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: Quasi 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: Quasi 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.