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(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(b(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(b(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(b(a(b(x1))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(b(a(b(x1))))))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(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(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(a(a(b(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(b(a(b(x1))))))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(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(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(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(b(a(a(b(x1)))))) → B(a(b(b(a(a(a(b(x1))))))))
B(a(a(a(b(b(b(x1))))))) → B(b(a(a(a(b(x1))))))
B(a(b(a(a(b(x1)))))) → B(a(a(a(b(x1)))))
B(b(a(a(b(x1))))) → B(b(x1))
B(a(a(a(b(a(b(x1))))))) → B(a(a(b(a(b(x1))))))
B(b(a(a(b(x1))))) → B(a(a(b(b(x1)))))
B(a(b(a(a(b(x1)))))) → B(b(a(a(a(b(x1))))))
B(a(a(a(b(b(b(x1))))))) → B(a(a(a(b(x1)))))
B(a(a(a(b(b(b(x1))))))) → B(b(b(a(a(a(b(x1)))))))
The TRS R consists of the following rules:
b(b(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(b(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(b(a(a(b(x1)))))) → B(a(b(b(a(a(a(b(x1))))))))
B(a(a(a(b(b(b(x1))))))) → B(b(a(a(a(b(x1))))))
B(a(b(a(a(b(x1)))))) → B(a(a(a(b(x1)))))
B(b(a(a(b(x1))))) → B(b(x1))
B(a(a(a(b(a(b(x1))))))) → B(a(a(b(a(b(x1))))))
B(b(a(a(b(x1))))) → B(a(a(b(b(x1)))))
B(a(b(a(a(b(x1)))))) → B(b(a(a(a(b(x1))))))
B(a(a(a(b(b(b(x1))))))) → B(a(a(a(b(x1)))))
B(a(a(a(b(b(b(x1))))))) → B(b(b(a(a(a(b(x1)))))))
The TRS R consists of the following rules:
b(b(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(b(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 3 SCCs with 6 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ SemLabProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(b(a(a(b(x1))))) → B(b(x1))
The TRS R consists of the following rules:
b(b(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(b(a(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.B: 0
a: 1
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
B.0(b.1(a.1(a.0(b.1(x1))))) → B.0(b.1(x1))
B.0(b.1(a.1(a.0(b.0(x1))))) → B.0(b.0(x1))
The TRS R consists of the following rules:
b.1(a.0(b.1(a.1(a.0(b.0(x1)))))) → b.1(a.0(b.0(b.1(a.1(a.1(a.0(b.0(x1))))))))
b.1(a.1(a.1(a.0(b.0(b.0(b.1(x1))))))) → b.0(b.0(b.1(a.1(a.1(a.0(b.1(x1)))))))
b.1(a.1(a.1(a.0(b.0(b.0(b.0(x1))))))) → b.0(b.0(b.1(a.1(a.1(a.0(b.0(x1)))))))
