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(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
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(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
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(x1)) → B(b(x1))
A(b(x1)) → B(b(b(x1)))
A(b(x1)) → B(b(b(b(x1))))
B(a(x1)) → A(a(x1))
B(a(x1)) → A(a(a(x1)))
B(a(x1)) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(x1) → 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(b(x1)) → B(b(x1))
A(b(x1)) → B(b(b(x1)))
A(b(x1)) → B(b(b(b(x1))))
B(a(x1)) → A(a(x1))
B(a(x1)) → A(a(a(x1)))
B(a(x1)) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → B(b(b(b(x1)))) at position [0] we obtained the following new rules:
A(b(a(x0))) → B(b(b(a(a(a(a(x0)))))))
A(b(y0)) → B(b(b(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → B(b(x1))
A(b(x1)) → B(b(b(x1)))
B(a(x1)) → A(a(x1))
B(a(x1)) → A(a(a(x1)))
A(b(a(x0))) → B(b(b(a(a(a(a(x0)))))))
B(a(x1)) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → B(b(b(x1))) at position [0] we obtained the following new rules:
A(b(y0)) → B(b(y0))
A(b(a(x0))) → B(b(a(a(a(a(x0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → B(b(x1))
B(a(x1)) → A(a(x1))
B(a(x1)) → A(a(a(x1)))
A(b(a(x0))) → B(b(b(a(a(a(a(x0)))))))
A(b(a(x0))) → B(b(a(a(a(a(x0))))))
B(a(x1)) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → B(b(x1)) at position [0] we obtained the following new rules:
A(b(x0)) → B(x0)
A(b(a(x0))) → B(a(a(a(a(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x0)) → B(x0)
A(b(a(x0))) → B(a(a(a(a(x0)))))
B(a(x1)) → A(a(x1))
B(a(x1)) → A(a(a(x1)))
A(b(a(x0))) → B(b(b(a(a(a(a(x0)))))))
B(a(x1)) → A(a(a(a(x1))))
A(b(a(x0))) → B(b(a(a(a(a(x0))))))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(x1)) → A(a(a(a(x1)))) at position [0] we obtained the following new rules:
B(a(b(x0))) → A(a(a(b(b(b(b(x0)))))))
B(a(y0)) → A(a(a(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x0)) → B(x0)
B(a(x1)) → A(a(x1))
A(b(a(x0))) → B(a(a(a(a(x0)))))
B(a(x1)) → A(a(a(x1)))
A(b(a(x0))) → B(b(b(a(a(a(a(x0)))))))
A(b(a(x0))) → B(b(a(a(a(a(x0))))))
B(a(b(x0))) → A(a(a(b(b(b(b(x0)))))))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(x1)) → A(a(a(x1))) at position [0] we obtained the following new rules:
B(a(b(x0))) → A(a(b(b(b(b(x0))))))
B(a(y0)) → A(a(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(x0))) → A(a(b(b(b(b(x0))))))
A(b(x0)) → B(x0)
A(b(a(x0))) → B(a(a(a(a(x0)))))
B(a(x1)) → A(a(x1))
A(b(a(x0))) → B(b(b(a(a(a(a(x0)))))))
A(b(a(x0))) → B(b(a(a(a(a(x0))))))
B(a(b(x0))) → A(a(a(b(b(b(b(x0)))))))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(x1) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(x1)) → A(a(x1)) at position [0] we obtained the following new rules:
B(a(b(x0))) → A(b(b(b(b(x0)))))
