Termination w.r.t. Q proof of TRS/various04.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f1 → g1
f1 → g2
f2 → g1
f2 → g2
g1 → h1
g1 → h2
g2 → h1
g2 → h2
h1 → i
h2 → i
e1(h1, h2, x, y, z) → e2(x, x, y, z, z)
e1(x1, x1, x, y, z) → e5(x1, x, y, z)
e2(f1, x, y, z, f2) → e3(x, y, x, y, y, z, y, z, x, y, z)
e2(x, x, y, z, z) → e6(x, y, z)
e2(i, x, y, z, i) → e6(x, y, z)
e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
e3(x, y, x, y, y, z, y, z, x, y, z) → e6(x, y, z)
e4(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) → e1(x1, x1, x, y, z)
e4(i, x1, i, x1, i, x1, i, x1, x, y, z) → e5(x1, x, y, z)
e4(x, x, x, x, x, x, x, x, x, x, x) → e6(x, x, x)
e5(i, x, y, z) → e6(x, y, z)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f1 → g1
f1 → g2
f2 → g1
f2 → g2
g1 → h1
g1 → h2
g2 → h1
g2 → h2
h1 → i
h2 → i
e1(h1, h2, x, y, z) → e2(x, x, y, z, z)
e1(x1, x1, x, y, z) → e5(x1, x, y, z)
e2(f1, x, y, z, f2) → e3(x, y, x, y, y, z, y, z, x, y, z)
e2(x, x, y, z, z) → e6(x, y, z)
e2(i, x, y, z, i) → e6(x, y, z)
e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
e3(x, y, x, y, y, z, y, z, x, y, z) → e6(x, y, z)
e4(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) → e1(x1, x1, x, y, z)
e4(i, x1, i, x1, i, x1, i, x1, x, y, z) → e5(x1, x, y, z)
e4(x, x, x, x, x, x, x, x, x, x, x) → e6(x, x, x)
e5(i, x, y, z) → e6(x, y, z)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F1 → G2
F2 → G1
E2(f1, x, y, z, f2) → E3(x, y, x, y, y, z, y, z, x, y, z)
G1 → H2
G2 → H1
E1(x1, x1, x, y, z) → E5(x1, x, y, z)
E1(h1, h2, x, y, z) → E2(x, x, y, z, z)
G1 → H1
F2 → G2
G2 → H2
E3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → E4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
E4(i, x1, i, x1, i, x1, i, x1, x, y, z) → E5(x1, x, y, z)
F1 → G1
E4(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) → E1(x1, x1, x, y, z)

The TRS R consists of the following rules:

f1 → g1
f1 → g2
f2 → g1
f2 → g2
g1 → h1
g1 → h2
g2 → h1
g2 → h2
h1 → i
h2 → i
e1(h1, h2, x, y, z) → e2(x, x, y, z, z)
e1(x1, x1, x, y, z) → e5(x1, x, y, z)
e2(f1, x, y, z, f2) → e3(x, y, x, y, y, z, y, z, x, y, z)
e2(x, x, y, z, z) → e6(x, y, z)
e2(i, x, y, z, i) → e6(x, y, z)
e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
e3(x, y, x, y, y, z, y, z, x, y, z) → e6(x, y, z)
e4(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) → e1(x1, x1, x, y, z)
e4(i, x1, i, x1, i, x1, i, x1, x, y, z) → e5(x1, x, y, z)
e4(x, x, x, x, x, x, x, x, x, x, x) → e6(x, x, x)
e5(i, x, y, z) → e6(x, y, z)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

F1 → G2
F2 → G1
E2(f1, x, y, z, f2) → E3(x, y, x, y, y, z, y, z, x, y, z)
G1 → H2
G2 → H1
E1(x1, x1, x, y, z) → E5(x1, x, y, z)
E1(h1, h2, x, y, z) → E2(x, x, y, z, z)
G1 → H1
F2 → G2
G2 → H2
E3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → E4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
E4(i, x1, i, x1, i, x1, i, x1, x, y, z) → E5(x1, x, y, z)
F1 → G1
E4(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) → E1(x1, x1, x, y, z)

