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__c -> a__f1(g1(c))
a__f1(g1(X)) -> g1(X)
mark1(c) -> a__c
mark1(f1(X)) -> a__f1(X)
mark1(g1(X)) -> g1(X)
a__c -> c
a__f1(X) -> f1(X)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a__c -> a__f1(g1(c))
a__f1(g1(X)) -> g1(X)
mark1(c) -> a__c
mark1(f1(X)) -> a__f1(X)
mark1(g1(X)) -> g1(X)
a__c -> c
a__f1(X) -> f1(X)

Q is empty.

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

MARK1(c) -> A__C
MARK1(f1(X)) -> A__F1(X)
A__C -> A__F1(g1(c))

The TRS R consists of the following rules:

a__c -> a__f1(g1(c))
a__f1(g1(X)) -> g1(X)
mark1(c) -> a__c
mark1(f1(X)) -> a__f1(X)
mark1(g1(X)) -> g1(X)
a__c -> c
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(c) -> A__C
MARK1(f1(X)) -> A__F1(X)
A__C -> A__F1(g1(c))

The TRS R consists of the following rules:

a__c -> a__f1(g1(c))
a__f1(g1(X)) -> g1(X)
mark1(c) -> a__c
mark1(f1(X)) -> a__f1(X)
mark1(g1(X)) -> g1(X)
a__c -> c
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 0 SCCs with 3 less nodes.