Termination w.r.t. Q proof of TRS/TRCSRExConc_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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MARK₁(h₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.

MARK₁(f₁(X)) -> MARK₁(X)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MARK₁(x₁) ) = x₁

POL( h₁(x₁) ) = x₁ + 1

POL( f₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MARK₁(f₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MARK₁(x₁) ) = x₁

POL( f₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ 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.