Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/TRCSRSLASHExConc_Zan97

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:

a__f₁(X) -> g₁(h₁(f₁(X)))
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(g₁(X)) -> g₁(X)
mark₁(h₁(X)) -> h₁(mark₁(X))
a__f₁(X) -> f₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__f₁(X) -> g₁(h₁(f₁(X)))
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(g₁(X)) -> g₁(X)
mark₁(h₁(X)) -> h₁(mark₁(X))
a__f₁(X) -> f₁(X)

Q is empty.

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

MARK₁(f₁(X)) -> A__F₁(mark₁(X))
MARK₁(h₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(X) -> g₁(h₁(f₁(X)))
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(g₁(X)) -> g₁(X)
mark₁(h₁(X)) -> h₁(mark₁(X))
a__f₁(X) -> f₁(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:

MARK₁(f₁(X)) -> A__F₁(mark₁(X))
MARK₁(h₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(X) -> g₁(h₁(f₁(X)))
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(g₁(X)) -> g₁(X)
mark₁(h₁(X)) -> h₁(mark₁(X))
a__f₁(X) -> f₁(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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

MARK₁(h₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(X) -> g₁(h₁(f₁(X)))
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(g₁(X)) -> g₁(X)
mark₁(h₁(X)) -> h₁(mark₁(X))
a__f₁(X) -> f₁(X)

Q is empty.
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.

MARK₁(f₁(X)) -> MARK₁(X)

Used argument filtering: MARK₁(x1) = x1
h₁(x1) = x1
f₁(x1) = f₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ QDPAfsSolverProof

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

MARK₁(h₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(X) -> g₁(h₁(f₁(X)))
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(g₁(X)) -> g₁(X)
mark₁(h₁(X)) -> h₁(mark₁(X))
a__f₁(X) -> f₁(X)

Q is empty.
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.

MARK₁(h₁(X)) -> MARK₁(X)

Used argument filtering: MARK₁(x1) = x1
h₁(x1) = h₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ PisEmptyProof

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

a__f₁(X) -> g₁(h₁(f₁(X)))
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(g₁(X)) -> g₁(X)
mark₁(h₁(X)) -> h₁(mark₁(X))
a__f₁(X) -> f₁(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.