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(x1) → x1
a(b(b(x1))) → b(b(b(a(c(a(x1))))))
c(b(x1)) → x1
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(a(c(a(x1))))))
c(b(x1)) → 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))) → C(a(x1))
A(b(b(x1))) → A(c(a(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(a(c(a(x1))))))
c(b(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(x1))) → A(x1)
A(b(b(x1))) → C(a(x1))
A(b(b(x1))) → A(c(a(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(a(c(a(x1))))))
c(b(x1)) → x1
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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(x1))) → A(x1)
A(b(b(x1))) → A(c(a(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(a(c(a(x1))))))
c(b(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(x1))) → A(c(a(x1))) at position [0] we obtained the following new rules:
A(b(b(b(b(x0))))) → A(c(b(b(b(a(c(a(x0))))))))
A(b(b(x0))) → A(c(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(x1))) → A(x1)
A(b(b(b(b(x0))))) → A(c(b(b(b(a(c(a(x0))))))))
A(b(b(x0))) → A(c(x0))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(a(c(a(x1))))))
c(b(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(b(x0))) → A(c(x0)) at position [0] we obtained the following new rules:
A(b(b(b(x0)))) → A(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(b(b(x0)))) → A(x0)
A(b(b(x1))) → A(x1)
A(b(b(b(b(x0))))) → A(c(b(b(b(a(c(a(x0))))))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(b(x1))) → b(b(b(a(c(a(x1))))))
c(b(x1)) → 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(x1) → x1
a(b(b(x1))) → b(b(b(a(c(a(x1))))))
c(b(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(b(a(x))) → a(c(a(b(b(b(x))))))
b(c(x)) → x
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(a(b(b(b(x))))))
b(c(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(b(x1))) → b(b(b(a(c(a(x1))))))
c(b(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(b(a(x))) → a(c(a(b(b(b(x))))))
b(c(x)) → x
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(b(a(x))) → a(c(a(b(b(b(x))))))
b(c(x)) → x
Q is empty.