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(b(b(x1)))) → a(b(a(a(x1))))
a(x1) → b(b(b(x1)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(x1)))) → a(b(a(a(x1))))
a(x1) → b(b(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:
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(b(a(a(x1))))
A(a(b(b(x1)))) → A(x1)
The TRS R consists of the following rules:
a(a(b(b(x1)))) → a(b(a(a(x1))))
a(x1) → b(b(b(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(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(b(a(a(x1))))
A(a(b(b(x1)))) → A(x1)
The TRS R consists of the following rules:
a(a(b(b(x1)))) → a(b(a(a(x1))))
a(x1) → b(b(b(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
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(x1)
The TRS R consists of the following rules:
a(a(b(b(x1)))) → a(b(a(a(x1))))
a(x1) → b(b(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(x1)))) → a(b(a(a(x1))))
a(x1) → b(b(b(x1)))
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(x1)))) → a(b(a(a(x1))))
a(x1) → b(b(b(x1)))
A(a(b(b(x1)))) → A(a(x1))
A(a(b(b(x1)))) → A(x1)
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → 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:
A1(x) → B(x)
B(b(a(a(x)))) → A1(b(a(x)))
B(b(a(a(x)))) → B(a(x))
A1(x) → B(b(x))
B(b(a(a(x)))) → A1(a(b(a(x))))
A1(x) → B(b(b(x)))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B(x)
B(b(a(a(x)))) → A1(b(a(x)))
B(b(a(a(x)))) → B(a(x))
A1(x) → B(b(x))
B(b(a(a(x)))) → A1(a(b(a(x))))
A1(x) → B(b(b(x)))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(a(a(x)))) → B(a(x)) at position [0] we obtained the following new rules:
B(b(a(a(x0)))) → B(b(b(b(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(x)))) → A1(b(a(x)))
A1(x) → B(x)
B(b(a(a(x0)))) → B(b(b(b(x0))))
A1(x) → B(b(x))
B(b(a(a(x)))) → A1(a(b(a(x))))
A1(x) → B(b(b(x)))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(x) → B(b(b(x))) at position [0] we obtained the following new rules:
A1(b(a(A(x0)))) → B(b(a(A(x0))))
A1(b(a(A(x0)))) → B(b(A(x0)))
A1(a(A(x0))) → B(A(x0))
A1(a(A(x0))) → B(a(A(x0)))
A1(b(a(a(x0)))) → B(b(a(a(b(a(x0))))))
A1(a(a(x0))) → B(a(a(b(a(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B(x)
B(b(a(a(x)))) → A1(b(a(x)))
A1(a(A(x0))) → B(a(A(x0)))
A1(b(a(a(x0)))) → B(b(a(a(b(a(x0))))))
A1(b(a(A(x0)))) → B(b(a(A(x0))))
A1(a(A(x0))) → B(A(x0))
A1(b(a(A(x0)))) → B(b(A(x0)))
B(b(a(a(x0)))) → B(b(b(b(x0))))
A1(x) → B(b(x))
B(b(a(a(x)))) → A1(a(b(a(x))))
A1(a(a(x0))) → B(a(a(b(a(x0)))))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → A(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(x)))) → A1(b(a(x)))
A1(x) → B(x)
A1(a(A(x0))) → B(a(A(x0)))
A1(b(a(a(x0)))) → B(b(a(a(b(a(x0))))))
A1(b(a(A(x0)))) → B(b(a(A(x0))))
B(b(a(a(x0)))) → B(b(b(b(x0))))
A1(x) → B(b(x))
B(b(a(a(x)))) → A1(a(b(a(x))))
A1(a(a(x0))) → B(a(a(b(a(x0)))))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(b(a(A(x0)))) → B(b(a(A(x0)))) at position [0] we obtained the following new rules:
A1(b(a(A(y0)))) → B(b(b(b(b(A(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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:
A1(x) → B(x)
B(b(a(a(x)))) → A1(b(a(x)))
A1(a(A(x0))) → B(a(A(x0)))
A1(b(a(a(x0)))) → B(b(a(a(b(a(x0))))))
B(b(a(a(x0)))) → B(b(b(b(x0))))
A1(x) → B(b(x))
B(b(a(a(x)))) → A1(a(b(a(x))))
A1(b(a(A(y0)))) → B(b(b(b(b(A(y0))))))
A1(a(a(x0))) → B(a(a(b(a(x0)))))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → A(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B(x)
B(b(a(a(x)))) → A1(b(a(x)))
A1(a(A(x0))) → B(a(A(x0)))
A1(b(a(a(x0)))) → B(b(a(a(b(a(x0))))))
B(b(a(a(x0)))) → B(b(b(b(x0))))
A1(x) → B(b(x))
B(b(a(a(x)))) → A1(a(b(a(x))))
A1(a(a(x0))) → B(a(a(b(a(x0)))))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(A(x0))) → B(a(A(x0))) at position [0] we obtained the following new rules:
A1(a(A(y0))) → B(b(b(b(A(y0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A1(a(A(y0))) → B(b(b(b(A(y0)))))
B(b(a(a(x)))) → A1(b(a(x)))
A1(x) → B(x)
A1(b(a(a(x0)))) → B(b(a(a(b(a(x0))))))
B(b(a(a(x0)))) → B(b(b(b(x0))))
A1(x) → B(b(x))
B(b(a(a(x)))) → A1(a(b(a(x))))
A1(a(a(x0))) → B(a(a(b(a(x0)))))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → A(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
B(b(a(a(x)))) → A1(b(a(x)))
A1(x) → B(x)
A1(b(a(a(x0)))) → B(b(a(a(b(a(x0))))))
B(b(a(a(x0)))) → B(b(b(b(x0))))
A1(x) → B(b(x))
B(b(a(a(x)))) → A1(a(b(a(x))))
A1(a(a(x0))) → B(a(a(b(a(x0)))))
The TRS R consists of the following rules:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → 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
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(a(b(b(x)))) → a(b(a(a(x))))
a(x) → b(b(b(x)))
A(a(b(b(x)))) → A(a(x))
A(a(b(b(x)))) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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:
a(a(b(b(x)))) → a(b(a(a(x))))
a(x) → b(b(b(x)))
A(a(b(b(x)))) → A(a(x))
A(a(b(b(x)))) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
b(b(a(A(x)))) → a(A(x))
b(b(a(A(x)))) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(a(b(b(x)))) → a(b(a(a(x))))
a(x) → b(b(b(x)))
A(a(b(b(x)))) → A(a(x))
A(a(b(b(x)))) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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:
a(a(b(b(x)))) → a(b(a(a(x))))
a(x) → b(b(b(x)))
A(a(b(b(x)))) → A(a(x))
A(a(b(b(x)))) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(x1)))) → a(b(a(a(x1))))
a(x1) → b(b(b(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(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(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(x1)))) → a(b(a(a(x1))))
a(x1) → b(b(b(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(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(a(a(x)))) → a(a(b(a(x))))
a(x) → b(b(b(x)))
Q is empty.