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(b(a(b(x)))) → b(a(b(a(a(b(x))))))
The set Q consists of the following terms:
a(b(a(b(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(a(b(x)))) → b(a(b(a(a(b(x))))))
The set Q consists of the following terms:
a(b(a(b(x0))))
We have reversed the following QTRS:
The set of rules R is
a(b(a(b(x)))) → b(a(b(a(a(b(x))))))
The set Q is {a(b(a(b(x0))))}.
We have obtained the following QTRS:
b(a(b(a(x)))) → b(a(a(b(a(b(x))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(a(x)))) → b(a(a(b(a(b(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:
A(b(a(b(x)))) → A(a(b(x)))
A(b(a(b(x)))) → A(b(a(a(b(x)))))
The TRS R consists of the following rules:
a(b(a(b(x)))) → b(a(b(a(a(b(x))))))
The set Q consists of the following terms:
a(b(a(b(x0))))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A(b(a(b(x)))) → A(a(b(x)))
A(b(a(b(x)))) → A(b(a(a(b(x)))))
The TRS R consists of the following rules:
a(b(a(b(x)))) → b(a(b(a(a(b(x))))))
The set Q consists of the following terms:
a(b(a(b(x0))))
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.