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(a(b(b(b(b(a(a(x1)))))))) → a(a(c(c(a(a(b(b(x1))))))))
a(a(c(c(x1)))) → c(c(c(c(a(a(x1))))))
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(b(x1))))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x1)))))))) → a(a(c(c(a(a(b(b(x1))))))))
a(a(c(c(x1)))) → c(c(c(c(a(a(x1))))))
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(b(x1))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(b(b(a(a(x1)))))))) → a(a(c(c(a(a(b(b(x1))))))))
a(a(c(c(x1)))) → c(c(c(c(a(a(x1))))))
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(b(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(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:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(b(b(a(a(x1)))))))) → a(a(c(c(a(a(b(b(x1))))))))
a(a(c(c(x1)))) → c(c(c(c(a(a(x1))))))
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(b(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(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:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(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:
A(a(c(c(x1)))) → A(a(x1))
A(a(c(c(x1)))) → C(a(a(x1)))
A(a(b(b(b(b(a(a(x1)))))))) → C(a(a(b(b(x1)))))
A(a(c(c(x1)))) → C(c(c(c(a(a(x1))))))
A(a(c(c(x1)))) → C(c(c(a(a(x1)))))
C(c(c(c(c(c(x1)))))) → C(c(b(b(x1))))
A(a(b(b(b(b(a(a(x1)))))))) → C(c(a(a(b(b(x1))))))
C(c(c(c(c(c(x1)))))) → C(b(b(x1)))
A(a(b(b(b(b(a(a(x1)))))))) → A(a(c(c(a(a(b(b(x1))))))))
A(a(b(b(b(b(a(a(x1)))))))) → A(c(c(a(a(b(b(x1)))))))
A(a(c(c(x1)))) → C(c(a(a(x1))))
A(a(c(c(x1)))) → A(x1)
A(a(b(b(b(b(a(a(x1)))))))) → A(a(b(b(x1))))
A(a(b(b(b(b(a(a(x1)))))))) → A(b(b(x1)))
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x1)))))))) → a(a(c(c(a(a(b(b(x1))))))))
a(a(c(c(x1)))) → c(c(c(c(a(a(x1))))))
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(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:
A(a(c(c(x1)))) → A(a(x1))
A(a(c(c(x1)))) → C(a(a(x1)))
A(a(b(b(b(b(a(a(x1)))))))) → C(a(a(b(b(x1)))))
A(a(c(c(x1)))) → C(c(c(c(a(a(x1))))))
A(a(c(c(x1)))) → C(c(c(a(a(x1)))))
C(c(c(c(c(c(x1)))))) → C(c(b(b(x1))))
A(a(b(b(b(b(a(a(x1)))))))) → C(c(a(a(b(b(x1))))))
C(c(c(c(c(c(x1)))))) → C(b(b(x1)))
A(a(b(b(b(b(a(a(x1)))))))) → A(a(c(c(a(a(b(b(x1))))))))
A(a(b(b(b(b(a(a(x1)))))))) → A(c(c(a(a(b(b(x1)))))))
A(a(c(c(x1)))) → C(c(a(a(x1))))
A(a(c(c(x1)))) → A(x1)
A(a(b(b(b(b(a(a(x1)))))))) → A(a(b(b(x1))))
A(a(b(b(b(b(a(a(x1)))))))) → A(b(b(x1)))
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x1)))))))) → a(a(c(c(a(a(b(b(x1))))))))
a(a(c(c(x1)))) → c(c(c(c(a(a(x1))))))
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(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 1 SCC with 9 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
A(a(c(c(x1)))) → A(a(x1))
A(a(b(b(b(b(a(a(x1)))))))) → A(a(c(c(a(a(b(b(x1))))))))
