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__g(X) → a__h(X)
a__c → d
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__c → c

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__g(X) → a__h(X)
a__c → d
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__c → c

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(X)
A__H(d) → A__G(c)
MARK(g(X)) → A__G(X)
MARK(c) → A__C
A__G(X) → A__H(X)

The TRS R consists of the following rules:

a__g(X) → a__h(X)
a__c → d
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__c → c

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(X)
A__H(d) → A__G(c)
MARK(g(X)) → A__G(X)
MARK(c) → A__C
A__G(X) → A__H(X)

The TRS R consists of the following rules:

a__g(X) → a__h(X)
a__c → d
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__c → c

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 3 less nodes.

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

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

A__H(d) → A__G(c)
A__G(X) → A__H(X)

The TRS R consists of the following rules:

a__g(X) → a__h(X)
a__c → d
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__c → c

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.


A__G(X) → A__H(X)
The remaining pairs can at least be oriented weakly.

A__H(d) → A__G(c)
Used ordering: Polynomial interpretation [25,35]:

POL(c) = 0   
POL(A__H(x1)) = 3/2 + (1/4)x_1   
POL(d) = 3   
POL(A__G(x1)) = 9/4 + (4)x_1   
The value of delta used in the strict ordering is 3/4.
The following usable rules [17] were oriented: none



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

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

A__H(d) → A__G(c)

The TRS R consists of the following rules:

a__g(X) → a__h(X)
a__c → d
a__h(d) → a__g(c)
mark(g(X)) → a__g(X)
mark(h(X)) → a__h(X)
mark(c) → a__c
mark(d) → d
a__g(X) → g(X)
a__h(X) → h(X)
a__c → c

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.