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:
b(b(x1)) → a(a(a(x1)))
b(a(b(x1))) → a(x1)
b(a(a(x1))) → b(a(b(x1)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(x1)) → a(a(a(x1)))
b(a(b(x1))) → a(x1)
b(a(a(x1))) → b(a(b(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:
B(a(a(x1))) → B(a(b(x1)))
B(a(a(x1))) → B(x1)
The TRS R consists of the following rules:
b(b(x1)) → a(a(a(x1)))
b(a(b(x1))) → a(x1)
b(a(a(x1))) → b(a(b(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(x1))) → B(a(b(x1)))
B(a(a(x1))) → B(x1)
The TRS R consists of the following rules:
b(b(x1)) → a(a(a(x1)))
b(a(b(x1))) → a(x1)
b(a(a(x1))) → b(a(b(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(a(x1))) → B(a(b(x1))) at position [0,0] we obtained the following new rules:
B(a(a(a(a(x0))))) → B(a(b(a(b(x0)))))
B(a(a(b(x0)))) → B(a(a(a(a(x0)))))
B(a(a(a(b(x0))))) → B(a(a(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(a(b(x0)))) → B(a(a(a(a(x0)))))
B(a(a(a(a(x0))))) → B(a(b(a(b(x0)))))
B(a(a(x1))) → B(x1)
B(a(a(a(b(x0))))) → B(a(a(x0)))
The TRS R consists of the following rules:
b(b(x1)) → a(a(a(x1)))
b(a(b(x1))) → a(x1)
b(a(a(x1))) → b(a(b(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(x1)) → a(a(a(x1)))
b(a(b(x1))) → a(x1)
b(a(a(x1))) → b(a(b(x1)))
B(a(a(b(x0)))) → B(a(a(a(a(x0)))))
B(a(a(a(a(x0))))) → B(a(b(a(b(x0)))))
B(a(a(x1))) → B(x1)
B(a(a(a(b(x0))))) → B(a(a(x0)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(x1)) → a(a(a(x1)))
b(a(b(x1))) → a(x1)
b(a(a(x1))) → b(a(b(x1)))
B(a(a(b(x0)))) → B(a(a(a(a(x0)))))
B(a(a(a(a(x0))))) → B(a(b(a(b(x0)))))
B(a(a(x1))) → B(x1)
B(a(a(a(b(x0))))) → B(a(a(x0)))
The set Q is empty.
We have obtained the following QTRS:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(x)))
b(a(a(B(x)))) → a(a(a(a(B(x)))))
a(a(a(a(B(x))))) → b(a(b(a(B(x)))))
a(a(B(x))) → B(x)
b(a(a(a(B(x))))) → a(a(B(x)))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(x)))
b(a(a(B(x)))) → a(a(a(a(B(x)))))
a(a(a(a(B(x))))) → b(a(b(a(B(x)))))
a(a(B(x))) → B(x)
b(a(a(a(B(x))))) → a(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:
B1(a(b(x))) → A(x)
A(a(a(a(B(x))))) → B1(a(B(x)))
B1(b(x)) → A(a(x))
A(a(b(x))) → B1(a(b(x)))
A(a(a(a(B(x))))) → B1(a(b(a(B(x)))))
B1(b(x)) → A(x)
A(a(a(a(B(x))))) → A(b(a(B(x))))
B1(a(a(B(x)))) → A(a(a(a(B(x)))))
B1(a(a(B(x)))) → A(a(a(B(x))))
B1(b(x)) → A(a(a(x)))
The TRS R consists of the following rules:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(x)))
b(a(a(B(x)))) → a(a(a(a(B(x)))))
a(a(a(a(B(x))))) → b(a(b(a(B(x)))))
a(a(B(x))) → B(x)
b(a(a(a(B(x))))) → a(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(x))) → A(x)
A(a(a(a(B(x))))) → B1(a(B(x)))
B1(b(x)) → A(a(x))
A(a(b(x))) → B1(a(b(x)))
A(a(a(a(B(x))))) → B1(a(b(a(B(x)))))
B1(b(x)) → A(x)
A(a(a(a(B(x))))) → A(b(a(B(x))))
B1(a(a(B(x)))) → A(a(a(a(B(x)))))
B1(a(a(B(x)))) → A(a(a(B(x))))
B1(b(x)) → A(a(a(x)))
The TRS R consists of the following rules:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(x)))
b(a(a(B(x)))) → a(a(a(a(B(x)))))
a(a(a(a(B(x))))) → b(a(b(a(B(x)))))
a(a(B(x))) → B(x)
b(a(a(a(B(x))))) → a(a(B(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
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(x))) → A(x)
B1(b(x)) → A(a(x))
