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 strictly oriented and are deleted.


MARK1(f1(X)) -> MARK1(X)
The remaining pairs can at least by weakly be oriented.

MARK1(h1(X)) -> MARK1(X)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
h1(x1)  =  x1
f1(x1)  =  f1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


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(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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


MARK1(h1(X)) -> MARK1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
h1(x1)  =  h1(x1)

Lexicographic Path Order [19].
Precedence:
h1 > MARK1


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.