Termination w.r.t. Q proof of TRS/Zantemaz11.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:

a₂(f, 0) -> a₂(s, 0)
a₂(d, 0) -> 0
a₂(d, a₂(s, x)) -> a₂(s, a₂(s, a₂(d, a₂(p, a₂(s, x)))))
a₂(f, a₂(s, x)) -> a₂(d, a₂(f, a₂(p, a₂(s, x))))
a₂(p, a₂(s, x)) -> x

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a₂(f, 0) -> a₂(s, 0)
a₂(d, 0) -> 0
a₂(d, a₂(s, x)) -> a₂(s, a₂(s, a₂(d, a₂(p, a₂(s, x)))))
a₂(f, a₂(s, x)) -> a₂(d, a₂(f, a₂(p, a₂(s, x))))
a₂(p, a₂(s, x)) -> x

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:

A₂(f, a₂(s, x)) -> A₂(p, a₂(s, x))
A₂(f, a₂(s, x)) -> A₂(d, a₂(f, a₂(p, a₂(s, x))))
A₂(d, a₂(s, x)) -> A₂(d, a₂(p, a₂(s, x)))
A₂(d, a₂(s, x)) -> A₂(s, a₂(d, a₂(p, a₂(s, x))))
A₂(d, a₂(s, x)) -> A₂(s, a₂(s, a₂(d, a₂(p, a₂(s, x)))))
A₂(f, a₂(s, x)) -> A₂(f, a₂(p, a₂(s, x)))
A₂(f, 0) -> A₂(s, 0)
A₂(d, a₂(s, x)) -> A₂(p, a₂(s, x))

The TRS R consists of the following rules:

a₂(f, 0) -> a₂(s, 0)
a₂(d, 0) -> 0
a₂(d, a₂(s, x)) -> a₂(s, a₂(s, a₂(d, a₂(p, a₂(s, x)))))
a₂(f, a₂(s, x)) -> a₂(d, a₂(f, a₂(p, a₂(s, x))))
a₂(p, a₂(s, x)) -> x

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

A₂(f, a₂(s, x)) -> A₂(p, a₂(s, x))
A₂(f, a₂(s, x)) -> A₂(d, a₂(f, a₂(p, a₂(s, x))))
A₂(d, a₂(s, x)) -> A₂(d, a₂(p, a₂(s, x)))
A₂(d, a₂(s, x)) -> A₂(s, a₂(d, a₂(p, a₂(s, x))))
A₂(d, a₂(s, x)) -> A₂(s, a₂(s, a₂(d, a₂(p, a₂(s, x)))))
A₂(f, a₂(s, x)) -> A₂(f, a₂(p, a₂(s, x)))
A₂(f, 0) -> A₂(s, 0)
A₂(d, a₂(s, x)) -> A₂(p, a₂(s, x))

The TRS R consists of the following rules:

a₂(f, 0) -> a₂(s, 0)
a₂(d, 0) -> 0
a₂(d, a₂(s, x)) -> a₂(s, a₂(s, a₂(d, a₂(p, a₂(s, x)))))
a₂(f, a₂(s, x)) -> a₂(d, a₂(f, a₂(p, a₂(s, x))))
a₂(p, a₂(s, x)) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

A₂(d, a₂(s, x)) -> A₂(d, a₂(p, a₂(s, x)))

The TRS R consists of the following rules:

a₂(f, 0) -> a₂(s, 0)
a₂(d, 0) -> 0
a₂(d, a₂(s, x)) -> a₂(s, a₂(s, a₂(d, a₂(p, a₂(s, x)))))
a₂(f, a₂(s, x)) -> a₂(d, a₂(f, a₂(p, a₂(s, x))))
a₂(p, a₂(s, x)) -> x

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.

A₂(d, a₂(s, x)) -> A₂(d, a₂(p, a₂(s, x)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( d ) = 1

POL( s ) = 3

POL( A₂(x₁, x₂) ) = max{0, x₂ - 3}

POL( p ) = max{0, -2}

POL( a₂(x₁, x₂) ) = max{0, 2x₁ + x₂ - 2}

The following usable rules [14] were oriented:

a₂(p, a₂(s, x)) -> x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a₂(f, 0) -> a₂(s, 0)
a₂(d, 0) -> 0
a₂(d, a₂(s, x)) -> a₂(s, a₂(s, a₂(d, a₂(p, a₂(s, x)))))
a₂(f, a₂(s, x)) -> a₂(d, a₂(f, a₂(p, a₂(s, x))))
a₂(p, a₂(s, x)) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

A₂(f, a₂(s, x)) -> A₂(f, a₂(p, a₂(s, x)))

The TRS R consists of the following rules:

a₂(f, 0) -> a₂(s, 0)
a₂(d, 0) -> 0
a₂(d, a₂(s, x)) -> a₂(s, a₂(s, a₂(d, a₂(p, a₂(s, x)))))
a₂(f, a₂(s, x)) -> a₂(d, a₂(f, a₂(p, a₂(s, x))))
a₂(p, a₂(s, x)) -> x

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.

A₂(f, a₂(s, x)) -> A₂(f, a₂(p, a₂(s, x)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( f ) = 1

POL( s ) = 3

POL( A₂(x₁, x₂) ) = max{0, x₂ - 3}

POL( p ) = max{0, -2}

POL( a₂(x₁, x₂) ) = max{0, 2x₁ + x₂ - 2}

The following usable rules [14] were oriented:

a₂(p, a₂(s, x)) -> x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a₂(f, 0) -> a₂(s, 0)
a₂(d, 0) -> 0
a₂(d, a₂(s, x)) -> a₂(s, a₂(s, a₂(d, a₂(p, a₂(s, x)))))
a₂(f, a₂(s, x)) -> a₂(d, a₂(f, a₂(p, a₂(s, x))))
a₂(p, a₂(s, x)) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.