B(a(x0)) → A(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(x0))) → A(a(b(b(b(b(x0))))))
A(b(x0)) → B(x0)
A(b(a(x0))) → B(a(a(a(a(x0)))))
B(a(x0)) → A(x0)
A(b(a(x0))) → B(b(b(a(a(a(a(x0)))))))
B(a(b(x0))) → A(b(b(b(b(x0)))))
A(b(a(x0))) → B(b(a(a(a(a(x0))))))
B(a(b(x0))) → A(a(a(b(b(b(b(x0)))))))
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(x1) → x1
B(a(b(x0))) → A(a(b(b(b(b(x0))))))
A(b(x0)) → B(x0)
A(b(a(x0))) → B(a(a(a(a(x0)))))
B(a(x0)) → A(x0)
A(b(a(x0))) → B(b(b(a(a(a(a(x0)))))))
B(a(b(x0))) → A(b(b(b(b(x0)))))
A(b(a(x0))) → B(b(a(a(a(a(x0))))))
B(a(b(x0))) → A(a(a(b(b(b(b(x0)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(x1) → x1
B(a(b(x0))) → A(a(b(b(b(b(x0))))))
A(b(x0)) → B(x0)
A(b(a(x0))) → B(a(a(a(a(x0)))))
B(a(x0)) → A(x0)
A(b(a(x0))) → B(b(b(a(a(a(a(x0)))))))
B(a(b(x0))) → A(b(b(b(b(x0)))))
A(b(a(x0))) → B(b(a(a(a(a(x0))))))
B(a(b(x0))) → A(a(a(b(b(b(b(x0)))))))
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(A(x)))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(A(x)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(A(x)))))))
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → b(b(b(b(x))))
b(a(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
B(a(b(x))) → A(a(b(b(b(b(x))))))
A(b(x)) → B(x)
A(b(a(x))) → B(a(a(a(a(x)))))
B(a(x)) → A(x)
A(b(a(x))) → B(b(b(a(a(a(a(x)))))))
B(a(b(x))) → A(b(b(b(b(x)))))
A(b(a(x))) → B(b(a(a(a(a(x))))))
B(a(b(x))) → A(a(a(b(b(b(b(x)))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(b(b(b(x))))
b(a(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
B(a(b(x))) → A(a(b(b(b(b(x))))))
A(b(x)) → B(x)
A(b(a(x))) → B(a(a(a(a(x)))))
B(a(x)) → A(x)
A(b(a(x))) → B(b(b(a(a(a(a(x)))))))
B(a(b(x))) → A(b(b(b(b(x)))))
A(b(a(x))) → B(b(a(a(a(a(x))))))
B(a(b(x))) → A(a(a(b(b(b(b(x)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(A(x)))))))
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → b(b(b(b(x))))
b(a(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
B(a(b(x))) → A(a(b(b(b(b(x))))))
A(b(x)) → B(x)
A(b(a(x))) → B(a(a(a(a(x)))))
B(a(x)) → A(x)
A(b(a(x))) → B(b(b(a(a(a(a(x)))))))
B(a(b(x))) → A(b(b(b(b(x)))))
A(b(a(x))) → B(b(a(a(a(a(x))))))
B(a(b(x))) → A(a(a(b(b(b(b(x)))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(b(b(b(x))))
b(a(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
B(a(b(x))) → A(a(b(b(b(b(x))))))
A(b(x)) → B(x)
A(b(a(x))) → B(a(a(a(a(x)))))
B(a(x)) → A(x)
A(b(a(x))) → B(b(b(a(a(a(a(x)))))))
B(a(b(x))) → A(b(b(b(b(x)))))
A(b(a(x))) → B(b(a(a(a(a(x))))))
B(a(b(x))) → A(a(a(b(b(b(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))) → A1(A(x))
B1(a(x)) → B1(x)
A1(b(A(x))) → A1(b(b(B(x))))
B1(a(B(x))) → B1(b(b(b(a(A(x))))))
B1(a(B(x))) → B1(b(A(x)))
B1(a(x)) → B1(b(b(x)))
A1(b(A(x))) → A1(a(a(a(b(B(x))))))
A1(b(A(x))) → A1(b(B(x)))
A1(b(A(x))) → A1(a(b(b(B(x)))))
