Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/AG01SLASHHASH3.35.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(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

The set Q consists of the following terms:

g₁(s₁(x0))
f₁(0)
f₁(s₁(x0))
g₁(0)

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(s₁(x)) -> F₁(x)
F₁(s₁(x)) -> G₁(x)

The TRS R consists of the following rules:

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

The set Q consists of the following terms:

g₁(s₁(x0))
f₁(0)
f₁(s₁(x0))
g₁(0)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

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

G₁(s₁(x)) -> F₁(x)
F₁(s₁(x)) -> G₁(x)

The TRS R consists of the following rules:

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

The set Q consists of the following terms:

g₁(s₁(x0))
f₁(0)
f₁(s₁(x0))
g₁(0)

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.

F₁(s₁(x)) -> G₁(x)

The remaining pairs can at least by weakly be oriented.

G₁(s₁(x)) -> F₁(x)

Used ordering: Combined order from the following AFS and order.
G₁(x1) = x1
s₁(x1) = s₁(x1)
F₁(x1) = F₁(x1)

Lexicographic Path Order [19].
Precedence:

[s₁, F_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

G₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

g₁(s₁(x)) -> f₁(x)
f₁(0) -> s₁(0)
f₁(s₁(x)) -> s₁(s₁(g₁(x)))
g₁(0) -> 0

The set Q consists of the following terms:

g₁(s₁(x0))
f₁(0)
f₁(s₁(x0))
g₁(0)

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.