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(b(x1))) → a(a(a(x1)))
a(a(a(x1))) → b(a(b(x1)))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(a(x1)))
a(a(a(x1))) → b(a(b(x1)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x1))) → a(a(a(x1)))
a(a(a(x1))) → b(a(b(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(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:
A(a(a(x1))) → B(x1)
A(a(a(x1))) → A(b(x1))
B(b(b(x1))) → A(x1)
A(a(a(x1))) → B(a(b(x1)))
B(b(b(x1))) → A(a(x1))
B(b(b(x1))) → A(a(a(x1)))
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(a(x1)))
a(a(a(x1))) → b(a(b(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(x1))) → B(x1)
A(a(a(x1))) → A(b(x1))
B(b(b(x1))) → A(x1)
A(a(a(x1))) → B(a(b(x1)))
B(b(b(x1))) → A(a(x1))
B(b(b(x1))) → A(a(a(x1)))
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(a(x1)))
a(a(a(x1))) → b(a(b(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
A(a(a(x1))) → B(x1)
A(a(a(x1))) → A(b(x1))
B(b(b(x1))) → A(x1)
B(b(b(x1))) → A(a(x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(B(x1)) = 2·x1
POL(a(x1)) = 2 + 2·x1
POL(b(x1)) = 2 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(b(x1))) → A(a(a(x1)))
A(a(a(x1))) → B(a(b(x1)))
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(a(x1)))
a(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 A(a(a(x1))) → B(a(b(x1))) at position [0] we obtained the following new rules:
A(a(a(b(b(x0))))) → B(a(a(a(a(x0)))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(a(b(b(x0))))) → B(a(a(a(a(x0)))))
B(b(b(x1))) → A(a(a(x1)))
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(a(x1)))
a(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x1))) → a(a(a(x1)))
a(a(a(x1))) → b(a(b(x1)))
A(a(a(b(b(x0))))) → B(a(a(a(a(x0)))))
B(b(b(x1))) → A(a(a(x1)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x1))) → a(a(a(x1)))
a(a(a(x1))) → b(a(b(x1)))
A(a(a(b(b(x0))))) → B(a(a(a(a(x0)))))
B(b(b(x1))) → A(a(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → a(a(A(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → a(a(A(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → a(a(A(x)))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
A(a(a(b(b(x))))) → B(a(a(a(a(x)))))
B(b(b(x))) → A(a(a(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
A(a(a(b(b(x))))) → B(a(a(a(a(x)))))
B(b(b(x))) → A(a(a(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → a(a(A(x)))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
A(a(a(b(b(x))))) → B(a(a(a(a(x)))))
B(b(b(x))) → A(a(a(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
A(a(a(b(b(x))))) → B(a(a(a(a(x)))))
B(b(b(x))) → A(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:
B1(b(a(a(A(x))))) → A1(a(a(B(x))))
A1(a(a(x))) → B1(a(b(x)))
B1(b(B(x))) → A1(A(x))
B1(b(b(x))) → A1(x)
B1(b(b(x))) → A1(a(x))
B1(b(B(x))) → A1(a(A(x)))
B1(b(a(a(A(x))))) → A1(a(B(x)))
B1(b(a(a(A(x))))) → A1(a(a(a(B(x)))))
B1(b(a(a(A(x))))) → A1(B(x))
B1(b(b(x))) → A1(a(a(x)))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → a(a(A(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(a(A(x))))) → A1(a(a(B(x))))
A1(a(a(x))) → B1(a(b(x)))
B1(b(B(x))) → A1(A(x))
B1(b(b(x))) → A1(x)
B1(b(b(x))) → A1(a(x))
B1(b(B(x))) → A1(a(A(x)))
B1(b(a(a(A(x))))) → A1(a(B(x)))
B1(b(a(a(A(x))))) → A1(a(a(a(B(x)))))
B1(b(a(a(A(x))))) → A1(B(x))
B1(b(b(x))) → A1(a(a(x)))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → 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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(a(A(x))))) → A1(a(a(B(x))))
A1(a(a(x))) → B1(a(b(x)))
B1(b(b(x))) → A1(x)
B1(b(b(x))) → A1(a(x))
B1(b(a(a(A(x))))) → A1(a(a(a(B(x)))))
B1(b(b(x))) → A1(a(a(x)))
A1(a(a(x))) → B1(x)
A1(a(a(x))) → A1(b(x))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → a(a(A(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
B1(b(a(a(A(x))))) → A1(a(a(B(x))))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2 + 2·x1
POL(A1(x1)) = 2·x1
POL(B(x1)) = 2 + 2·x1
POL(B1(x1)) = 2·x1
POL(a(x1)) = 2·x1
