Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/HMSLASHt013.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

  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(0, s₁(x0))
-₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(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:

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)))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(0, s₁(x0))
-₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(x0))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(0, s₁(x0))
-₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(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

  ↳ Non-Overlap Check

    ↳ 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)))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(0, s₁(x0))
-₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
-¹₂(x1, x2) = -¹₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(0, s₁(x0))
-₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(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

  ↳ Non-Overlap Check

    ↳ 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)))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(0, s₁(x0))
-₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
G₁(x1) = G₁(x1)
s₁(x1) = s₁(x1)
F₁(x1) = F₁(x1)
g₁(x1) = g₁(x1)
f₁(x1) = x1
0 = 0
-₂(x1, x2) = x1

Lexicographic Path Order [19].
Precedence:

0 > [G₁, s₁, g_1] > F₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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)))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(0, s₁(x0))
-₂(s₁(x0), s₁(x1))
f₁(0)
f₁(s₁(x0))
g₁(0)
g₁(s₁(x0))

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