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) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
b(a(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) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
b(a(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(x1) → B(c(x1))
A(b(b(x1))) → A(x1)
A(b(b(x1))) → B(a(a(x1)))
A(b(b(x1))) → A(a(x1))
A(b(b(x1))) → B(b(a(a(x1))))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
b(a(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(x1) → B(c(x1))
A(b(b(x1))) → A(x1)
A(b(b(x1))) → B(a(a(x1)))
A(b(b(x1))) → A(a(x1))
A(b(b(x1))) → B(b(a(a(x1))))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
b(a(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 3 less nodes.
↳ 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(a(x1))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
b(a(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(a(x1)) at position [0] we obtained the following new rules:
A(b(b(b(b(x0))))) → A(b(b(a(a(x0)))))
A(b(b(x0))) → A(b(c(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(b(b(x0))))) → A(b(b(a(a(x0)))))
A(b(b(x0))) → A(b(c(x0)))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
b(a(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
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ 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(b(b(a(a(x0)))))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
b(a(x1)) → 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
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
b(a(x1)) → x1
A(b(b(x1))) → A(x1)
A(b(b(b(b(x0))))) → A(b(b(a(a(x0)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
b(a(x1)) → x1
A(b(b(x1))) → A(x1)
A(b(b(b(b(x0))))) → A(b(b(a(a(x0)))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
a(b(x)) → x
b(b(A(x))) → A(x)
b(b(b(b(A(x))))) → a(a(b(b(A(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
a(b(x)) → x
b(b(A(x))) → A(x)
b(b(b(b(A(x))))) → a(a(b(b(A(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
a(b(x)) → x
b(b(A(x))) → A(x)
b(b(b(b(A(x))))) → a(a(b(b(A(x)))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(c(x))
a(b(b(x))) → b(b(a(a(x))))
b(a(x)) → x
A(b(b(x))) → A(x)
A(b(b(b(b(x))))) → A(b(b(a(a(x)))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(c(x))
a(b(b(x))) → b(b(a(a(x))))
b(a(x)) → x
A(b(b(x))) → A(x)
A(b(b(b(b(x))))) → A(b(b(a(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:
B(b(a(x))) → B(b(x))
B(b(a(x))) → B(x)
B(b(b(b(A(x))))) → A1(a(b(b(A(x)))))
A1(x) → B(x)
B(b(b(b(A(x))))) → A1(b(b(A(x))))
B(b(a(x))) → A1(b(b(x)))
B(b(a(x))) → A1(a(b(b(x))))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
a(b(x)) → x
b(b(A(x))) → A(x)
b(b(b(b(A(x))))) → a(a(b(b(A(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(x))) → B(b(x))
B(b(a(x))) → B(x)
B(b(b(b(A(x))))) → A1(a(b(b(A(x)))))
A1(x) → B(x)
B(b(b(b(A(x))))) → A1(b(b(A(x))))
B(b(a(x))) → A1(b(b(x)))
B(b(a(x))) → A1(a(b(b(x))))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
a(b(x)) → x
b(b(A(x))) → A(x)
b(b(b(b(A(x))))) → a(a(b(b(A(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
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
b(a(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
a(b(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) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
a(b(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(c(x1))
a(b(b(x1))) → b(b(a(a(x1))))
b(a(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
a(b(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) → c(b(x))
b(b(a(x))) → a(a(b(b(x))))
a(b(x)) → x
Q is empty.