Termination w.r.t. Q proof of TRS/HMt013.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:

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
f₁(0) -> 0
f₁(s₁(x)) -> -₂(s₁(x), g₁(f₁(x)))
g₁(0) -> s₁(0)
g₁(s₁(x)) -> -₂(s₁(x), f₁(g₁(x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
f₁(0) -> 0
f₁(s₁(x)) -> -₂(s₁(x), g₁(f₁(x)))
g₁(0) -> s₁(0)
g₁(s₁(x)) -> -₂(s₁(x), f₁(g₁(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:

F₁(s₁(x)) -> -¹₂(s₁(x), g₁(f₁(x)))
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
G₁(s₁(x)) -> G₁(x)
G₁(s₁(x)) -> F₁(g₁(x))
G₁(s₁(x)) -> -¹₂(s₁(x), f₁(g₁(x)))
F₁(s₁(x)) -> G₁(f₁(x))
F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
f₁(0) -> 0
f₁(s₁(x)) -> -₂(s₁(x), g₁(f₁(x)))
g₁(0) -> s₁(0)
g₁(s₁(x)) -> -₂(s₁(x), f₁(g₁(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:

F₁(s₁(x)) -> -¹₂(s₁(x), g₁(f₁(x)))
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
G₁(s₁(x)) -> G₁(x)
G₁(s₁(x)) -> F₁(g₁(x))
G₁(s₁(x)) -> -¹₂(s₁(x), f₁(g₁(x)))
F₁(s₁(x)) -> G₁(f₁(x))
F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
f₁(0) -> 0
f₁(s₁(x)) -> -₂(s₁(x), g₁(f₁(x)))
g₁(0) -> s₁(0)
g₁(s₁(x)) -> -₂(s₁(x), f₁(g₁(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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
f₁(0) -> 0
f₁(s₁(x)) -> -₂(s₁(x), g₁(f₁(x)))
g₁(0) -> s₁(0)
g₁(s₁(x)) -> -₂(s₁(x), f₁(g₁(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.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(-¹₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
f₁(0) -> 0
f₁(s₁(x)) -> -₂(s₁(x), g₁(f₁(x)))
g₁(0) -> s₁(0)
g₁(s₁(x)) -> -₂(s₁(x), f₁(g₁(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:

G₁(s₁(x)) -> G₁(x)
G₁(s₁(x)) -> F₁(g₁(x))
F₁(s₁(x)) -> G₁(f₁(x))
F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
f₁(0) -> 0
f₁(s₁(x)) -> -₂(s₁(x), g₁(f₁(x)))
g₁(0) -> s₁(0)
g₁(s₁(x)) -> -₂(s₁(x), f₁(g₁(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.

G₁(s₁(x)) -> G₁(x)
G₁(s₁(x)) -> F₁(g₁(x))
F₁(s₁(x)) -> F₁(x)

The remaining pairs can at least be oriented weakly.

F₁(s₁(x)) -> G₁(f₁(x))

Used ordering: Polynomial interpretation [21]:

POL(-₂(x₁, x₂)) = 2·x₁
POL(0) = 2
POL(F₁(x₁)) = 2·x₁
POL(G₁(x₁)) = 2 + 2·x₁
POL(f₁(x₁)) = 2·x₁
POL(g₁(x₁)) = 1 + 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented:

g₁(0) -> s₁(0)
f₁(0) -> 0
-₂(x, 0) -> x
g₁(s₁(x)) -> -₂(s₁(x), f₁(g₁(x)))
f₁(s₁(x)) -> -₂(s₁(x), g₁(f₁(x)))
-₂(s₁(x), s₁(y)) -> -₂(x, y)
-₂(0, s₁(y)) -> 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

F₁(s₁(x)) -> G₁(f₁(x))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(0, s₁(y)) -> 0
-₂(s₁(x), s₁(y)) -> -₂(x, y)
f₁(0) -> 0
f₁(s₁(x)) -> -₂(s₁(x), g₁(f₁(x)))
g₁(0) -> s₁(0)
g₁(s₁(x)) -> -₂(s₁(x), f₁(g₁(x)))

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.