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.

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 contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

A__H1(d) -> A__G1(c)
Used argument filtering: A__H1(x1)  =  x1
d  =  d
A__G1(x1)  =  x1
c  =  c
Used ordering: Quasi Precedence: d > c


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
QDP
              ↳ DependencyGraphProof

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

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 contains 0 SCCs with 1 less node.