A1(b(A(x))) → B1(b(B(x)))
A1(b(A(x))) → A1(B(x))
B1(a(B(x))) → B1(a(a(A(x))))
A1(b(x)) → A1(x)
A1(b(x)) → A1(a(x))
B1(a(x)) → B1(b(x))
B1(a(B(x))) → B1(A(x))
B1(a(B(x))) → B1(b(b(a(A(x)))))
B1(a(x)) → B1(b(b(b(x))))
A1(b(A(x))) → A1(a(a(a(B(x)))))
B1(a(B(x))) → B1(b(b(A(x))))
B1(a(B(x))) → A1(a(A(x)))
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(x)) → A1(a(a(a(x))))
B1(a(B(x))) → B1(b(b(b(A(x)))))
B1(a(B(x))) → B1(a(A(x)))
A1(b(A(x))) → A1(a(a(b(b(B(x))))))
B1(a(B(x))) → B1(b(a(A(x))))
A1(b(A(x))) → A1(a(b(B(x))))
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(A(x))) → A1(a(a(a(b(b(B(x)))))))
A1(b(x)) → A1(a(a(x)))
B1(a(B(x))) → B1(b(a(a(A(x)))))
B1(a(B(x))) → B1(b(b(a(a(A(x))))))
A1(b(A(x))) → A1(a(B(x)))
B1(a(B(x))) → B1(b(b(b(a(a(A(x)))))))
A1(b(A(x))) → B1(B(x))
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(A(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(B(x))) → A1(A(x))
B1(a(x)) → B1(x)
A1(b(A(x))) → A1(b(b(B(x))))
B1(a(B(x))) → B1(b(b(b(a(A(x))))))
B1(a(B(x))) → B1(b(A(x)))
B1(a(x)) → B1(b(b(x)))
A1(b(A(x))) → A1(a(a(a(b(B(x))))))
A1(b(A(x))) → A1(b(B(x)))
A1(b(A(x))) → A1(a(b(b(B(x)))))
A1(b(A(x))) → B1(b(B(x)))
A1(b(A(x))) → A1(B(x))
B1(a(B(x))) → B1(a(a(A(x))))
A1(b(x)) → A1(x)
A1(b(x)) → A1(a(x))
B1(a(x)) → B1(b(x))
B1(a(B(x))) → B1(A(x))
B1(a(B(x))) → B1(b(b(a(A(x)))))
B1(a(x)) → B1(b(b(b(x))))
A1(b(A(x))) → A1(a(a(a(B(x)))))
B1(a(B(x))) → B1(b(b(A(x))))
B1(a(B(x))) → A1(a(A(x)))
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(x)) → A1(a(a(a(x))))
B1(a(B(x))) → B1(b(b(b(A(x)))))
B1(a(B(x))) → B1(a(A(x)))
A1(b(A(x))) → A1(a(a(b(b(B(x))))))
B1(a(B(x))) → B1(b(a(A(x))))
A1(b(A(x))) → A1(a(b(B(x))))
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(A(x))) → A1(a(a(a(b(b(B(x)))))))
A1(b(x)) → A1(a(a(x)))
B1(a(B(x))) → B1(b(a(a(A(x)))))
B1(a(B(x))) → B1(b(b(a(a(A(x))))))
A1(b(A(x))) → A1(a(B(x)))
B1(a(B(x))) → B1(b(b(b(a(a(A(x)))))))
A1(b(A(x))) → B1(B(x))
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(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 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
A1(b(A(x))) → A1(b(b(B(x))))
B1(a(B(x))) → B1(b(b(b(a(A(x))))))
B1(a(B(x))) → B1(b(A(x)))
B1(a(x)) → B1(b(b(x)))
A1(b(A(x))) → A1(a(a(a(b(B(x))))))
A1(b(A(x))) → A1(b(B(x)))
A1(b(A(x))) → A1(a(b(b(B(x)))))
A1(b(A(x))) → B1(b(B(x)))
B1(a(B(x))) → B1(a(a(A(x))))
A1(b(x)) → A1(a(x))
A1(b(x)) → A1(x)
B1(a(x)) → B1(b(x))
B1(a(B(x))) → B1(b(b(a(A(x)))))
B1(a(x)) → B1(b(b(b(x))))
A1(b(A(x))) → A1(a(a(a(B(x)))))
B1(a(B(x))) → B1(b(b(A(x))))
B1(a(B(x))) → A1(a(A(x)))
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(x)) → A1(a(a(a(x))))
B1(a(B(x))) → B1(b(b(b(A(x)))))
B1(a(B(x))) → B1(a(A(x)))
A1(b(A(x))) → A1(a(a(b(b(B(x))))))
B1(a(B(x))) → B1(b(a(A(x))))
A1(b(A(x))) → A1(a(b(B(x))))
