Termination w.r.t. Q proof of TRS/various23.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:

g₂(0, f₂(x, x)) -> x
g₂(x, s₁(y)) -> g₂(f₂(x, y), 0)
g₂(s₁(x), y) -> g₂(f₂(x, y), 0)
g₂(f₂(x, y), 0) -> f₂(g₂(x, 0), g₂(y, 0))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

g₂(0, f₂(x, x)) -> x
g₂(x, s₁(y)) -> g₂(f₂(x, y), 0)
g₂(s₁(x), y) -> g₂(f₂(x, y), 0)
g₂(f₂(x, y), 0) -> f₂(g₂(x, 0), g₂(y, 0))

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:

G₂(x, s₁(y)) -> G₂(f₂(x, y), 0)
G₂(f₂(x, y), 0) -> G₂(y, 0)
G₂(s₁(x), y) -> G₂(f₂(x, y), 0)
G₂(f₂(x, y), 0) -> G₂(x, 0)

The TRS R consists of the following rules:

g₂(0, f₂(x, x)) -> x
g₂(x, s₁(y)) -> g₂(f₂(x, y), 0)
g₂(s₁(x), y) -> g₂(f₂(x, y), 0)
g₂(f₂(x, y), 0) -> f₂(g₂(x, 0), g₂(y, 0))

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:

G₂(x, s₁(y)) -> G₂(f₂(x, y), 0)
G₂(f₂(x, y), 0) -> G₂(y, 0)
G₂(s₁(x), y) -> G₂(f₂(x, y), 0)
G₂(f₂(x, y), 0) -> G₂(x, 0)

The TRS R consists of the following rules:

g₂(0, f₂(x, x)) -> x
g₂(x, s₁(y)) -> g₂(f₂(x, y), 0)
g₂(s₁(x), y) -> g₂(f₂(x, y), 0)
g₂(f₂(x, y), 0) -> f₂(g₂(x, 0), g₂(y, 0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

G₂(f₂(x, y), 0) -> G₂(y, 0)
G₂(s₁(x), y) -> G₂(f₂(x, y), 0)
G₂(f₂(x, y), 0) -> G₂(x, 0)

The TRS R consists of the following rules:

g₂(0, f₂(x, x)) -> x
g₂(x, s₁(y)) -> g₂(f₂(x, y), 0)
g₂(s₁(x), y) -> g₂(f₂(x, y), 0)
g₂(f₂(x, y), 0) -> f₂(g₂(x, 0), g₂(y, 0))

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.

G₂(f₂(x, y), 0) -> G₂(y, 0)
G₂(f₂(x, y), 0) -> G₂(x, 0)

The remaining pairs can at least be oriented weakly.

G₂(s₁(x), y) -> G₂(f₂(x, y), 0)

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

POL( G₂(x₁, x₂) ) = max{0, x₁ + x₂ - 2}

POL( f₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( 0 ) = 2

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

G₂(s₁(x), y) -> G₂(f₂(x, y), 0)

The TRS R consists of the following rules:

g₂(0, f₂(x, x)) -> x
g₂(x, s₁(y)) -> g₂(f₂(x, y), 0)
g₂(s₁(x), y) -> g₂(f₂(x, y), 0)
g₂(f₂(x, y), 0) -> f₂(g₂(x, 0), g₂(y, 0))

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.