Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/SK90SLASH2.16.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:

f₁(0) -> 1
f₁(s₁(x)) -> g₂(x, s₁(x))
g₂(0, y) -> y
g₂(s₁(x), y) -> g₂(x, +₂(y, s₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
g₂(s₁(x), y) -> g₂(x, s₁(+₂(y, x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(0) -> 1
f₁(s₁(x)) -> g₂(x, s₁(x))
g₂(0, y) -> y
g₂(s₁(x), y) -> g₂(x, +₂(y, s₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
g₂(s₁(x), y) -> g₂(x, s₁(+₂(y, 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:

+¹₂(x, s₁(y)) -> +¹₂(x, y)
G₂(s₁(x), y) -> +¹₂(y, s₁(x))
G₂(s₁(x), y) -> G₂(x, +₂(y, s₁(x)))
G₂(s₁(x), y) -> G₂(x, s₁(+₂(y, x)))
F₁(s₁(x)) -> G₂(x, s₁(x))
G₂(s₁(x), y) -> +¹₂(y, x)

The TRS R consists of the following rules:

f₁(0) -> 1
f₁(s₁(x)) -> g₂(x, s₁(x))
g₂(0, y) -> y
g₂(s₁(x), y) -> g₂(x, +₂(y, s₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
g₂(s₁(x), y) -> g₂(x, s₁(+₂(y, 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:

+¹₂(x, s₁(y)) -> +¹₂(x, y)
G₂(s₁(x), y) -> +¹₂(y, s₁(x))
G₂(s₁(x), y) -> G₂(x, +₂(y, s₁(x)))
G₂(s₁(x), y) -> G₂(x, s₁(+₂(y, x)))
F₁(s₁(x)) -> G₂(x, s₁(x))
G₂(s₁(x), y) -> +¹₂(y, x)

The TRS R consists of the following rules:

f₁(0) -> 1
f₁(s₁(x)) -> g₂(x, s₁(x))
g₂(0, y) -> y
g₂(s₁(x), y) -> g₂(x, +₂(y, s₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
g₂(s₁(x), y) -> g₂(x, s₁(+₂(y, 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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

+¹₂(x, s₁(y)) -> +¹₂(x, y)

The TRS R consists of the following rules:

f₁(0) -> 1
f₁(s₁(x)) -> g₂(x, s₁(x))
g₂(0, y) -> y
g₂(s₁(x), y) -> g₂(x, +₂(y, s₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
g₂(s₁(x), y) -> g₂(x, s₁(+₂(y, 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 strictly oriented and are deleted.

+¹₂(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(x2)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[+^1₁, s_1]

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:

f₁(0) -> 1
f₁(s₁(x)) -> g₂(x, s₁(x))
g₂(0, y) -> y
g₂(s₁(x), y) -> g₂(x, +₂(y, s₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
g₂(s₁(x), y) -> g₂(x, s₁(+₂(y, 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:

G₂(s₁(x), y) -> G₂(x, +₂(y, s₁(x)))
G₂(s₁(x), y) -> G₂(x, s₁(+₂(y, x)))

The TRS R consists of the following rules:

f₁(0) -> 1
f₁(s₁(x)) -> g₂(x, s₁(x))
g₂(0, y) -> y
g₂(s₁(x), y) -> g₂(x, +₂(y, s₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
g₂(s₁(x), y) -> g₂(x, s₁(+₂(y, 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 strictly oriented and are deleted.

G₂(s₁(x), y) -> G₂(x, +₂(y, s₁(x)))
G₂(s₁(x), y) -> G₂(x, s₁(+₂(y, x)))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
G₂(x1, x2) = G₂(x1, x2)
s₁(x1) = s₁(x1)
+₂(x1, x2) = +₂(x1, x2)
0 = 0

Lexicographic Path Order [19].
Precedence:

G₂ > +₂ > s₁
0 > s₁

The following usable rules [14] were oriented:

+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

f₁(0) -> 1
f₁(s₁(x)) -> g₂(x, s₁(x))
g₂(0, y) -> y
g₂(s₁(x), y) -> g₂(x, +₂(y, s₁(x)))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
g₂(s₁(x), y) -> g₂(x, s₁(+₂(y, 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.