Termination w.r.t. Q proof of TRS/SK904.51.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:

f₁(a) -> g₁(h₁(a))
h₁(g₁(x)) -> g₁(h₁(f₁(x)))
k₃(x, h₁(x), a) -> h₁(x)
k₃(f₁(x), y, x) -> f₁(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(a) -> g₁(h₁(a))
h₁(g₁(x)) -> g₁(h₁(f₁(x)))
k₃(x, h₁(x), a) -> h₁(x)
k₃(f₁(x), y, x) -> f₁(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:

F₁(a) -> H₁(a)
H₁(g₁(x)) -> F₁(x)
H₁(g₁(x)) -> H₁(f₁(x))

The TRS R consists of the following rules:

f₁(a) -> g₁(h₁(a))
h₁(g₁(x)) -> g₁(h₁(f₁(x)))
k₃(x, h₁(x), a) -> h₁(x)
k₃(f₁(x), y, x) -> f₁(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:

F₁(a) -> H₁(a)
H₁(g₁(x)) -> F₁(x)
H₁(g₁(x)) -> H₁(f₁(x))

The TRS R consists of the following rules:

f₁(a) -> g₁(h₁(a))
h₁(g₁(x)) -> g₁(h₁(f₁(x)))
k₃(x, h₁(x), a) -> h₁(x)
k₃(f₁(x), y, x) -> f₁(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

          ↳ QDPOrderProof

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

H₁(g₁(x)) -> H₁(f₁(x))

The TRS R consists of the following rules:

f₁(a) -> g₁(h₁(a))
h₁(g₁(x)) -> g₁(h₁(f₁(x)))
k₃(x, h₁(x), a) -> h₁(x)
k₃(f₁(x), y, x) -> f₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

H₁(g₁(x)) -> H₁(f₁(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(H₁(x₁)) = 2·x₁
POL(a) = 2
POL(f₁(x₁)) = 1 + x₁
POL(g₁(x₁)) = 2 + 2·x₁
POL(h₁(x₁)) = 0

The following usable rules [14] were oriented:

f₁(a) -> g₁(h₁(a))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f₁(a) -> g₁(h₁(a))
h₁(g₁(x)) -> g₁(h₁(f₁(x)))
k₃(x, h₁(x), a) -> h₁(x)
k₃(f₁(x), y, x) -> f₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.