Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/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.

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 strictly oriented and are deleted.

A__H₁(d) -> A__G₁(c)

The remaining pairs can at least by weakly be oriented.

A__G₁(X) -> A__H₁(X)

Used ordering: Combined order from the following AFS and order.
A__H₁(x1) = A__H₁(x1)
d = d
A__G₁(x1) = A__G₁(x1)
c = c

Lexicographic Path Order [19].
Precedence:

[A_{_H₁}, A_{_{G_1]}} > c
d > c

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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 [13,14,18] contains 0 SCCs with 1 less node.