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__g1(X) -> a__h1(X)
a__c -> d
a__h1(d) -> a__g1(c)
mark1(g1(X)) -> a__g1(X)
mark1(h1(X)) -> a__h1(X)
mark1(c) -> a__c
mark1(d) -> d
a__g1(X) -> g1(X)
a__h1(X) -> h1(X)
a__c -> c

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__g1(X) -> a__h1(X)
a__c -> d
a__h1(d) -> a__g1(c)
mark1(g1(X)) -> a__g1(X)
mark1(h1(X)) -> a__h1(X)
mark1(c) -> a__c
mark1(d) -> d
a__g1(X) -> g1(X)
a__h1(X) -> h1(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:

MARK1(c) -> A__C
MARK1(h1(X)) -> A__H1(X)
MARK1(g1(X)) -> A__G1(X)
A__H1(d) -> A__G1(c)
A__G1(X) -> A__H1(X)

The TRS R consists of the following rules:

a__g1(X) -> a__h1(X)
a__c -> d
a__h1(d) -> a__g1(c)
mark1(g1(X)) -> a__g1(X)
mark1(h1(X)) -> a__h1(X)
mark1(c) -> a__c
mark1(d) -> d
a__g1(X) -> g1(X)
a__h1(X) -> h1(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:

MARK1(c) -> A__C
MARK1(h1(X)) -> A__H1(X)
MARK1(g1(X)) -> A__G1(X)
A__H1(d) -> A__G1(c)
A__G1(X) -> A__H1(X)

The TRS R consists of the following rules:

a__g1(X) -> a__h1(X)
a__c -> d
a__h1(d) -> a__g1(c)
mark1(g1(X)) -> a__g1(X)
mark1(h1(X)) -> a__h1(X)
mark1(c) -> a__c
mark1(d) -> d
a__g1(X) -> g1(X)
a__h1(X) -> h1(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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

A__H1(d) -> A__G1(c)
A__G1(X) -> A__H1(X)

The TRS R consists of the following rules:

a__g1(X) -> a__h1(X)
a__c -> d
a__h1(d) -> a__g1(c)
mark1(g1(X)) -> a__g1(X)
mark1(h1(X)) -> a__h1(X)
mark1(c) -> a__c
mark1(d) -> d
a__g1(X) -> g1(X)
a__h1(X) -> h1(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__H1(d) -> A__G1(c)
A__G1(X) -> A__H1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(A__G1(x1)) = 2 + 2·x1   
POL(A__H1(x1)) = 1 + 2·x1   
POL(c) = 0   
POL(d) = 2   

The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a__g1(X) -> a__h1(X)
a__c -> d
a__h1(d) -> a__g1(c)
mark1(g1(X)) -> a__g1(X)
mark1(h1(X)) -> a__h1(X)
mark1(c) -> a__c
mark1(d) -> d
a__g1(X) -> g1(X)
a__h1(X) -> h1(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.