YES
Termination w.r.t. Q proof of ../tpdb/TRS/Rubio/aoto.trs
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:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(X))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(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:
F(g(f(X))) → F(g(X))
F(f(X)) → F(g(f(g(f(X)))))
F(f(X)) → F(g(f(X)))
The TRS R consists of the following rules:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(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:
F(g(f(X))) → F(g(X))
F(f(X)) → F(g(f(g(f(X)))))
F(f(X)) → F(g(f(X)))
The TRS R consists of the following rules:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
F(g(f(X))) → F(g(X))
The TRS R consists of the following rules:
f(f(X)) → f(g(f(g(f(X)))))
f(g(f(X))) → f(g(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
Q DP problem:
The TRS P consists of the following rules:
F(g(f(X))) → F(g(X))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.
The following dependency pairs can be deleted:
F(g(f(X))) → F(g(X))
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(F(x1)) = 2·x1
POL(f(x1)) = 2·x1
POL(g(x1)) = 2·x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.