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

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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

      ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
QUOT₃(x1, x2, x3) = x1
0 = 0
s₁(x1) = s₁(x1)
DIV₂(x1, x2) = x1

Lexicographic Path Order [19].
Precedence:

0 > s₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))

The remaining pairs can at least by weakly be oriented.

DIV₂(x, y) -> QUOT₃(x, y, y)

Used ordering: Combined order from the following AFS and order.
QUOT₃(x1, x2, x3) = QUOT₂(x1, x2)
0 = 0
s₁(x1) = s
DIV₂(x1, x2) = DIV₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

[QUOT₂, 0, DIV_2] > s

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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 [13,14,18] contains 0 SCCs with 1 less node.