A(a(b(x))) → B1(a(b(x)))
A(a(a(a(B(x))))) → B1(a(b(a(B(x)))))
B1(b(x)) → A(x)
B1(a(a(B(x)))) → A(a(a(a(B(x)))))
B1(a(a(B(x)))) → A(a(a(B(x))))
B1(b(x)) → A(a(a(x)))
The TRS R consists of the following rules:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(x)))
b(a(a(B(x)))) → a(a(a(a(B(x)))))
a(a(a(a(B(x))))) → b(a(b(a(B(x)))))
a(a(B(x))) → B(x)
b(a(a(a(B(x))))) → a(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(B(x)))) → A(a(a(B(x)))) at position [0] we obtained the following new rules:
B1(a(a(B(x0)))) → A(B(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(x))) → A(x)
B1(b(x)) → A(a(x))
A(a(b(x))) → B1(a(b(x)))
A(a(a(a(B(x))))) → B1(a(b(a(B(x)))))
B1(b(x)) → A(x)
B1(a(a(B(x)))) → A(a(a(a(B(x)))))
B1(a(a(B(x0)))) → A(B(x0))
B1(b(x)) → A(a(a(x)))
The TRS R consists of the following rules:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(x)))
b(a(a(B(x)))) → a(a(a(a(B(x)))))
a(a(a(a(B(x))))) → b(a(b(a(B(x)))))
a(a(B(x))) → B(x)
b(a(a(a(B(x))))) → a(a(B(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(x))) → A(x)
B1(b(x)) → A(a(x))
A(a(b(x))) → B1(a(b(x)))
A(a(a(a(B(x))))) → B1(a(b(a(B(x)))))
B1(b(x)) → A(x)
B1(a(a(B(x)))) → A(a(a(a(B(x)))))
B1(b(x)) → A(a(a(x)))
The TRS R consists of the following rules:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(x)))
b(a(a(B(x)))) → a(a(a(a(B(x)))))
a(a(a(a(B(x))))) → b(a(b(a(B(x)))))
a(a(B(x))) → B(x)
b(a(a(a(B(x))))) → a(a(B(x)))
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
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(x)))
b(a(a(B(x)))) → a(a(a(a(B(x)))))
a(a(a(a(B(x))))) → b(a(b(a(B(x)))))
a(a(B(x))) → B(x)
b(a(a(a(B(x))))) → a(a(B(x)))
The set Q is empty.
We have obtained the following QTRS:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
b(a(a(x))) → b(a(b(x)))
B(a(a(b(x)))) → B(a(a(a(a(x)))))
B(a(a(a(a(x))))) → B(a(b(a(b(x)))))
B(a(a(x))) → B(x)
B(a(a(a(b(x))))) → B(a(a(x)))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
b(a(a(x))) → b(a(b(x)))
B(a(a(b(x)))) → B(a(a(a(a(x)))))
B(a(a(a(a(x))))) → B(a(b(a(b(x)))))
B(a(a(x))) → B(x)
B(a(a(a(b(x))))) → B(a(a(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(x)))
b(a(a(B(x)))) → a(a(a(a(B(x)))))
a(a(a(a(B(x))))) → b(a(b(a(B(x)))))
a(a(B(x))) → B(x)
b(a(a(a(B(x))))) → a(a(B(x)))
The set Q is empty.
We have obtained the following QTRS:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
b(a(a(x))) → b(a(b(x)))
B(a(a(b(x)))) → B(a(a(a(a(x)))))
B(a(a(a(a(x))))) → B(a(b(a(b(x)))))
B(a(a(x))) → B(x)
B(a(a(a(b(x))))) → B(a(a(x)))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
b(a(a(x))) → b(a(b(x)))
B(a(a(b(x)))) → B(a(a(a(a(x)))))
B(a(a(a(a(x))))) → B(a(b(a(b(x)))))
B(a(a(x))) → B(x)
B(a(a(a(b(x))))) → B(a(a(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(x1)) → a(a(a(x1)))
b(a(b(x1))) → a(x1)
b(a(a(x1))) → b(a(b(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(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:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(x1)) → a(a(a(x1)))
b(a(b(x1))) → a(x1)
b(a(a(x1))) → b(a(b(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(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:
b(b(x)) → a(a(a(x)))
b(a(b(x))) → a(x)
a(a(b(x))) → b(a(b(x)))
Q is empty.