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(A(x))) → A1(a(a(a(b(b(B(x)))))))
A1(b(x)) → A1(a(a(x)))
B1(a(B(x))) → B1(b(a(a(A(x)))))
B1(a(B(x))) → B1(b(b(a(a(A(x))))))
A1(b(A(x))) → A1(a(B(x)))
B1(a(B(x))) → B1(b(b(b(a(a(A(x)))))))
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(A(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(B(x))) → B1(b(A(x))) at position [0] we obtained the following new rules:
B1(a(B(x0))) → B1(B(x0))
B1(a(B(y0))) → B1(A(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
A1(b(A(x))) → A1(b(b(B(x))))
B1(a(B(x))) → B1(b(b(b(a(A(x))))))
B1(a(x)) → B1(b(b(x)))
A1(b(A(x))) → A1(a(a(a(b(B(x))))))
A1(b(A(x))) → A1(a(b(b(B(x)))))
A1(b(A(x))) → A1(b(B(x)))
A1(b(A(x))) → B1(b(B(x)))
B1(a(B(x))) → B1(a(a(A(x))))
A1(b(x)) → A1(x)
A1(b(x)) → A1(a(x))
B1(a(x)) → B1(b(x))
B1(a(B(y0))) → B1(A(y0))
B1(a(x)) → B1(b(b(b(x))))
B1(a(B(x))) → B1(b(b(a(A(x)))))
A1(b(A(x))) → A1(a(a(a(B(x)))))
B1(a(B(x))) → B1(b(b(A(x))))
B1(a(B(x))) → A1(a(A(x)))
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(x)) → A1(a(a(a(x))))
B1(a(B(x))) → B1(b(b(b(A(x)))))
B1(a(B(x))) → B1(a(A(x)))
A1(b(A(x))) → A1(a(a(b(b(B(x))))))
B1(a(B(x))) → B1(b(a(A(x))))
A1(b(A(x))) → A1(a(b(B(x))))
A1(b(A(x))) → A1(a(a(b(B(x)))))
B1(a(B(x0))) → B1(B(x0))
A1(b(A(x))) → A1(a(a(a(b(b(B(x)))))))
A1(b(x)) → A1(a(a(x)))
B1(a(B(x))) → B1(b(a(a(A(x)))))
A1(b(A(x))) → A1(a(B(x)))
B1(a(B(x))) → B1(b(b(a(a(A(x))))))
B1(a(B(x))) → B1(b(b(b(a(a(A(x)))))))
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ 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(x)) → B1(x)
A1(b(A(x))) → A1(b(b(B(x))))
B1(a(B(x))) → B1(b(b(b(a(A(x))))))
B1(a(x)) → B1(b(b(x)))
A1(b(A(x))) → A1(a(a(a(b(B(x))))))
A1(b(A(x))) → A1(b(B(x)))
A1(b(A(x))) → A1(a(b(b(B(x)))))
A1(b(A(x))) → B1(b(B(x)))
B1(a(B(x))) → B1(a(a(A(x))))
A1(b(x)) → A1(a(x))
A1(b(x)) → A1(x)
B1(a(x)) → B1(b(x))
B1(a(B(x))) → B1(b(b(a(A(x)))))
B1(a(x)) → B1(b(b(b(x))))
A1(b(A(x))) → A1(a(a(a(B(x)))))
B1(a(B(x))) → B1(b(b(A(x))))
B1(a(B(x))) → A1(a(A(x)))
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(x)) → A1(a(a(a(x))))
B1(a(B(x))) → B1(b(b(b(A(x)))))
B1(a(B(x))) → B1(a(A(x)))
A1(b(A(x))) → A1(a(a(b(b(B(x))))))
B1(a(B(x))) → B1(b(a(A(x))))
A1(b(A(x))) → A1(a(b(B(x))))
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(A(x))) → A1(a(a(a(b(b(B(x)))))))
A1(b(x)) → A1(a(a(x)))
B1(a(B(x))) → B1(b(a(a(A(x)))))
B1(a(B(x))) → B1(b(b(a(a(A(x))))))
A1(b(A(x))) → A1(a(B(x)))
B1(a(B(x))) → B1(b(b(b(a(a(A(x)))))))
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(A(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(B(x))) → A1(a(A(x))) at position [0] we obtained the following new rules:
B1(a(B(y0))) → A1(A(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ 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(B(y0))) → A1(A(y0))
B1(a(x)) → B1(x)
