Termination w.r.t. Q proof of TRS/SK902.25.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:

fib₁(0) -> 0
fib₁(s₁(0)) -> s₁(0)
fib₁(s₁(s₁(x))) -> +₂(fib₁(s₁(x)), fib₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

fib₁(0) -> 0
fib₁(s₁(0)) -> s₁(0)
fib₁(s₁(s₁(x))) -> +₂(fib₁(s₁(x)), fib₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))

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)
FIB₁(s₁(s₁(x))) -> FIB₁(s₁(x))
FIB₁(s₁(s₁(x))) -> FIB₁(x)
FIB₁(s₁(s₁(x))) -> +¹₂(fib₁(s₁(x)), fib₁(x))

The TRS R consists of the following rules:

fib₁(0) -> 0
fib₁(s₁(0)) -> s₁(0)
fib₁(s₁(s₁(x))) -> +₂(fib₁(s₁(x)), fib₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))

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)
FIB₁(s₁(s₁(x))) -> FIB₁(s₁(x))
FIB₁(s₁(s₁(x))) -> FIB₁(x)
FIB₁(s₁(s₁(x))) -> +¹₂(fib₁(s₁(x)), fib₁(x))

The TRS R consists of the following rules:

fib₁(0) -> 0
fib₁(s₁(0)) -> s₁(0)
fib₁(s₁(s₁(x))) -> +₂(fib₁(s₁(x)), fib₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))

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 1 less node.

↳ 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:

fib₁(0) -> 0
fib₁(s₁(0)) -> s₁(0)
fib₁(s₁(s₁(x))) -> +₂(fib₁(s₁(x)), fib₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))

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)) = 2·x₂
POL(s₁(x₁)) = 1 + 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:

fib₁(0) -> 0
fib₁(s₁(0)) -> s₁(0)
fib₁(s₁(s₁(x))) -> +₂(fib₁(s₁(x)), fib₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))

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:

FIB₁(s₁(s₁(x))) -> FIB₁(s₁(x))
FIB₁(s₁(s₁(x))) -> FIB₁(x)

The TRS R consists of the following rules:

fib₁(0) -> 0
fib₁(s₁(0)) -> s₁(0)
fib₁(s₁(s₁(x))) -> +₂(fib₁(s₁(x)), fib₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))

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.

FIB₁(s₁(s₁(x))) -> FIB₁(s₁(x))
FIB₁(s₁(s₁(x))) -> FIB₁(x)

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

POL(FIB₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·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:

fib₁(0) -> 0
fib₁(s₁(0)) -> s₁(0)
fib₁(s₁(s₁(x))) -> +₂(fib₁(s₁(x)), fib₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))

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.