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__f(g(c))
a__f(g(X)) → g(X)
mark(c) → a__c
mark(f(X)) → a__f(X)
mark(g(X)) → g(X)
a__c → c
a__f(X) → f(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__c → a__f(g(c))
a__f(g(X)) → g(X)
mark(c) → a__c
mark(f(X)) → a__f(X)
mark(g(X)) → g(X)
a__c → c
a__f(X) → f(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:

MARK(c) → A__C
MARK(f(X)) → A__F(X)
A__C → A__F(g(c))

The TRS R consists of the following rules:

a__c → a__f(g(c))
a__f(g(X)) → g(X)
mark(c) → a__c
mark(f(X)) → a__f(X)
mark(g(X)) → g(X)
a__c → c
a__f(X) → f(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

MARK(c) → A__C
MARK(f(X)) → A__F(X)
A__C → A__F(g(c))

The TRS R consists of the following rules:

a__c → a__f(g(c))
a__f(g(X)) → g(X)
mark(c) → a__c
mark(f(X)) → a__f(X)
mark(g(X)) → g(X)
a__c → c
a__f(X) → f(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK(c) → A__C
MARK(f(X)) → A__F(X)
A__C → A__F(g(c))

The TRS R consists of the following rules:

a__c → a__f(g(c))
a__f(g(X)) → g(X)
mark(c) → a__c
mark(f(X)) → a__f(X)
mark(g(X)) → g(X)
a__c → c
a__f(X) → f(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.