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:

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

  ↳ 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)))

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(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)))

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(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)))

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.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( QUOT₃(x₁, ..., x₃) ) = max{0, x₁ - 2}

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ 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)))

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.