POL(b(x1)) = 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → B1(a(b(x)))
B1(b(b(x))) → A1(a(x))
B1(b(b(x))) → A1(x)
B1(b(a(a(A(x))))) → A1(a(a(a(B(x)))))
B1(b(b(x))) → A1(a(a(x)))
A1(a(a(x))) → A1(b(x))
A1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → a(a(A(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
B1(b(b(x))) → A1(a(x))
B1(b(b(x))) → A1(x)
A1(a(a(x))) → A1(b(x))
A1(a(a(x))) → B1(x)
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(A1(x1)) = 2 + 2·x1
POL(B(x1)) = 2·x1
POL(B1(x1)) = 2 + 2·x1
POL(a(x1)) = 1 + 2·x1
POL(b(x1)) = 1 + 2·x1
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(x))) → B1(a(b(x)))
B1(b(a(a(A(x))))) → A1(a(a(a(B(x)))))
B1(b(b(x))) → A1(a(a(x)))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → a(a(A(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(a(x))) → B1(a(b(x))) at position [0] we obtained the following new rules:
A1(a(a(b(B(x0))))) → B1(a(a(a(A(x0)))))
A1(a(a(b(a(a(A(x0))))))) → B1(a(a(a(a(a(B(x0)))))))
A1(a(a(b(b(x0))))) → B1(a(a(a(a(x0)))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(b(B(x0))))) → B1(a(a(a(A(x0)))))
A1(a(a(b(a(a(A(x0))))))) → B1(a(a(a(a(a(B(x0)))))))
B1(b(a(a(A(x))))) → A1(a(a(a(B(x)))))
B1(b(b(x))) → A1(a(a(x)))
A1(a(a(b(b(x0))))) → B1(a(a(a(a(x0)))))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → a(a(A(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(a(a(A(x))))) → A1(a(a(a(B(x))))) at position [0] we obtained the following new rules:
B1(b(a(a(A(y0))))) → A1(b(a(b(B(y0)))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(b(B(x0))))) → B1(a(a(a(A(x0)))))
A1(a(a(b(a(a(A(x0))))))) → B1(a(a(a(a(a(B(x0)))))))
B1(b(a(a(A(y0))))) → A1(b(a(b(B(y0)))))
B1(b(b(x))) → A1(a(a(x)))
A1(a(a(b(b(x0))))) → B1(a(a(a(a(x0)))))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → 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 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(b(B(x0))))) → B1(a(a(a(A(x0)))))
A1(a(a(b(a(a(A(x0))))))) → B1(a(a(a(a(a(B(x0)))))))
B1(b(b(x))) → A1(a(a(x)))
A1(a(a(b(b(x0))))) → B1(a(a(a(a(x0)))))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → a(a(A(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(a(b(B(x0))))) → B1(a(a(a(A(x0))))) at position [0] we obtained the following new rules:
A1(a(a(b(B(y0))))) → B1(b(a(b(A(y0)))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(b(a(a(A(x0))))))) → B1(a(a(a(a(a(B(x0)))))))
A1(a(a(b(B(y0))))) → B1(b(a(b(A(y0)))))
B1(b(b(x))) → A1(a(a(x)))
A1(a(a(b(b(x0))))) → B1(a(a(a(a(x0)))))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → 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 1 less node.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(b(a(a(A(x0))))))) → B1(a(a(a(a(a(B(x0)))))))
B1(b(b(x))) → A1(a(a(x)))
A1(a(a(b(b(x0))))) → B1(a(a(a(a(x0)))))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → a(a(A(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(a(b(a(a(A(x0))))))) → B1(a(a(a(a(a(B(x0))))))) at position [0] we obtained the following new rules:
A1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(b(a(B(y0)))))))
A1(a(a(b(a(a(A(y0))))))) → B1(a(a(b(a(b(B(y0)))))))
A1(a(a(b(a(a(A(y0))))))) → B1(b(a(b(a(a(B(y0)))))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(a(b(a(a(A(y0))))))) → B1(a(a(b(a(b(B(y0)))))))
A1(a(a(b(a(a(A(y0))))))) → B1(b(a(b(a(a(B(y0)))))))
A1(a(a(b(a(a(A(y0))))))) → B1(a(b(a(b(a(B(y0)))))))
B1(b(b(x))) → A1(a(a(x)))
A1(a(a(b(b(x0))))) → B1(a(a(a(a(x0)))))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → 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 3 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(x))) → A1(a(a(x)))
A1(a(a(b(b(x0))))) → B1(a(a(a(a(x0)))))
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
b(b(a(a(A(x))))) → a(a(a(a(B(x)))))
b(b(B(x))) → 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
b(b(b(x1))) → a(a(a(x1)))
a(a(a(x1))) → b(a(b(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(b(x))) → a(a(a(x)))
a(a(a(x))) → b(a(b(x)))
Q is empty.