b.1(a.0(b.1(a.1(a.0(b.1(x1)))))) → b.1(a.0(b.0(b.1(a.1(a.1(a.0(b.1(x1))))))))
b.1(a.1(a.1(a.0(b.1(a.0(b.0(x1))))))) → b.1(a.1(a.0(b.1(a.0(b.0(x1))))))
b.0(b.1(a.1(a.0(b.1(x1))))) → b.1(a.1(a.0(b.0(b.1(x1)))))
b.1(a.1(a.1(a.0(b.1(a.0(b.1(x1))))))) → b.1(a.1(a.0(b.1(a.0(b.1(x1))))))
b.0(b.1(a.1(a.0(b.0(x1))))) → b.1(a.1(a.0(b.0(b.0(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ SemLabProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B.0(b.1(a.1(a.0(b.1(x1))))) → B.0(b.1(x1))
B.0(b.1(a.1(a.0(b.0(x1))))) → B.0(b.0(x1))
The TRS R consists of the following rules:
b.1(a.0(b.1(a.1(a.0(b.0(x1)))))) → b.1(a.0(b.0(b.1(a.1(a.1(a.0(b.0(x1))))))))
b.1(a.1(a.1(a.0(b.0(b.0(b.1(x1))))))) → b.0(b.0(b.1(a.1(a.1(a.0(b.1(x1)))))))
b.1(a.1(a.1(a.0(b.0(b.0(b.0(x1))))))) → b.0(b.0(b.1(a.1(a.1(a.0(b.0(x1)))))))
b.1(a.0(b.1(a.1(a.0(b.1(x1)))))) → b.1(a.0(b.0(b.1(a.1(a.1(a.0(b.1(x1))))))))
b.1(a.1(a.1(a.0(b.1(a.0(b.0(x1))))))) → b.1(a.1(a.0(b.1(a.0(b.0(x1))))))
b.0(b.1(a.1(a.0(b.1(x1))))) → b.1(a.1(a.0(b.0(b.1(x1)))))
b.1(a.1(a.1(a.0(b.1(a.0(b.1(x1))))))) → b.1(a.1(a.0(b.1(a.0(b.1(x1))))))
b.0(b.1(a.1(a.0(b.0(x1))))) → b.1(a.1(a.0(b.0(b.0(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:
B.0(b.1(a.1(a.0(b.1(x1))))) → B.0(b.1(x1))
B.0(b.1(a.1(a.0(b.0(x1))))) → B.0(b.0(x1))
Used ordering: POLO with Polynomial interpretation [25]:
POL(B.0(x1)) = x1
POL(a.0(x1)) = 1 + x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ SemLabProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b.1(a.0(b.1(a.1(a.0(b.0(x1)))))) → b.1(a.0(b.0(b.1(a.1(a.1(a.0(b.0(x1))))))))
b.1(a.1(a.1(a.0(b.0(b.0(b.1(x1))))))) → b.0(b.0(b.1(a.1(a.1(a.0(b.1(x1)))))))
b.1(a.1(a.1(a.0(b.0(b.0(b.0(x1))))))) → b.0(b.0(b.1(a.1(a.1(a.0(b.0(x1)))))))
b.1(a.0(b.1(a.1(a.0(b.1(x1)))))) → b.1(a.0(b.0(b.1(a.1(a.1(a.0(b.1(x1))))))))
b.1(a.1(a.1(a.0(b.1(a.0(b.0(x1))))))) → b.1(a.1(a.0(b.1(a.0(b.0(x1))))))
b.0(b.1(a.1(a.0(b.1(x1))))) → b.1(a.1(a.0(b.0(b.1(x1)))))
b.1(a.1(a.1(a.0(b.1(a.0(b.1(x1))))))) → b.1(a.1(a.0(b.1(a.0(b.1(x1))))))
b.0(b.1(a.1(a.0(b.0(x1))))) → b.1(a.1(a.0(b.0(b.0(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPToSRSProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(a(a(a(b(b(b(x1))))))) → B(a(a(a(b(x1)))))
The TRS R consists of the following rules:
b(b(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QDP
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(b(a(b(x1))))))
B(a(a(a(b(b(b(x1))))))) → B(a(a(a(b(x1)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(b(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(b(a(b(x1))))))
B(a(a(a(b(b(b(x1))))))) → B(a(a(a(b(x1)))))
The set Q is empty.
