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__f(X) → g(h(f(X)))
mark(f(X)) → a__f(mark(X))
mark(g(X)) → g(X)
mark(h(X)) → h(mark(X))
a__f(X) → f(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__f(X) → g(h(f(X)))
mark(f(X)) → a__f(mark(X))
mark(g(X)) → g(X)
mark(h(X)) → h(mark(X))
a__f(X) → f(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:

MARK(f(X)) → A__F(mark(X))
MARK(h(X)) → MARK(X)
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(X) → g(h(f(X)))
mark(f(X)) → a__f(mark(X))
mark(g(X)) → g(X)
mark(h(X)) → h(mark(X))
a__f(X) → f(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

MARK(f(X)) → A__F(mark(X))
MARK(h(X)) → MARK(X)
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(X) → g(h(f(X)))
mark(f(X)) → a__f(mark(X))
mark(g(X)) → g(X)
mark(h(X)) → h(mark(X))
a__f(X) → f(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

MARK(f(X)) → A__F(mark(X))
MARK(h(X)) → MARK(X)
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(X) → g(h(f(X)))
mark(f(X)) → a__f(mark(X))
mark(g(X)) → g(X)
mark(h(X)) → h(mark(X))
a__f(X) → f(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

MARK(h(X)) → MARK(X)
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(X) → g(h(f(X)))
mark(f(X)) → a__f(mark(X))
mark(g(X)) → g(X)
mark(h(X)) → h(mark(X))
a__f(X) → f(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.


MARK(h(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

MARK(f(X)) → MARK(X)
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
h(x1)  =  h(x1)
f(x1)  =  x1

Recursive path order with status [2].
Quasi-Precedence:
[MARK1, h1]

Status:
h1: multiset
MARK1: multiset


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ QDPOrderProof

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

MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(X) → g(h(f(X)))
mark(f(X)) → a__f(mark(X))
mark(g(X)) → g(X)
mark(h(X)) → h(mark(X))
a__f(X) → f(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.


MARK(f(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
f(x1)  =  f(x1)

Recursive path order with status [2].
Quasi-Precedence:
[MARK1, f1]

Status:
f1: multiset
MARK1: multiset


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ PisEmptyProof

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

a__f(X) → g(h(f(X)))
mark(f(X)) → a__f(mark(X))
mark(g(X)) → g(X)
mark(h(X)) → h(mark(X))
a__f(X) → f(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.