Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/SK90SLASH2.53.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:

f₂(x, y) -> g₂(x, y)
g₂(h₁(x), y) -> h₁(f₂(x, y))
g₂(h₁(x), y) -> h₁(g₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₂(x, y) -> g₂(x, y)
g₂(h₁(x), y) -> h₁(f₂(x, y))
g₂(h₁(x), y) -> h₁(g₂(x, y))

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:

G₂(h₁(x), y) -> G₂(x, y)
F₂(x, y) -> G₂(x, y)
G₂(h₁(x), y) -> F₂(x, y)

The TRS R consists of the following rules:

f₂(x, y) -> g₂(x, y)
g₂(h₁(x), y) -> h₁(f₂(x, y))
g₂(h₁(x), y) -> h₁(g₂(x, y))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

G₂(h₁(x), y) -> G₂(x, y)
F₂(x, y) -> G₂(x, y)
G₂(h₁(x), y) -> F₂(x, y)

The TRS R consists of the following rules:

f₂(x, y) -> g₂(x, y)
g₂(h₁(x), y) -> h₁(f₂(x, y))
g₂(h₁(x), y) -> h₁(g₂(x, y))

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.

G₂(h₁(x), y) -> G₂(x, y)
G₂(h₁(x), y) -> F₂(x, y)

The remaining pairs can at least by weakly be oriented.

F₂(x, y) -> G₂(x, y)

Used ordering: Combined order from the following AFS and order.
G₂(x1, x2) = G₁(x1)
h₁(x1) = h₁(x1)
F₂(x1, x2) = F₁(x1)

Lexicographic Path Order [19].
Precedence:

[G₁, F_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F₂(x, y) -> G₂(x, y)

The TRS R consists of the following rules:

f₂(x, y) -> g₂(x, y)
g₂(h₁(x), y) -> h₁(f₂(x, y))
g₂(h₁(x), y) -> h₁(g₂(x, y))

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.