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(a(b(x1))) → b(b(a(a(x1))))
b(b(x1)) → c(a(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(a(b(x1))) → b(b(a(a(x1))))
b(b(x1)) → c(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(a(b(x1))) → B(b(a(a(x1))))
A(a(b(x1))) → B(a(a(x1)))
B(b(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(b(x1)) → c(a(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:
A(a(b(x1))) → B(b(a(a(x1))))
A(a(b(x1))) → B(a(a(x1)))
B(b(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(b(x1)) → c(a(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(b(x1))) → B(a(a(x1))) at position [0] we obtained the following new rules:
A(a(b(a(b(x0))))) → B(a(b(b(a(a(x0))))))
A(a(b(b(x0)))) → B(b(b(a(a(x0)))))
A(a(b(y0))) → B(a(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(x1))) → B(b(a(a(x1))))
A(a(b(a(b(x0))))) → B(a(b(b(a(a(x0))))))
B(b(x1)) → A(x1)
A(a(b(b(x0)))) → B(b(b(a(a(x0)))))
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
A(a(b(y0))) → B(a(y0))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(b(x1)) → c(a(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:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(b(x1)) → c(a(x1))
A(a(b(x1))) → B(b(a(a(x1))))
A(a(b(a(b(x0))))) → B(a(b(b(a(a(x0))))))
B(b(x1)) → A(x1)
A(a(b(b(x0)))) → B(b(b(a(a(x0)))))
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
A(a(b(y0))) → B(a(y0))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(b(x1)) → c(a(x1))
A(a(b(x1))) → B(b(a(a(x1))))
A(a(b(a(b(x0))))) → B(a(b(b(a(a(x0))))))
B(b(x1)) → A(x1)
A(a(b(b(x0)))) → B(b(b(a(a(x0)))))
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
A(a(b(y0))) → B(a(y0))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → a(B(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → 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(A(x))) → A1(b(B(x)))
B1(b(a(A(x)))) → B1(b(B(x)))
B1(a(b(a(A(x))))) → A1(B(x))
B1(a(A(x))) → A1(a(b(B(x))))
B1(a(b(a(A(x))))) → A1(b(b(a(B(x)))))
B1(b(x)) → A1(c(x))
B1(a(A(x))) → B1(B(x))
B1(b(a(A(x)))) → A1(b(b(B(x))))
B1(a(a(x))) → A1(b(b(x)))
B1(a(b(a(A(x))))) → A1(a(b(b(a(B(x))))))
B1(a(b(a(A(x))))) → B1(a(B(x)))
B1(a(a(x))) → A1(a(b(b(x))))
B1(b(a(A(x)))) → B1(B(x))
B1(a(a(x))) → B1(b(x))
B1(b(a(A(x)))) → A1(a(b(b(B(x)))))
B1(a(a(x))) → B1(x)
B1(a(A(x))) → A1(B(x))
B1(a(b(a(A(x))))) → B1(b(a(B(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → 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
Q DP problem:
The TRS P consists of the following rules:
B1(a(A(x))) → A1(b(B(x)))
B1(b(a(A(x)))) → B1(b(B(x)))
B1(a(b(a(A(x))))) → A1(B(x))
B1(a(A(x))) → A1(a(b(B(x))))
B1(a(b(a(A(x))))) → A1(b(b(a(B(x)))))
B1(b(x)) → A1(c(x))
B1(a(A(x))) → B1(B(x))
B1(b(a(A(x)))) → A1(b(b(B(x))))
B1(a(a(x))) → A1(b(b(x)))
B1(a(b(a(A(x))))) → A1(a(b(b(a(B(x))))))
B1(a(b(a(A(x))))) → B1(a(B(x)))
B1(a(a(x))) → A1(a(b(b(x))))
B1(b(a(A(x)))) → B1(B(x))
B1(a(a(x))) → B1(b(x))
B1(b(a(A(x)))) → A1(a(b(b(B(x)))))
B1(a(a(x))) → B1(x)
B1(a(A(x))) → A1(B(x))
B1(a(b(a(A(x))))) → B1(b(a(B(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → 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 13 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(A(x)))) → B1(b(B(x)))
B1(a(b(a(A(x))))) → B1(a(B(x)))
B1(a(a(x))) → B1(b(x))
B1(a(a(x))) → B1(x)
B1(a(b(a(A(x))))) → B1(b(a(B(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → a(B(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(b(a(A(x))))) → B1(b(a(B(x)))) at position [0] we obtained the following new rules:
B1(a(b(a(A(y0))))) → B1(b(B(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(a(A(x))))) → B1(a(B(x)))
B1(b(a(A(x)))) → B1(b(B(x)))
B1(a(a(x))) → B1(b(x))
B1(a(b(a(A(y0))))) → B1(b(B(y0)))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → a(B(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(b(a(A(x))))) → B1(a(B(x))) at position [0] we obtained the following new rules:
B1(a(b(a(A(y0))))) → B1(B(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(A(x)))) → B1(b(B(x)))
B1(a(b(a(A(y0))))) → B1(B(y0))
B1(a(a(x))) → B1(b(x))
B1(a(b(a(A(y0))))) → B1(b(B(y0)))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → 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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(A(x)))) → B1(b(B(x)))
B1(a(a(x))) → B1(b(x))
B1(a(b(a(A(y0))))) → B1(b(B(y0)))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → a(B(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(a(A(x)))) → B1(b(B(x))) at position [0] we obtained the following new rules:
B1(b(a(A(x0)))) → B1(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(A(x0)))) → B1(A(x0))
B1(a(a(x))) → B1(b(x))
B1(a(b(a(A(y0))))) → B1(b(B(y0)))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → 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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(x))) → B1(b(x))
B1(a(b(a(A(y0))))) → B1(b(B(y0)))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → a(B(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(b(a(A(y0))))) → B1(b(B(y0))) at position [0] we obtained the following new rules:
B1(a(b(a(A(x0))))) → B1(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(x))) → B1(b(x))
B1(a(b(a(A(x0))))) → B1(A(x0))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → 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
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(x))) → B1(b(x))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
b(a(A(x))) → a(a(b(B(x))))
b(a(b(a(A(x))))) → a(a(b(b(a(B(x))))))
b(B(x)) → A(x)
b(b(a(A(x)))) → a(a(b(b(B(x)))))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
b(a(A(x))) → 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
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(b(x1)) → c(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(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(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(b(x1)) → c(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(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(a(a(x))) → a(a(b(b(x))))
b(b(x)) → a(c(x))
Q is empty.