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:

f(x, y) → g1(x, x, y)
f(x, y) → g1(y, x, x)
f(x, y) → g2(x, y, y)
f(x, y) → g2(y, y, x)
g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)
h(x, x) → x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(x, y) → g1(x, x, y)
f(x, y) → g1(y, x, x)
f(x, y) → g2(x, y, y)
f(x, y) → g2(y, y, x)
g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)
h(x, x) → x

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(x, y) → G2(y, y, x)
F(x, y) → G2(x, y, y)
G2(x, y, y) → H(x, y)
G2(y, y, x) → H(x, y)
G1(y, x, x) → H(x, y)
F(x, y) → G1(y, x, x)
F(x, y) → G1(x, x, y)
G1(x, x, y) → H(x, y)

The TRS R consists of the following rules:

f(x, y) → g1(x, x, y)
f(x, y) → g1(y, x, x)
f(x, y) → g2(x, y, y)
f(x, y) → g2(y, y, x)
g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)
h(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:

F(x, y) → G2(y, y, x)
F(x, y) → G2(x, y, y)
G2(x, y, y) → H(x, y)
G2(y, y, x) → H(x, y)
G1(y, x, x) → H(x, y)
F(x, y) → G1(y, x, x)
F(x, y) → G1(x, x, y)
G1(x, x, y) → H(x, y)

The TRS R consists of the following rules:

f(x, y) → g1(x, x, y)
f(x, y) → g1(y, x, x)
f(x, y) → g2(x, y, y)
f(x, y) → g2(y, y, x)
g1(x, x, y) → h(x, y)
g1(y, x, x) → h(x, y)
g2(x, y, y) → h(x, y)
g2(y, y, x) → h(x, y)
h(x, x) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 8 less nodes.