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.