Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/Ste92SLASHmotivation.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 strictly oriented and are deleted.

G₁(g₁(x)) -> G₁(h₁(g₁(x)))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
G₁(x1) = G₁(x1)
g₁(x1) = g
h₁(x1) = h
f₂(x1, x2) = f₁(x2)

Lexicographic Path Order [19].
Precedence:

[G₁, g] > [h, f_1]

The following usable rules [14] were oriented:

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