A(a(b(b(b(b(a(a(x1)))))))) → A(c(c(a(a(b(b(x1)))))))
A(a(c(c(x1)))) → A(x1)
A(a(b(b(b(b(a(a(x1)))))))) → A(a(b(b(x1))))
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x1)))))))) → a(a(c(c(a(a(b(b(x1))))))))
a(a(c(c(x1)))) → c(c(c(c(a(a(x1))))))
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(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(b(b(b(b(a(a(x1)))))))) → A(c(c(a(a(b(b(x1)))))))
A(a(c(c(x1)))) → A(x1)
A(a(b(b(b(b(a(a(x1)))))))) → A(a(b(b(x1))))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(a(x1)) = 2 + x1
POL(b(x1)) = x1
POL(c(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
Q DP problem:
The TRS P consists of the following rules:
A(a(c(c(x1)))) → A(a(x1))
A(a(b(b(b(b(a(a(x1)))))))) → A(a(c(c(a(a(b(b(x1))))))))
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x1)))))))) → a(a(c(c(a(a(b(b(x1))))))))
a(a(c(c(x1)))) → c(c(c(c(a(a(x1))))))
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(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
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x1)))))))) → a(a(c(c(a(a(b(b(x1))))))))
a(a(c(c(x1)))) → c(c(c(c(a(a(x1))))))
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(b(x1))))))
A(a(c(c(x1)))) → A(a(x1))
A(a(b(b(b(b(a(a(x1)))))))) → A(a(c(c(a(a(b(b(x1))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(b(b(a(a(x1)))))))) → a(a(c(c(a(a(b(b(x1))))))))
a(a(c(c(x1)))) → c(c(c(c(a(a(x1))))))
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(b(x1))))))
A(a(c(c(x1)))) → A(a(x1))
A(a(b(b(b(b(a(a(x1)))))))) → A(a(c(c(a(a(b(b(x1))))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(a(A(x))))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(a(A(x))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(a(A(x))))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(b(b(b(a(a(x)))))))) → a(a(c(c(a(a(b(b(x))))))))
a(a(c(c(x)))) → c(c(c(c(a(a(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
A(a(c(c(x)))) → A(a(x))
A(a(b(b(b(b(a(a(x)))))))) → A(a(c(c(a(a(b(b(x))))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → a(a(c(c(a(a(b(b(x))))))))
a(a(c(c(x)))) → c(c(c(c(a(a(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
A(a(c(c(x)))) → A(a(x))
A(a(b(b(b(b(a(a(x)))))))) → A(a(c(c(a(a(b(b(x))))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(a(A(x))))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(b(b(b(b(a(a(x)))))))) → a(a(c(c(a(a(b(b(x))))))))
a(a(c(c(x)))) → c(c(c(c(a(a(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
A(a(c(c(x)))) → A(a(x))
A(a(b(b(b(b(a(a(x)))))))) → A(a(c(c(a(a(b(b(x))))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → a(a(c(c(a(a(b(b(x))))))))
a(a(c(c(x)))) → c(c(c(c(a(a(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
A(a(c(c(x)))) → A(a(x))
A(a(b(b(b(b(a(a(x)))))))) → A(a(c(c(a(a(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:
C(c(a(a(x)))) → C(c(x))
