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

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

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

G2 → H2
F2 → G1
G2 → H1
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)
E1(h1, h2, x, y, z) → E2(x, x, y, z, z)
E1(x1, x1, x, y, z) → E5(x1, x, y, z)
F2 → G2
E2(f1, x, y, z, f2) → E3(x, y, x, y, y, z, y, z, x, y, z)
G1 → H1
G1 → H2
F1 → G2
F1 → G1
E3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → E4(x1, x1, x2, x2, x3, x3, x4, x4, 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 [15,17,22] contains 1 SCC with 10 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ ForwardInstantiation

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

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

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.
By forward instantiating [14] the rule E1(h1, h2, x, y, z) → E2(x, x, y, z, z) we obtained the following new rules:

E1(h1, h2, f1, x1, f2) → E2(f1, f1, x1, f2, f2)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ ForwardInstantiation

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

E1(h1, h2, f1, x1, f2) → E2(f1, f1, x1, f2, f2)
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)
E3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → E4(x1, x1, x2, x2, x3, x3, x4, x4, 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.
By forward instantiating [14] the rule E4(g1, x1, g2, x1, g1, x1, g2, x1, x, y, z) → E1(x1, x1, x, y, z) we obtained the following new rules:

E4(g1, x0, g2, x0, g1, x0, g2, x0, f1, x2, f2) → E1(x0, x0, f1, x2, f2)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ ForwardInstantiation

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

E1(h1, h2, f1, x1, f2) → E2(f1, f1, x1, f2, f2)
E4(g1, x0, g2, x0, g1, x0, g2, x0, f1, x2, f2) → E1(x0, x0, f1, x2, f2)
E2(f1, x, y, z, f2) → E3(x, y, x, y, y, z, y, z, 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)

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.
By forward instantiating [14] the rule E3(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) → E4(x1, x1, x2, x2, x3, x3, x4, x4, x, y, z) we obtained the following new rules:

E3(x0, x0, x1, x1, x2, x2, x3, x3, f1, x5, f2) → E4(x0, x0, x1, x1, x2, x2, x3, x3, f1, x5, f2)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ ForwardInstantiation
                    ↳ QDP
                      ↳ ForwardInstantiation

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

E1(h1, h2, f1, x1, f2) → E2(f1, f1, x1, f2, f2)
E3(x0, x0, x1, x1, x2, x2, x3, x3, f1, x5, f2) → E4(x0, x0, x1, x1, x2, x2, x3, x3, f1, x5, f2)
E4(g1, x0, g2, x0, g1, x0, g2, x0, f1, x2, f2) → E1(x0, x0, f1, x2, f2)
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.
By forward instantiating [14] the rule E2(f1, x, y, z, f2) → E3(x, y, x, y, y, z, y, z, x, y, z) we obtained the following new rules:

E2(f1, f1, x1, f2, f2) → E3(f1, x1, f1, x1, x1, f2, x1, f2, f1, x1, f2)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ ForwardInstantiation
            ↳ QDP
              ↳ ForwardInstantiation
                ↳ QDP
                  ↳ ForwardInstantiation
                    ↳ QDP
                      ↳ ForwardInstantiation
                        ↳ QDP

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

E1(h1, h2, f1, x1, f2) → E2(f1, f1, x1, f2, f2)
E3(x0, x0, x1, x1, x2, x2, x3, x3, f1, x5, f2) → E4(x0, x0, x1, x1, x2, x2, x3, x3, f1, x5, f2)
E2(f1, f1, x1, f2, f2) → E3(f1, x1, f1, x1, x1, f2, x1, f2, f1, x1, f2)
E4(g1, x0, g2, x0, g1, x0, g2, x0, f1, x2, f2) → E1(x0, x0, f1, x2, f2)

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.