Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/SK90SLASH2.25.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

  ↳ Non-Overlap Check

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.

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:

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))

The set Q consists of the following terms:

fib₁(0)
fib₁(s₁(0))
fib₁(s₁(s₁(x0)))
+₂(x0, 0)
+₂(x0, s₁(x1))

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))

The set Q consists of the following terms:

fib₁(0)
fib₁(s₁(0))
fib₁(s₁(s₁(x0)))
+₂(x0, 0)
+₂(x0, s₁(x1))

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:

+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))

The set Q consists of the following terms:

fib₁(0)
fib₁(s₁(0))
fib₁(s₁(s₁(x0)))
+₂(x0, 0)
+₂(x0, s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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))

The set Q consists of the following terms:

fib₁(0)
fib₁(s₁(0))
fib₁(s₁(s₁(x0)))
+₂(x0, 0)
+₂(x0, s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

+¹₂(x, s₁(y)) -> +¹₂(x, y)

Used argument filtering: +¹₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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))

The set Q consists of the following terms:

fib₁(0)
fib₁(s₁(0))
fib₁(s₁(s₁(x0)))
+₂(x0, 0)
+₂(x0, s₁(x1))

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

                ↳ QDPAfsSolverProof

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))

The set Q consists of the following terms:

fib₁(0)
fib₁(s₁(0))
fib₁(s₁(s₁(x0)))
+₂(x0, 0)
+₂(x0, s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: FIB₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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))

The set Q consists of the following terms:

fib₁(0)
fib₁(s₁(0))
fib₁(s₁(s₁(x0)))
+₂(x0, 0)
+₂(x0, s₁(x1))

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