MAYBE
Termination w.r.t. Q proof of ../tpdb/SRS/Waldmann/jw5.srs
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(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(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:
B(a(a(a(a(b(b(x1))))))) → B(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
B(a(b(b(x1)))) → B(b(b(a(b(x1)))))
B(a(a(b(b(x1))))) → B(a(a(b(x1))))
B(a(b(b(x1)))) → B(b(a(b(x1))))
B(a(b(b(x1)))) → B(a(b(x1)))
B(a(a(b(b(x1))))) → B(b(a(a(b(x1)))))
B(a(a(b(b(x1))))) → B(a(b(b(a(a(b(x1)))))))
B(a(a(a(a(b(b(x1))))))) → B(b(a(a(a(a(b(x1)))))))
B(a(a(a(b(b(x1)))))) → B(a(a(a(b(x1)))))
B(a(a(a(a(b(b(x1))))))) → B(a(a(a(a(b(x1))))))
B(a(a(a(b(b(x1)))))) → B(b(a(a(a(b(x1))))))
B(a(a(a(b(b(x1)))))) → B(a(a(b(b(a(a(a(b(x1)))))))))
The TRS R consists of the following rules:
b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(a(b(b(x1))))))) → B(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
B(a(b(b(x1)))) → B(b(b(a(b(x1)))))
B(a(a(b(b(x1))))) → B(a(a(b(x1))))
B(a(b(b(x1)))) → B(b(a(b(x1))))
B(a(b(b(x1)))) → B(a(b(x1)))
B(a(a(b(b(x1))))) → B(b(a(a(b(x1)))))
B(a(a(b(b(x1))))) → B(a(b(b(a(a(b(x1)))))))
B(a(a(a(a(b(b(x1))))))) → B(b(a(a(a(a(b(x1)))))))
B(a(a(a(b(b(x1)))))) → B(a(a(a(b(x1)))))
B(a(a(a(a(b(b(x1))))))) → B(a(a(a(a(b(x1))))))
B(a(a(a(b(b(x1)))))) → B(b(a(a(a(b(x1))))))
B(a(a(a(b(b(x1)))))) → B(a(a(b(b(a(a(a(b(x1)))))))))
The TRS R consists of the following rules:
b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(a(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 4 SCCs with 8 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPToSRSProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(a(b(b(x1)))) → B(a(b(x1)))
The TRS R consists of the following rules:
b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QDP
↳ QDP
↳ QDP
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
B(a(b(b(x1)))) → B(a(b(x1)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
B(a(b(b(x1)))) → B(a(b(x1)))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
b(b(a(B(x)))) → b(a(B(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ QDP
↳ QDP
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
b(b(a(B(x)))) → b(a(B(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
b(b(a(B(x)))) → b(a(B(x)))
The set Q is empty.
We have obtained the following QTRS:
b(a(b(b(x)))) → b(b(b(a(b(x)))))
b(a(a(b(b(x))))) → b(a(b(b(a(a(b(x)))))))
b(a(a(a(b(b(x)))))) → b(a(a(b(b(a(a(a(b(x)))))))))
b(a(a(a(a(b(b(x))))))) → b(a(a(a(b(b(a(a(a(a(b(x)))))))))))
B(a(b(b(x)))) → B(a(b(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ QDP
↳ QDP
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(a(b(x)))))
b(a(a(b(b(x))))) → b(a(b(b(a(a(b(x)))))))
b(a(a(a(b(b(x)))))) → b(a(a(b(b(a(a(a(b(x)))))))))
b(a(a(a(a(b(b(x))))))) → b(a(a(a(b(b(a(a(a(a(b(x)))))))))))
B(a(b(b(x)))) → B(a(b(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
b(b(a(B(x)))) → b(a(B(x)))
The set Q is empty.
We have obtained the following QTRS:
b(a(b(b(x)))) → b(b(b(a(b(x)))))
b(a(a(b(b(x))))) → b(a(b(b(a(a(b(x)))))))
b(a(a(a(b(b(x)))))) → b(a(a(b(b(a(a(a(b(x)))))))))
b(a(a(a(a(b(b(x))))))) → b(a(a(a(b(b(a(a(a(a(b(x)))))))))))
B(a(b(b(x)))) → B(a(b(x)))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDP
↳ QDP
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(b(b(x)))) → b(b(b(a(b(x)))))
b(a(a(b(b(x))))) → b(a(b(b(a(a(b(x)))))))
b(a(a(a(b(b(x)))))) → b(a(a(b(b(a(a(a(b(x)))))))))
b(a(a(a(a(b(b(x))))))) → b(a(a(a(b(b(a(a(a(a(b(x)))))))))))
B(a(b(b(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:
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(a(b(x))))) → B1(a(a(b(b(a(b(x)))))))
B1(b(a(a(a(a(b(x))))))) → B1(b(a(a(a(b(x))))))
B1(b(a(a(a(a(b(x))))))) → B1(a(a(a(a(b(b(a(a(a(b(x)))))))))))
B1(b(a(a(a(b(x)))))) → B1(a(a(b(x))))
B1(b(a(a(a(a(b(x))))))) → B1(a(a(a(b(x)))))
B1(b(a(a(a(b(x)))))) → B1(a(a(a(b(b(a(a(b(x)))))))))
B1(b(a(b(x)))) → B1(b(b(x)))
B1(b(a(a(b(x))))) → B1(a(b(x)))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(x)))) → B1(a(b(b(b(x)))))
B1(b(a(b(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
b(b(a(B(x)))) → b(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(a(b(x))))) → B1(a(a(b(b(a(b(x)))))))
