Termination w.r.t. Q proof of TRS/TRCSREx6_GM04

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.

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(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 [13,14,18] contains 0 SCCs with 3 less nodes.