Termination w.r.t. Q proof of TRS/nontermin/AG01#4.22.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:

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

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

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

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
quot(x0, 0, s(x1))

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
quot(x0, 0, s(x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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)) → QUOT(x, s(z), s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
quot(x0, 0, s(x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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)) → QUOT(x, s(z), s(z))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
quot(x0, 0, s(x1))

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)) → QUOT(x, s(z), s(z))

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

Lexicographic path order with status [19].
Precedence:

s₁ > QUOT₂
0 > QUOT₂

Status:

QUOT₂: [2,1]
s₁: multiset
0: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
quot(x0, 0, s(x1))

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.