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__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(X)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK(h(X)) → A__H(mark(X))
MARK(h(X)) → MARK(X)
A__F(f(X)) → A__C(f(g(f(X))))
A__H(X) → A__C(d(X))
MARK(c(X)) → A__C(X)
MARK(f(X)) → A__F(mark(X))
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(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:

MARK(h(X)) → A__H(mark(X))
MARK(h(X)) → MARK(X)
A__F(f(X)) → A__C(f(g(f(X))))
A__H(X) → A__C(d(X))
MARK(c(X)) → A__C(X)
MARK(f(X)) → A__F(mark(X))
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

MARK(h(X)) → MARK(X)
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MARK(f(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

MARK(h(X)) → MARK(X)
Used ordering: Polynomial interpretation [25,35]:

POL(MARK(x1)) = (2)x_1   
POL(f(x1)) = 4 + (2)x_1   
POL(h(x1)) = (2)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof

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

MARK(h(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MARK(h(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MARK(x1)) = (2)x_1   
POL(h(x1)) = 4 + (4)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] 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__f(f(X)) → a__c(f(g(f(X))))
a__c(X) → d(X)
a__h(X) → a__c(d(X))
mark(f(X)) → a__f(mark(X))
mark(c(X)) → a__c(X)
mark(h(X)) → a__h(mark(X))
mark(g(X)) → g(X)
mark(d(X)) → d(X)
a__f(X) → f(X)
a__c(X) → c(X)
a__h(X) → h(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.