Termination w.r.t. Q proof of TRS/various05.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
e1₅(h1, h2, x, y, z) -> e2₅(x, x, y, z, z)
e2₅(f1, x, y, z, f2) -> e3₁₁(x, y, x, y, y, z, y, z, x, y, z)
e3₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) -> e4₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
e4₁₁(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) -> e1₅(x1, x1, 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
e1₅(h1, h2, x, y, z) -> e2₅(x, x, y, z, z)
e2₅(f1, x, y, z, f2) -> e3₁₁(x, y, x, y, y, z, y, z, x, y, z)
e3₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) -> e4₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
e4₁₁(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) -> e1₅(x1, x1, 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:

E1₅(h1, h2, x, y, z) -> E2₅(x, x, y, z, z)
F1 -> G2
F2 -> G2
F2 -> G1
E3₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) -> E4₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
F1 -> G1
E4₁₁(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) -> E1₅(x1, x1, x, y, z)
E2₅(f1, x, y, z, f2) -> E3₁₁(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
e1₅(h1, h2, x, y, z) -> e2₅(x, x, y, z, z)
e2₅(f1, x, y, z, f2) -> e3₁₁(x, y, x, y, y, z, y, z, x, y, z)
e3₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) -> e4₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
e4₁₁(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) -> e1₅(x1, x1, x, y, z)

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:

E1₅(h1, h2, x, y, z) -> E2₅(x, x, y, z, z)
F1 -> G2
F2 -> G2
F2 -> G1
E3₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) -> E4₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
F1 -> G1
E4₁₁(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) -> E1₅(x1, x1, x, y, z)
E2₅(f1, x, y, z, f2) -> E3₁₁(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
e1₅(h1, h2, x, y, z) -> e2₅(x, x, y, z, z)
e2₅(f1, x, y, z, f2) -> e3₁₁(x, y, x, y, y, z, y, z, x, y, z)
e3₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) -> e4₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
e4₁₁(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) -> e1₅(x1, x1, 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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

E1₅(h1, h2, x, y, z) -> E2₅(x, x, y, z, z)
E3₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) -> E4₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
E4₁₁(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) -> E1₅(x1, x1, x, y, z)
E2₅(f1, x, y, z, f2) -> E3₁₁(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
e1₅(h1, h2, x, y, z) -> e2₅(x, x, y, z, z)
e2₅(f1, x, y, z, f2) -> e3₁₁(x, y, x, y, y, z, y, z, x, y, z)
e3₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) -> e4₁₁(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z)
e4₁₁(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) -> e1₅(x1, x1, x, y, z)

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