Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/TRCSRSLASHEx1_Zan97

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: 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.