) -> g23(y, y, x)
g13(x, x, y) -> h2(x, y)
g13(y, x, x) -> h2(x, y)
g23(x, y, y) -> h2(x, y)
g23(y, y, x) -> h2(x, y)
h2(x, 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:

G23(x, y, y) -> H2(x, y)
F2(x, y) -> G13(y, x, x)
F2(x, y) -> G13(x, x, y)
G13(x, x, y) -> H2(x, y)
G13(y, x, x) -> H2(x, y)
F2(x, y) -> G23(y, y, x)
F2(x, y) -> G23(x, y, y)
G23(y, y, x) -> H2(x, y)

The TRS R consists of the following rules:

f2(x, y) -> g13(x, x, y)
f2(x, y) -> g13(y, x, x)
f2(x, y) -> g23(x, y, y)
f2(x, y) -> g23(y, y, x)
g13(x, x, y) -> h2(x, y)
g13(y, x, x) -> h2(x, y)
g23(x, y, y) -> h2(x, y)
g23(y, y, x) -> h2(x, y)
h2(x, x) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 8 less nodes.