We have obtained the following QTRS:
b(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(a(a(b(x))))))
b(b(b(a(a(a(B(x))))))) → b(a(a(a(B(x)))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(a(a(b(x))))))
b(b(b(a(a(a(B(x))))))) → b(a(a(a(B(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(a(a(b(x))))))
b(b(b(a(a(a(B(x))))))) → b(a(a(a(B(x)))))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(b(x))))) → b(a(a(b(b(x)))))
b(a(a(a(b(b(b(x))))))) → b(b(b(a(a(a(b(x)))))))
b(a(b(a(a(b(x)))))) → b(a(b(b(a(a(a(b(x))))))))
b(a(a(a(b(a(b(x))))))) → b(a(a(b(a(b(x))))))
B(a(a(a(b(b(b(x))))))) → B(a(a(a(b(x)))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(a(b(x))))) → b(a(a(b(b(x)))))
b(a(a(a(b(b(b(x))))))) → b(b(b(a(a(a(b(x)))))))
b(a(b(a(a(b(x)))))) → b(a(b(b(a(a(a(b(x))))))))
b(a(a(a(b(a(b(x))))))) → b(a(a(b(a(b(x))))))
B(a(a(a(b(b(b(x))))))) → B(a(a(a(b(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(a(a(b(x))))))
b(b(b(a(a(a(B(x))))))) → b(a(a(a(B(x)))))
The set Q is empty.
We have obtained the following QTRS:
b(b(a(a(b(x))))) → b(a(a(b(b(x)))))
b(a(a(a(b(b(b(x))))))) → b(b(b(a(a(a(b(x)))))))
b(a(b(a(a(b(x)))))) → b(a(b(b(a(a(a(b(x))))))))
b(a(a(a(b(a(b(x))))))) → b(a(a(b(a(b(x))))))
B(a(a(a(b(b(b(x))))))) → B(a(a(a(b(x)))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q restricted rewrite system:
The TRS R consists of the following rules:
b(b(a(a(b(x))))) → b(a(a(b(b(x)))))
b(a(a(a(b(b(b(x))))))) → b(b(b(a(a(a(b(x)))))))
b(a(b(a(a(b(x)))))) → b(a(b(b(a(a(a(b(x))))))))
b(a(a(a(b(a(b(x))))))) → b(a(a(b(a(b(x))))))
B(a(a(a(b(b(b(x))))))) → 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:
B1(b(b(a(a(a(b(x))))))) → B1(b(b(x)))
B1(b(b(a(a(a(b(x))))))) → B1(a(a(a(b(b(b(x)))))))
B1(a(a(b(a(b(x)))))) → B1(b(a(b(x))))
B1(a(a(b(a(b(x)))))) → B1(a(a(a(b(b(a(b(x))))))))
B1(a(b(a(a(a(b(x))))))) → B1(a(b(a(a(b(x))))))
B1(a(b(a(a(a(b(x))))))) → B1(a(a(b(x))))
B1(a(a(b(b(x))))) → B1(a(a(b(x))))
B1(a(a(b(b(x))))) → B1(b(a(a(b(x)))))
B1(b(b(a(a(a(b(x))))))) → B1(b(x))
The TRS R consists of the following rules:
b(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(a(a(b(x))))))
b(b(b(a(a(a(B(x))))))) → b(a(a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(a(a(a(b(x))))))) → B1(b(b(x)))
B1(b(b(a(a(a(b(x))))))) → B1(a(a(a(b(b(b(x)))))))
B1(a(a(b(a(b(x)))))) → B1(b(a(b(x))))
B1(a(a(b(a(b(x)))))) → B1(a(a(a(b(b(a(b(x))))))))
B1(a(b(a(a(a(b(x))))))) → B1(a(b(a(a(b(x))))))
B1(a(b(a(a(a(b(x))))))) → B1(a(a(b(x))))
B1(a(a(b(b(x))))) → B1(a(a(b(x))))
B1(a(a(b(b(x))))) → B1(b(a(a(b(x)))))
B1(b(b(a(a(a(b(x))))))) → B1(b(x))
The TRS R consists of the following rules:
b(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(a(a(b(x))))))
b(b(b(a(a(a(B(x))))))) → b(a(a(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 3 SCCs with 5 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ SemLabProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(b(b(a(a(a(b(x))))))) → B1(b(b(x)))
B1(b(b(a(a(a(b(x))))))) → B1(b(x))
The TRS R consists of the following rules:
b(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(a(a(b(x))))))
b(b(b(a(a(a(B(x))))))) → b(a(a(a(B(x)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R.
