Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/nonterminSLASHAG01SLASHHASH4.22.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:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, s₁(z), s₁(z)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, s₁(z), s₁(z)))

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:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, s₁(z), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
quot₃(x0, 0, s₁(x1))

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

QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, s₁(z), s₁(z))

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, s₁(z), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
quot₃(x0, 0, s₁(x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, s₁(z), s₁(z))

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, s₁(z), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
quot₃(x0, 0, s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)

Used argument filtering: QUOT₃(x1, x2, x3) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

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

QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, s₁(z), s₁(z))

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, s₁(z), s₁(z)))

The set Q consists of the following terms:

quot₃(0, s₁(x0), s₁(x1))
quot₃(s₁(x0), s₁(x1), x2)
quot₃(x0, 0, s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.