Termination w.r.t. Q proof of TRS/various03.trs

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:

f₄(g₁(x), h₁(x), y, x) -> f₄(y, y, y, x)
f₄(x, y, z, 0) -> 2
g₁(0) -> 2
h₁(0) -> 2

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₄(g₁(x), h₁(x), y, x) -> f₄(y, y, y, x)
f₄(x, y, z, 0) -> 2
g₁(0) -> 2
h₁(0) -> 2

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:

F₄(g₁(x), h₁(x), y, x) -> F₄(y, y, y, x)

The TRS R consists of the following rules:

f₄(g₁(x), h₁(x), y, x) -> f₄(y, y, y, x)
f₄(x, y, z, 0) -> 2
g₁(0) -> 2
h₁(0) -> 2

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:

F₄(g₁(x), h₁(x), y, x) -> F₄(y, y, y, x)

The TRS R consists of the following rules:

f₄(g₁(x), h₁(x), y, x) -> f₄(y, y, y, x)
f₄(x, y, z, 0) -> 2
g₁(0) -> 2
h₁(0) -> 2

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.