Interpretation over the domain with elements from 0 to 1.B: 0
a: 1
B1: 0
b: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:
B1.0(b.0(b.1(a.1(a.1(a.0(b.1(x))))))) → B1.0(b.1(x))
B1.0(b.0(b.1(a.1(a.1(a.0(b.0(x))))))) → B1.0(b.0(b.0(x)))
B1.0(b.0(b.1(a.1(a.1(a.0(b.1(x))))))) → B1.0(b.0(b.1(x)))
B1.0(b.0(b.1(a.1(a.1(a.0(b.0(x))))))) → B1.0(b.0(x))
The TRS R consists of the following rules:
b.0(b.0(b.1(a.1(a.1(a.0(B.1(x))))))) → b.1(a.1(a.1(a.0(B.1(x)))))
b.0(b.0(b.1(a.1(a.1(a.0(b.1(x))))))) → b.1(a.1(a.1(a.0(b.0(b.0(b.1(x)))))))
b.1(a.1(a.0(b.0(b.0(x))))) → b.0(b.1(a.1(a.0(b.0(x)))))
b.1(a.0(b.1(a.1(a.1(a.0(b.0(x))))))) → b.1(a.0(b.1(a.1(a.0(b.0(x))))))
b.1(a.1(a.0(b.0(b.1(x))))) → b.0(b.1(a.1(a.0(b.1(x)))))
b.1(a.1(a.0(b.1(a.0(b.0(x)))))) → b.1(a.1(a.1(a.0(b.0(b.1(a.0(b.0(x))))))))
b.0(b.0(b.1(a.1(a.1(a.0(B.0(x))))))) → b.1(a.1(a.1(a.0(B.0(x)))))
b.1(a.1(a.0(b.1(a.0(b.1(x)))))) → b.1(a.1(a.1(a.0(b.0(b.1(a.0(b.1(x))))))))
b.1(a.0(b.1(a.1(a.1(a.0(b.1(x))))))) → b.1(a.0(b.1(a.1(a.0(b.1(x))))))
b.0(b.0(b.1(a.1(a.1(a.0(b.0(x))))))) → b.1(a.1(a.1(a.0(b.0(b.0(b.0(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ SemLabProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1.0(b.0(b.1(a.1(a.1(a.0(b.1(x))))))) → B1.0(b.1(x))
B1.0(b.0(b.1(a.1(a.1(a.0(b.0(x))))))) → B1.0(b.0(b.0(x)))
B1.0(b.0(b.1(a.1(a.1(a.0(b.1(x))))))) → B1.0(b.0(b.1(x)))
B1.0(b.0(b.1(a.1(a.1(a.0(b.0(x))))))) → B1.0(b.0(x))
The TRS R consists of the following rules:
b.0(b.0(b.1(a.1(a.1(a.0(B.1(x))))))) → b.1(a.1(a.1(a.0(B.1(x)))))
b.0(b.0(b.1(a.1(a.1(a.0(b.1(x))))))) → b.1(a.1(a.1(a.0(b.0(b.0(b.1(x)))))))
b.1(a.1(a.0(b.0(b.0(x))))) → b.0(b.1(a.1(a.0(b.0(x)))))
b.1(a.0(b.1(a.1(a.1(a.0(b.0(x))))))) → b.1(a.0(b.1(a.1(a.0(b.0(x))))))
b.1(a.1(a.0(b.0(b.1(x))))) → b.0(b.1(a.1(a.0(b.1(x)))))
b.1(a.1(a.0(b.1(a.0(b.0(x)))))) → b.1(a.1(a.1(a.0(b.0(b.1(a.0(b.0(x))))))))
b.0(b.0(b.1(a.1(a.1(a.0(B.0(x))))))) → b.1(a.1(a.1(a.0(B.0(x)))))
b.1(a.1(a.0(b.1(a.0(b.1(x)))))) → b.1(a.1(a.1(a.0(b.0(b.1(a.0(b.1(x))))))))
b.1(a.0(b.1(a.1(a.1(a.0(b.1(x))))))) → b.1(a.0(b.1(a.1(a.0(b.1(x))))))
b.0(b.0(b.1(a.1(a.1(a.0(b.0(x))))))) → b.1(a.1(a.1(a.0(b.0(b.0(b.0(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:
