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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

f(x, y, w, w, a) → g1(x, x, y, w)
f(x, y, w, a, a) → g1(y, x, x, w)
f(x, y, a, a, w) → g2(x, y, y, w)
f(x, y, a, w, w) → g2(y, y, x, w)
g1(x, x, y, a) → h(x, y)
g1(y, x, x, a) → h(x, y)
g2(x, y, y, a) → h(x, y)
g2(y, y, x, a) → 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.