B1(b(a(a(a(a(b(x))))))) → B1(b(a(a(a(b(x))))))
B1(b(a(a(a(a(b(x))))))) → B1(a(a(a(a(b(b(a(a(a(b(x)))))))))))
B1(b(a(a(a(b(x)))))) → B1(a(a(b(x))))
B1(b(a(a(a(a(b(x))))))) → B1(a(a(a(b(x)))))
B1(b(a(a(a(b(x)))))) → B1(a(a(a(b(b(a(a(b(x)))))))))
B1(b(a(b(x)))) → B1(b(b(x)))
B1(b(a(a(b(x))))) → B1(a(b(x)))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(x)))) → B1(a(b(b(b(x)))))
B1(b(a(b(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
b(b(a(B(x)))) → b(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 7 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(a(a(a(b(x))))))) → B1(b(a(a(a(b(x))))))
B1(b(a(b(x)))) → B1(b(b(x)))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(x)))) → B1(b(x))
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
b(b(a(B(x)))) → b(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(a(b(x)))) → B1(b(b(x))) at position [0] we obtained the following new rules:
B1(b(a(b(b(a(a(b(x0)))))))) → B1(b(b(a(a(b(b(a(b(x0)))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(a(b(b(b(x0))))))
B1(b(a(b(b(a(a(a(a(b(x0)))))))))) → B1(b(b(a(a(a(a(b(b(a(a(a(b(x0)))))))))))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(a(b(b(b(x0)))))))
B1(b(a(b(a(a(b(x0))))))) → B1(b(a(a(b(b(a(b(x0))))))))
B1(b(a(b(b(a(a(a(b(x0))))))))) → B1(b(b(a(a(a(b(b(a(a(b(x0)))))))))))
B1(b(a(b(a(a(a(b(x0)))))))) → B1(b(a(a(a(b(b(a(a(b(x0))))))))))
B1(b(a(b(a(a(a(a(b(x0))))))))) → B1(b(a(a(a(a(b(b(a(a(a(b(x0))))))))))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(a(B(x0)))))) → B1(b(a(B(x0))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(a(a(a(b(x))))))) → B1(b(a(a(a(b(x))))))
B1(b(a(b(b(a(a(a(a(b(x0)))))))))) → B1(b(b(a(a(a(a(b(b(a(a(a(b(x0)))))))))))))
B1(b(a(b(a(a(b(x0))))))) → B1(b(a(a(b(b(a(b(x0))))))))
B1(b(a(b(b(a(a(a(b(x0))))))))) → B1(b(b(a(a(a(b(b(a(a(b(x0)))))))))))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(a(a(a(a(b(x0))))))))) → B1(b(a(a(a(a(b(b(a(a(a(b(x0))))))))))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(b(b(a(a(b(x0)))))))) → B1(b(b(a(a(b(b(a(b(x0)))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(a(b(b(b(x0))))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(a(b(b(b(x0)))))))
B1(b(a(b(a(a(a(b(x0)))))))) → B1(b(a(a(a(b(b(a(a(b(x0))))))))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(a(B(x0)))))
B1(b(a(b(a(B(x0)))))) → B1(b(a(B(x0))))
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
b(b(a(B(x)))) → b(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
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(a(a(a(b(x)))))) → B1(b(a(a(b(x)))))
B1(b(a(a(a(a(b(x))))))) → B1(b(a(a(a(b(x))))))
B1(b(a(b(b(a(a(a(a(b(x0)))))))))) → B1(b(b(a(a(a(a(b(b(a(a(a(b(x0)))))))))))))
B1(b(a(b(a(a(b(x0))))))) → B1(b(a(a(b(b(a(b(x0))))))))
B1(b(a(b(b(a(a(a(b(x0))))))))) → B1(b(b(a(a(a(b(b(a(a(b(x0)))))))))))
B1(b(a(a(b(x))))) → B1(b(a(b(x))))
B1(b(a(b(a(a(a(a(b(x0))))))))) → B1(b(a(a(a(a(b(b(a(a(a(b(x0))))))))))))
B1(b(a(b(x)))) → B1(b(x))
B1(b(a(b(b(a(a(b(x0)))))))) → B1(b(b(a(a(b(b(a(b(x0)))))))))
B1(b(a(b(a(b(x0)))))) → B1(b(a(b(b(b(x0))))))
B1(b(a(b(b(a(b(x0))))))) → B1(b(b(a(b(b(b(x0)))))))
B1(b(a(b(a(a(a(b(x0)))))))) → B1(b(a(a(a(b(b(a(a(b(x0))))))))))
B1(b(a(b(b(a(B(x0))))))) → B1(b(b(a(B(x0)))))
The TRS R consists of the following rules:
b(b(a(b(x)))) → b(a(b(b(b(x)))))
b(b(a(a(b(x))))) → b(a(a(b(b(a(b(x)))))))
b(b(a(a(a(b(x)))))) → b(a(a(a(b(b(a(a(b(x)))))))))
b(b(a(a(a(a(b(x))))))) → b(a(a(a(a(b(b(a(a(a(b(x)))))))))))
b(b(a(B(x)))) → b(a(B(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(a(a(b(b(x1))))) → B(a(a(b(x1))))
The TRS R consists of the following rules:
b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(b(b(x1)))))) → B(a(a(a(b(x1)))))
The TRS R consists of the following rules:
b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(a(b(b(x1))))))) → B(a(a(a(a(b(x1))))))
The TRS R consists of the following rules:
b(a(b(b(x1)))) → b(b(b(a(b(x1)))))
b(a(a(b(b(x1))))) → b(a(b(b(a(a(b(x1)))))))
b(a(a(a(b(b(x1)))))) → b(a(a(b(b(a(a(a(b(x1)))))))))
b(a(a(a(a(b(b(x1))))))) → b(a(a(a(b(b(a(a(a(a(b(x1)))))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.