Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/AProVESLASHIJCAR

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:

div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))

Q is empty.

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

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

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

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))

Q is empty.
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
DIV₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))

Q is empty.
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(x, 0, s₁(z)) -> DIV₂(x, s₁(z))

Used argument filtering: QUOT₃(x1, x2, x3) = x2
0 = 0
DIV₂(x1, x2) = x2
s₁(x1) = s
Used ordering: Quasi Precedence: 0 > s

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPAfsSolverProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

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

DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))

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