B1.0(b.0(b.1(a.1(a.1(a.0(b.1(x))))))) → B1.0(b.1(x))
B1.0(b.0(b.1(a.1(a.1(a.0(b.0(x))))))) → B1.0(b.0(b.0(x)))
B1.0(b.0(b.1(a.1(a.1(a.0(b.1(x))))))) → B1.0(b.0(b.1(x)))
B1.0(b.0(b.1(a.1(a.1(a.0(b.0(x))))))) → B1.0(b.0(x))
Used ordering: POLO with Polynomial interpretation [25]:
POL(B.0(x1)) = x1
POL(B.1(x1)) = x1
POL(B1.0(x1)) = x1
POL(a.0(x1)) = 1 + x1
POL(a.1(x1)) = x1
POL(b.0(x1)) = x1
POL(b.1(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ SemLabProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b.0(b.0(b.1(a.1(a.1(a.0(B.1(x))))))) → b.1(a.1(a.1(a.0(B.1(x)))))
b.0(b.0(b.1(a.1(a.1(a.0(b.1(x))))))) → b.1(a.1(a.1(a.0(b.0(b.0(b.1(x)))))))
b.1(a.1(a.0(b.0(b.0(x))))) → b.0(b.1(a.1(a.0(b.0(x)))))
b.1(a.0(b.1(a.1(a.1(a.0(b.0(x))))))) → b.1(a.0(b.1(a.1(a.0(b.0(x))))))
b.1(a.1(a.0(b.0(b.1(x))))) → b.0(b.1(a.1(a.0(b.1(x)))))
b.1(a.1(a.0(b.1(a.0(b.0(x)))))) → b.1(a.1(a.1(a.0(b.0(b.1(a.0(b.0(x))))))))
b.0(b.0(b.1(a.1(a.1(a.0(B.0(x))))))) → b.1(a.1(a.1(a.0(B.0(x)))))
b.1(a.1(a.0(b.1(a.0(b.1(x)))))) → b.1(a.1(a.1(a.0(b.0(b.1(a.0(b.1(x))))))))
b.1(a.0(b.1(a.1(a.1(a.0(b.1(x))))))) → b.1(a.0(b.1(a.1(a.0(b.1(x))))))
b.0(b.0(b.1(a.1(a.1(a.0(b.0(x))))))) → b.1(a.1(a.1(a.0(b.0(b.0(b.0(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(a(a(b(b(x))))) → B1(a(a(b(x))))
The TRS R consists of the following rules:
b(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(a(a(b(x))))))
b(b(b(a(a(a(B(x))))))) → b(a(a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B1(a(b(a(a(a(b(x))))))) → B1(a(b(a(a(b(x))))))
The TRS R consists of the following rules:
b(a(a(b(b(x))))) → b(b(a(a(b(x)))))
b(b(b(a(a(a(b(x))))))) → b(a(a(a(b(b(b(x)))))))
b(a(a(b(a(b(x)))))) → b(a(a(a(b(b(a(b(x))))))))
b(a(b(a(a(a(b(x))))))) → b(a(b(a(a(b(x))))))
b(b(b(a(a(a(B(x))))))) → b(a(a(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
Q DP problem:
The TRS P consists of the following rules:
B(a(b(a(a(b(x1)))))) → B(a(b(b(a(a(a(b(x1))))))))
The TRS R consists of the following rules:
b(b(a(a(b(x1))))) → b(a(a(b(b(x1)))))
b(a(a(a(b(b(b(x1))))))) → b(b(b(a(a(a(b(x1)))))))
b(a(b(a(a(b(x1)))))) → b(a(b(b(a(a(a(b(x1))))))))
b(a(a(a(b(a(b(x1))))))) → b(a(a(b(a(b(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.