Termination w.r.t. Q proof of TRS/AProVEIJCAR

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:

div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(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:

QUOT(x, 0, s(z)) → DIV(x, s(z))
QUOT(s(x), s(y), z) → QUOT(x, y, z)
DIV(x, y) → QUOT(x, y, y)

The TRS R consists of the following rules:

div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

QUOT(x, 0, s(z)) → DIV(x, s(z))
QUOT(s(x), s(y), z) → QUOT(x, y, z)
DIV(x, y) → QUOT(x, y, y)

The TRS R consists of the following rules:

div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

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

QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → DIV(x, s(z))
DIV(x, y) → QUOT(x, y, y)

The TRS R consists of the following rules:

div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(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 oriented strictly and are deleted.

QUOT(s(x), s(y), z) → QUOT(x, y, z)

The remaining pairs can at least be oriented weakly.

QUOT(x, 0, s(z)) → DIV(x, s(z))
DIV(x, y) → QUOT(x, y, y)

Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2, x3) = QUOT(x1, x3)
s(x1) = s(x1)
0 = 0
DIV(x1, x2) = DIV(x1, x2)

Recursive path order with status [2].
Quasi-Precedence:

[QUOT₂, s₁, 0, DIV_2]

Status:

QUOT₂: [1,2]
0: multiset
s₁: [1]
DIV₂: [1,2]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

QUOT(x, 0, s(z)) → DIV(x, s(z))
DIV(x, y) → QUOT(x, y, y)

The TRS R consists of the following rules:

div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(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 oriented strictly and are deleted.

QUOT(x, 0, s(z)) → DIV(x, s(z))

The remaining pairs can at least be oriented weakly.

DIV(x, y) → QUOT(x, y, y)

Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2, x3) = x2
0 = 0
s(x1) = s
DIV(x1, x2) = x2

Recursive path order with status [2].
Quasi-Precedence:

0 > s

Status:

0: multiset
s: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

DIV(x, y) → QUOT(x, y, y)

The TRS R consists of the following rules:

div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(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.