Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(x1)) → x1
a(b(b(x1))) → b(b(a(b(a(x1)))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(x1)) → x1
a(b(b(x1))) → b(b(a(b(a(x1)))))
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(b(x1))) → A(x1)
A(b(b(x1))) → A(b(a(x1)))
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(x1)) → x1
a(b(b(x1))) → b(b(a(b(a(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(x1))) → A(x1)
A(b(b(x1))) → A(b(a(x1)))
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(x1)) → x1
a(b(b(x1))) → b(b(a(b(a(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(x1))) → A(b(a(x1))) at position [0,0] we obtained the following new rules:
A(b(b(a(x0)))) → A(b(x0))
A(b(b(b(x0)))) → A(b(x0))
A(b(b(b(b(x0))))) → A(b(b(b(a(b(a(x0)))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(x1))) → A(x1)
A(b(b(b(x0)))) → A(b(x0))
A(b(b(a(x0)))) → A(b(x0))
A(b(b(b(b(x0))))) → A(b(b(b(a(b(a(x0)))))))
The TRS R consists of the following rules:
a(a(x1)) → x1
a(b(x1)) → x1
a(b(b(x1))) → b(b(a(b(a(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → x1
a(b(x1)) → x1
a(b(b(x1))) → b(b(a(b(a(x1)))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → x
b(a(x)) → x
b(b(a(x))) → a(b(a(b(b(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → x
b(a(x)) → x
b(b(a(x))) → a(b(a(b(b(x)))))
Q is empty.