Termination w.r.t. Q proof of TRS/Ste92motivation.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:

g₁(h₁(g₁(x))) -> g₁(x)
g₁(g₁(x)) -> g₁(h₁(g₁(x)))
h₁(h₁(x)) -> h₁(f₂(h₁(x), x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g₁(h₁(g₁(x))) -> g₁(x)
g₁(g₁(x)) -> g₁(h₁(g₁(x)))
h₁(h₁(x)) -> h₁(f₂(h₁(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(g₁(x)) -> H₁(g₁(x))
G₁(g₁(x)) -> G₁(h₁(g₁(x)))
H₁(h₁(x)) -> H₁(f₂(h₁(x), x))

The TRS R consists of the following rules:

g₁(h₁(g₁(x))) -> g₁(x)
g₁(g₁(x)) -> g₁(h₁(g₁(x)))
h₁(h₁(x)) -> h₁(f₂(h₁(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(g₁(x)) -> H₁(g₁(x))
G₁(g₁(x)) -> G₁(h₁(g₁(x)))
H₁(h₁(x)) -> H₁(f₂(h₁(x), x))

The TRS R consists of the following rules:

g₁(h₁(g₁(x))) -> g₁(x)
g₁(g₁(x)) -> g₁(h₁(g₁(x)))
h₁(h₁(x)) -> h₁(f₂(h₁(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

          ↳ QDPOrderProof

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

G₁(g₁(x)) -> G₁(h₁(g₁(x)))

The TRS R consists of the following rules:

g₁(h₁(g₁(x))) -> g₁(x)
g₁(g₁(x)) -> g₁(h₁(g₁(x)))
h₁(h₁(x)) -> h₁(f₂(h₁(x), 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.

G₁(g₁(x)) -> G₁(h₁(g₁(x)))

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

POL(G₁(x₁)) = 2·x₁ + x₁²
POL(f₂(x₁, x₂)) = 0
POL(g₁(x₁)) = 3 + 3·x₁ + x₁²
POL(h₁(x₁)) = 1

The following usable rules [14] were oriented:

h₁(h₁(x)) -> h₁(f₂(h₁(x), x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ PisEmptyProof

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

g₁(h₁(g₁(x))) -> g₁(x)
g₁(g₁(x)) -> g₁(h₁(g₁(x)))
h₁(h₁(x)) -> h₁(f₂(h₁(x), 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.