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)
MARK1(f1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MARK1(x1)) = 3·x1 + 3·x12   
POL(f1(x1)) = 3 + 3·x1 + 3·x12   
POL(h1(x1)) = 3 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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.