YES
Termination w.r.t. Q proof of ../tpdb/TRS/Rubio/p266.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(a(b(f(X))))
f(a(g(X))) → b(X)
b(X) → a(X)
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(X)) → f(a(b(f(X))))
f(a(g(X))) → b(X)
b(X) → a(X)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(X)) → f(a(b(f(X))))
f(a(g(X))) → b(X)
b(X) → a(X)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
f(a(g(X))) → b(X)
Used ordering:
Polynomial interpretation [25]:
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(f(x1)) = 1 + 2·x1
POL(g(x1)) = 2·x1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(X)) → f(a(b(f(X))))
b(X) → a(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(f(X)) → B(f(X))
F(f(X)) → F(a(b(f(X))))
The TRS R consists of the following rules:
f(f(X)) → f(a(b(f(X))))
b(X) → a(X)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(f(X)) → B(f(X))
F(f(X)) → F(a(b(f(X))))
The TRS R consists of the following rules:
f(f(X)) → f(a(b(f(X))))
b(X) → a(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 2 less nodes.