A1(b(A(x))) → A1(b(b(B(x))))
B1(a(B(x))) → B1(b(b(b(a(A(x))))))
B1(a(x)) → B1(b(b(x)))
A1(b(A(x))) → A1(a(a(a(b(B(x))))))
A1(b(A(x))) → A1(a(b(b(B(x)))))
A1(b(A(x))) → A1(b(B(x)))
A1(b(A(x))) → B1(b(B(x)))
B1(a(B(x))) → B1(a(a(A(x))))
A1(b(x)) → A1(x)
A1(b(x)) → A1(a(x))
B1(a(x)) → B1(b(x))
B1(a(x)) → B1(b(b(b(x))))
B1(a(B(x))) → B1(b(b(a(A(x)))))
A1(b(A(x))) → A1(a(a(a(B(x)))))
B1(a(B(x))) → B1(b(b(A(x))))
A1(b(A(x))) → A1(a(a(B(x))))
A1(b(x)) → A1(a(a(a(x))))
B1(a(B(x))) → B1(b(b(b(A(x)))))
B1(a(B(x))) → B1(a(A(x)))
A1(b(A(x))) → A1(a(a(b(b(B(x))))))
B1(a(B(x))) → B1(b(a(A(x))))
A1(b(A(x))) → A1(a(b(B(x))))
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(A(x))) → A1(a(a(a(b(b(B(x)))))))
A1(b(x)) → A1(a(a(x)))
B1(a(B(x))) → B1(b(a(a(A(x)))))
A1(b(A(x))) → A1(a(B(x)))
B1(a(B(x))) → B1(b(b(a(a(A(x))))))
B1(a(B(x))) → B1(b(b(b(a(a(A(x)))))))
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(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 2 SCCs with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(a(x)) → B1(x)
B1(a(B(x))) → B1(b(b(b(A(x)))))
B1(a(B(x))) → B1(b(b(b(a(A(x))))))
B1(a(B(x))) → B1(a(A(x)))
B1(a(x)) → B1(b(b(x)))
B1(a(B(x))) → B1(b(a(A(x))))
B1(a(B(x))) → B1(a(a(A(x))))
B1(a(x)) → B1(b(x))
B1(a(x)) → B1(b(b(b(x))))
B1(a(B(x))) → B1(b(b(a(A(x)))))
B1(a(B(x))) → B1(b(a(a(A(x)))))
B1(a(B(x))) → B1(b(b(A(x))))
B1(a(B(x))) → B1(b(b(a(a(A(x))))))
B1(a(B(x))) → B1(b(b(b(a(a(A(x)))))))
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(A(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(b(x)) → A1(a(a(a(x))))
A1(b(A(x))) → A1(b(b(B(x))))
A1(b(A(x))) → A1(a(a(a(b(B(x))))))
A1(b(A(x))) → A1(b(B(x)))
A1(b(A(x))) → A1(a(a(b(b(B(x))))))
A1(b(A(x))) → A1(a(b(b(B(x)))))
A1(b(A(x))) → A1(a(b(B(x))))
A1(b(x)) → A1(a(x))
A1(b(x)) → A1(x)
A1(b(A(x))) → A1(a(a(b(B(x)))))
A1(b(A(x))) → A1(a(a(a(b(b(B(x)))))))
A1(b(x)) → A1(a(a(x)))
A1(b(A(x))) → A1(a(a(a(B(x)))))
A1(b(A(x))) → A1(a(B(x)))
A1(b(A(x))) → A1(a(a(B(x))))
The TRS R consists of the following rules:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
b(a(B(x))) → b(b(b(b(a(A(x))))))
b(A(x)) → B(x)
a(b(A(x))) → a(a(a(a(B(x)))))
a(B(x)) → A(x)
a(b(A(x))) → a(a(a(a(b(b(B(x)))))))
b(a(B(x))) → b(b(b(b(A(x)))))
a(b(A(x))) → a(a(a(a(b(B(x))))))
b(a(B(x))) → b(b(b(b(a(a(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(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(x1) → x1
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
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:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → b(b(b(b(x1))))
b(a(x1)) → a(a(a(a(x1))))
a(x1) → x1
b(x1) → x1
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
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:
b(a(x)) → b(b(b(b(x))))
a(b(x)) → a(a(a(a(x))))
a(x) → x
b(x) → x
Q is empty.