Termination w.r.t. Q proof of TRS/secret05cime4.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:

g₁(X) -> u₃(h₁(X), h₁(X), X)
u₃(d, c₁(Y), X) -> k₁(Y)
h₁(d) -> c₁(a)
h₁(d) -> c₁(b)
f₃(k₁(a), k₁(b), X) -> f₃(X, X, X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g₁(X) -> u₃(h₁(X), h₁(X), X)
u₃(d, c₁(Y), X) -> k₁(Y)
h₁(d) -> c₁(a)
h₁(d) -> c₁(b)
f₃(k₁(a), k₁(b), X) -> f₃(X, X, X)

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:

G₁(X) -> H₁(X)
F₃(k₁(a), k₁(b), X) -> F₃(X, X, X)
G₁(X) -> U₃(h₁(X), h₁(X), X)

The TRS R consists of the following rules:

g₁(X) -> u₃(h₁(X), h₁(X), X)
u₃(d, c₁(Y), X) -> k₁(Y)
h₁(d) -> c₁(a)
h₁(d) -> c₁(b)
f₃(k₁(a), k₁(b), X) -> f₃(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(X) -> H₁(X)
F₃(k₁(a), k₁(b), X) -> F₃(X, X, X)
G₁(X) -> U₃(h₁(X), h₁(X), X)

The TRS R consists of the following rules:

g₁(X) -> u₃(h₁(X), h₁(X), X)
u₃(d, c₁(Y), X) -> k₁(Y)
h₁(d) -> c₁(a)
h₁(d) -> c₁(b)
f₃(k₁(a), k₁(b), X) -> f₃(X, X, X)

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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

F₃(k₁(a), k₁(b), X) -> F₃(X, X, X)

The TRS R consists of the following rules:

g₁(X) -> u₃(h₁(X), h₁(X), X)
u₃(d, c₁(Y), X) -> k₁(Y)
h₁(d) -> c₁(a)
h₁(d) -> c₁(b)
f₃(k₁(a), k₁(b), X) -> f₃(X, X, X)

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