Termination w.r.t. Q proof of TRS/SK902.26.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) -> 0
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> p₁(h₁(g₁(x)))
g₁(0) -> pair₂(s₁(0), s₁(0))
g₁(s₁(x)) -> h₁(g₁(x))
h₁(x) -> pair₂(+₂(p₁(x), q₁(x)), p₁(x))
p₁(pair₂(x, y)) -> x
q₁(pair₂(x, y)) -> y
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
f₁(s₁(s₁(x))) -> +₂(p₁(g₁(x)), q₁(g₁(x)))
g₁(s₁(x)) -> pair₂(+₂(p₁(g₁(x)), q₁(g₁(x))), p₁(g₁(x)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(0) -> 0
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> p₁(h₁(g₁(x)))
g₁(0) -> pair₂(s₁(0), s₁(0))
g₁(s₁(x)) -> h₁(g₁(x))
h₁(x) -> pair₂(+₂(p₁(x), q₁(x)), p₁(x))
p₁(pair₂(x, y)) -> x
q₁(pair₂(x, y)) -> y
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
f₁(s₁(s₁(x))) -> +₂(p₁(g₁(x)), q₁(g₁(x)))
g₁(s₁(x)) -> pair₂(+₂(p₁(g₁(x)), q₁(g₁(x))), p₁(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:

+¹₂(x, s₁(y)) -> +¹₂(x, y)
H₁(x) -> Q₁(x)
H₁(x) -> P₁(x)
G₁(s₁(x)) -> G₁(x)
F₁(s₁(s₁(x))) -> +¹₂(p₁(g₁(x)), q₁(g₁(x)))
F₁(s₁(s₁(x))) -> G₁(x)
G₁(s₁(x)) -> P₁(g₁(x))
F₁(s₁(s₁(x))) -> H₁(g₁(x))
F₁(s₁(s₁(x))) -> P₁(g₁(x))
G₁(s₁(x)) -> H₁(g₁(x))
G₁(s₁(x)) -> +¹₂(p₁(g₁(x)), q₁(g₁(x)))
F₁(s₁(s₁(x))) -> P₁(h₁(g₁(x)))
F₁(s₁(s₁(x))) -> Q₁(g₁(x))
H₁(x) -> +¹₂(p₁(x), q₁(x))
G₁(s₁(x)) -> Q₁(g₁(x))

The TRS R consists of the following rules:

f₁(0) -> 0
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> p₁(h₁(g₁(x)))
g₁(0) -> pair₂(s₁(0), s₁(0))
g₁(s₁(x)) -> h₁(g₁(x))
h₁(x) -> pair₂(+₂(p₁(x), q₁(x)), p₁(x))
p₁(pair₂(x, y)) -> x
q₁(pair₂(x, y)) -> y
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
f₁(s₁(s₁(x))) -> +₂(p₁(g₁(x)), q₁(g₁(x)))
g₁(s₁(x)) -> pair₂(+₂(p₁(g₁(x)), q₁(g₁(x))), p₁(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:

+¹₂(x, s₁(y)) -> +¹₂(x, y)
H₁(x) -> Q₁(x)
H₁(x) -> P₁(x)
G₁(s₁(x)) -> G₁(x)
F₁(s₁(s₁(x))) -> +¹₂(p₁(g₁(x)), q₁(g₁(x)))
F₁(s₁(s₁(x))) -> G₁(x)
G₁(s₁(x)) -> P₁(g₁(x))
F₁(s₁(s₁(x))) -> H₁(g₁(x))
F₁(s₁(s₁(x))) -> P₁(g₁(x))
G₁(s₁(x)) -> H₁(g₁(x))
G₁(s₁(x)) -> +¹₂(p₁(g₁(x)), q₁(g₁(x)))
F₁(s₁(s₁(x))) -> P₁(h₁(g₁(x)))
F₁(s₁(s₁(x))) -> Q₁(g₁(x))
H₁(x) -> +¹₂(p₁(x), q₁(x))
G₁(s₁(x)) -> Q₁(g₁(x))

The TRS R consists of the following rules:

f₁(0) -> 0
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> p₁(h₁(g₁(x)))
g₁(0) -> pair₂(s₁(0), s₁(0))
g₁(s₁(x)) -> h₁(g₁(x))
h₁(x) -> pair₂(+₂(p₁(x), q₁(x)), p₁(x))
p₁(pair₂(x, y)) -> x
q₁(pair₂(x, y)) -> y
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
f₁(s₁(s₁(x))) -> +₂(p₁(g₁(x)), q₁(g₁(x)))
g₁(s₁(x)) -> pair₂(+₂(p₁(g₁(x)), q₁(g₁(x))), p₁(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 13 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) -> 0
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> p₁(h₁(g₁(x)))
g₁(0) -> pair₂(s₁(0), s₁(0))
g₁(s₁(x)) -> h₁(g₁(x))
h₁(x) -> pair₂(+₂(p₁(x), q₁(x)), p₁(x))
p₁(pair₂(x, y)) -> x
q₁(pair₂(x, y)) -> y
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
f₁(s₁(s₁(x))) -> +₂(p₁(g₁(x)), q₁(g₁(x)))
g₁(s₁(x)) -> pair₂(+₂(p₁(g₁(x)), q₁(g₁(x))), p₁(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(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)) = 3·x₂
POL(s₁(x₁)) = 1 + 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:

f₁(0) -> 0
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> p₁(h₁(g₁(x)))
g₁(0) -> pair₂(s₁(0), s₁(0))
g₁(s₁(x)) -> h₁(g₁(x))
h₁(x) -> pair₂(+₂(p₁(x), q₁(x)), p₁(x))
p₁(pair₂(x, y)) -> x
q₁(pair₂(x, y)) -> y
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
f₁(s₁(s₁(x))) -> +₂(p₁(g₁(x)), q₁(g₁(x)))
g₁(s₁(x)) -> pair₂(+₂(p₁(g₁(x)), q₁(g₁(x))), p₁(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)

The TRS R consists of the following rules:

f₁(0) -> 0
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> p₁(h₁(g₁(x)))
g₁(0) -> pair₂(s₁(0), s₁(0))
g₁(s₁(x)) -> h₁(g₁(x))
h₁(x) -> pair₂(+₂(p₁(x), q₁(x)), p₁(x))
p₁(pair₂(x, y)) -> x
q₁(pair₂(x, y)) -> y
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
f₁(s₁(s₁(x))) -> +₂(p₁(g₁(x)), q₁(g₁(x)))
g₁(s₁(x)) -> pair₂(+₂(p₁(g₁(x)), q₁(g₁(x))), p₁(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)

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

POL(G₁(x₁)) = 3·x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ 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) -> 0
f₁(s₁(0)) -> s₁(0)
f₁(s₁(s₁(x))) -> p₁(h₁(g₁(x)))
g₁(0) -> pair₂(s₁(0), s₁(0))
g₁(s₁(x)) -> h₁(g₁(x))
h₁(x) -> pair₂(+₂(p₁(x), q₁(x)), p₁(x))
p₁(pair₂(x, y)) -> x
q₁(pair₂(x, y)) -> y
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
f₁(s₁(s₁(x))) -> +₂(p₁(g₁(x)), q₁(g₁(x)))
g₁(s₁(x)) -> pair₂(+₂(p₁(g₁(x)), q₁(g₁(x))), p₁(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.