The TRS R consists of the following rules:

f1 → g1
f1 → g2
f2 → g1
f2 → g2
g1 → h1
g1 → h2
g2 → h1
g2 → h2
h1 → i
h2 → i
e1(h1, h2, x, y, z) → e2(x, x, y, z, z)
e1(x1, x1, x, y, z) → e5(x1, x, y, z)
e2(f1, x, y, z, f2) → e3(x, y, x, y, y, z, y, z, x, y, z)
e2(x, x, y, z, z) → e6(x, y, z)
e2(i, x, y, z, i) → e6(x, y, z)
e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
e3(x, y, x, y, y, z, y, z, x, y, z) → e6(x, y, z)
e4(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) → e1(x1, x1, x, y, z)
e4(i, x1, i, x1, i, x1, i, x1, x, y, z) → e5(x1, x, y, z)
e4(x, x, x, x, x, x, x, x, x, x, x) → e6(x, x, x)
e5(i, x, y, z) → e6(x, y, z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F1 → G2
F2 → G1
E2(f1, x, y, z, f2) → E3(x, y, x, y, y, z, y, z, x, y, z)
G1 → H2
E1(x1, x1, x, y, z) → E5(x1, x, y, z)
G2 → H1
E1(h1, h2, x, y, z) → E2(x, x, y, z, z)
G1 → H1
F2 → G2
E3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → E4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
G2 → H2
F1 → G1
E4(i, x1, i, x1, i, x1, i, x1, x, y, z) → E5(x1, x, y, z)
E4(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) → E1(x1, x1, x, y, z)

The TRS R consists of the following rules:

f1 → g1
f1 → g2
f2 → g1
f2 → g2
g1 → h1
g1 → h2
g2 → h1
g2 → h2
h1 → i
h2 → i
e1(h1, h2, x, y, z) → e2(x, x, y, z, z)
e1(x1, x1, x, y, z) → e5(x1, x, y, z)
e2(f1, x, y, z, f2) → e3(x, y, x, y, y, z, y, z, x, y, z)
e2(x, x, y, z, z) → e6(x, y, z)
e2(i, x, y, z, i) → e6(x, y, z)
e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
e3(x, y, x, y, y, z, y, z, x, y, z) → e6(x, y, z)
e4(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) → e1(x1, x1, x, y, z)
e4(i, x1, i, x1, i, x1, i, x1, x, y, z) → e5(x1, x, y, z)
e4(x, x, x, x, x, x, x, x, x, x, x) → e6(x, x, x)
e5(i, x, y, z) → e6(x, y, z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

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

E1(h1, h2, x, y, z) → E2(x, x, y, z, z)
E3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → E4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
E4(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) → E1(x1, x1, x, y, z)
E2(f1, x, y, z, f2) → E3(x, y, x, y, y, z, y, z, x, y, z)

The TRS R consists of the following rules:

f1 → g1
f1 → g2
f2 → g1
f2 → g2
g1 → h1
g1 → h2
g2 → h1
g2 → h2
h1 → i
h2 → i
e1(h1, h2, x, y, z) → e2(x, x, y, z, z)
e1(x1, x1, x, y, z) → e5(x1, x, y, z)
e2(f1, x, y, z, f2) → e3(x, y, x, y, y, z, y, z, x, y, z)
e2(x, x, y, z, z) → e6(x, y, z)
e2(i, x, y, z, i) → e6(x, y, z)
e3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → e4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
e3(x, y, x, y, y, z, y, z, x, y, z) → e6(x, y, z)
e4(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) → e1(x1, x1, x, y, z)
e4(i, x1, i, x1, i, x1, i, x1, x, y, z) → e5(x1, x, y, z)
e4(x, x, x, x, x, x, x, x, x, x, x) → e6(x, x, x)
e5(i, x, y, z) → e6(x, y, z)

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