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__c → a__f(g(c))
a__f(g(X)) → g(X)
mark(c) → a__c
mark(f(X)) → a__f(X)
mark(g(X)) → g(X)
a__c → c
a__f(X) → f(X)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a__c → a__f(g(c))
a__f(g(X)) → g(X)
mark(c) → a__c
mark(f(X)) → a__f(X)
mark(g(X)) → g(X)
a__c → c
a__f(X) → f(X)
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
MARK(f(X)) → A__F(X)
MARK(c) → A__C
A__C → A__F(g(c))
The TRS R consists of the following rules:
a__c → a__f(g(c))
a__f(g(X)) → g(X)
mark(c) → a__c
mark(f(X)) → a__f(X)
mark(g(X)) → g(X)
a__c → c
a__f(X) → f(X)
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:
MARK(f(X)) → A__F(X)
MARK(c) → A__C
A__C → A__F(g(c))
The TRS R consists of the following rules:
a__c → a__f(g(c))
a__f(g(X)) → g(X)
mark(c) → a__c
mark(f(X)) → a__f(X)
mark(g(X)) → g(X)
a__c → c
a__f(X) → f(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 3 less nodes.