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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK(c) → A__C
MARK(h(X)) → A__H(X)
MARK(g(X)) → A__G(X)
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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

MARK(c) → A__C
MARK(h(X)) → A__H(X)
MARK(g(X)) → A__G(X)
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 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(h(X)) → A__H(X)
MARK(c) → A__C
MARK(g(X)) → A__G(X)
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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13].


The following pairs can be oriented strictly and are deleted.


A__H(d) → A__G(c)
A__G(X) → A__H(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A__H(x1)  =  A__H(x1)
d  =  d
A__G(x1)  =  A__G(x1)
c  =  c

Recursive Path Order [2].
Precedence:
d > AG1 > AH1
d > c > AH1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.