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:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

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:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(s(x)) → -¹(s(x), g(f(x)))
-¹(s(x), s(y)) → -¹(x, y)
G(s(x)) → G(x)
G(s(x)) → F(g(x))
G(s(x)) → -¹(s(x), f(g(x)))
F(s(x)) → G(f(x))
F(s(x)) → F(x)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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:

F(s(x)) → -¹(s(x), g(f(x)))
-¹(s(x), s(y)) → -¹(x, y)
G(s(x)) → G(x)
G(s(x)) → F(g(x))
G(s(x)) → -¹(s(x), f(g(x)))
F(s(x)) → G(f(x))
F(s(x)) → F(x)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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

              ↳ DependencyGraphProof

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

F(s(x)) → -¹(s(x), g(f(x)))
-¹(s(x), s(y)) → -¹(x, y)
F(s(x)) → G(f(x))
G(s(x)) → -¹(s(x), f(g(x)))
G(s(x)) → F(g(x))
G(s(x)) → G(x)
F(s(x)) → F(x)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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.

-¹(s(x), s(y)) → -¹(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
-¹(x1, x2) = x2
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

G(s(x)) → G(x)
G(s(x)) → F(g(x))
F(s(x)) → G(f(x))
F(s(x)) → F(x)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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.

G(s(x)) → G(x)
F(s(x)) → G(f(x))
F(s(x)) → F(x)

The remaining pairs can at least be oriented weakly.

G(s(x)) → F(g(x))

Used ordering: Combined order from the following AFS and order.
G(x1) = x1
s(x1) = s(x1)
F(x1) = x1
g(x1) = g(x1)
f(x1) = x1
-(x1, x2) = x1
0 = 0

Lexicographic Path Order [19].
Precedence:

[s₁, g_1]

The following usable rules [14] were oriented:

g(s(x)) → -(s(x), f(g(x)))
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(s(x)) → -(s(x), g(f(x)))
-(x, 0) → x
f(0) → 0
g(0) → s(0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

G(s(x)) → F(g(x))

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

The set Q consists of the following terms:

-(x0, 0)
-(0, s(x0))
-(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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.