A1(a(b(b(b(b(a(a(x)))))))) → A1(a(c(c(a(a(x))))))
C(c(a(a(x)))) → C(c(c(c(x))))
A1(a(b(b(b(b(a(a(x)))))))) → C(a(a(x)))
A1(a(b(b(b(b(a(a(x)))))))) → C(c(a(a(x))))
C(c(a(a(x)))) → A1(a(c(c(c(c(x))))))
A1(a(b(b(b(b(a(a(x)))))))) → A1(c(c(a(a(x)))))
C(c(c(c(c(c(x)))))) → C(c(b(b(x))))
A1(a(b(b(b(b(a(A(x)))))))) → C(a(A(x)))
A1(a(b(b(b(b(a(A(x)))))))) → C(c(a(A(x))))
C(c(c(c(c(c(x)))))) → C(b(b(x)))
C(c(a(a(x)))) → C(c(c(x)))
A1(a(b(b(b(b(a(A(x)))))))) → A1(c(c(a(A(x)))))
A1(a(b(b(b(b(a(A(x)))))))) → A1(a(c(c(a(A(x))))))
C(c(a(a(x)))) → A1(c(c(c(c(x)))))
C(c(a(a(x)))) → C(x)
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(a(A(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
C(c(a(a(x)))) → C(c(x))
A1(a(b(b(b(b(a(a(x)))))))) → A1(a(c(c(a(a(x))))))
C(c(a(a(x)))) → C(c(c(c(x))))
A1(a(b(b(b(b(a(a(x)))))))) → C(a(a(x)))
A1(a(b(b(b(b(a(a(x)))))))) → C(c(a(a(x))))
C(c(a(a(x)))) → A1(a(c(c(c(c(x))))))
A1(a(b(b(b(b(a(a(x)))))))) → A1(c(c(a(a(x)))))
C(c(c(c(c(c(x)))))) → C(c(b(b(x))))
A1(a(b(b(b(b(a(A(x)))))))) → C(a(A(x)))
A1(a(b(b(b(b(a(A(x)))))))) → C(c(a(A(x))))
C(c(c(c(c(c(x)))))) → C(b(b(x)))
C(c(a(a(x)))) → C(c(c(x)))
A1(a(b(b(b(b(a(A(x)))))))) → A1(c(c(a(A(x)))))
A1(a(b(b(b(b(a(A(x)))))))) → A1(a(c(c(a(A(x))))))
C(c(a(a(x)))) → A1(c(c(c(c(x)))))
C(c(a(a(x)))) → C(x)
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(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 5 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
C(c(a(a(x)))) → C(c(x))
C(c(a(a(x)))) → A1(a(c(c(c(c(x))))))
A1(a(b(b(b(b(a(a(x)))))))) → A1(c(c(a(a(x)))))
C(c(a(a(x)))) → C(c(c(c(x))))
A1(a(b(b(b(b(a(a(x)))))))) → A1(a(c(c(a(a(x))))))
C(c(a(a(x)))) → C(c(c(x)))
A1(a(b(b(b(b(a(A(x)))))))) → A1(c(c(a(A(x)))))
A1(a(b(b(b(b(a(A(x)))))))) → A1(a(c(c(a(A(x))))))
C(c(a(a(x)))) → C(x)
C(c(a(a(x)))) → A1(c(c(c(c(x)))))
A1(a(b(b(b(b(a(a(x)))))))) → C(c(a(a(x))))
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(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:
A1(a(b(b(b(b(a(A(x)))))))) → A1(c(c(a(A(x)))))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 1 + x1
POL(A1(x1)) = x1
POL(C(x1)) = x1
POL(a(x1)) = 2·x1
POL(b(x1)) = x1
POL(c(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
C(c(a(a(x)))) → C(c(x))
C(c(a(a(x)))) → A1(a(c(c(c(c(x))))))
A1(a(b(b(b(b(a(a(x)))))))) → A1(c(c(a(a(x)))))
A1(a(b(b(b(b(a(a(x)))))))) → A1(a(c(c(a(a(x))))))
C(c(a(a(x)))) → C(c(c(c(x))))
C(c(a(a(x)))) → C(c(c(x)))
A1(a(b(b(b(b(a(A(x)))))))) → A1(a(c(c(a(A(x))))))
C(c(a(a(x)))) → A1(c(c(c(c(x)))))
C(c(a(a(x)))) → C(x)
A1(a(b(b(b(b(a(a(x)))))))) → C(c(a(a(x))))
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(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:
C(c(a(a(x)))) → C(c(x))
C(c(a(a(x)))) → A1(a(c(c(c(c(x))))))
A1(a(b(b(b(b(a(a(x)))))))) → A1(c(c(a(a(x)))))
C(c(a(a(x)))) → C(c(c(c(x))))
C(c(a(a(x)))) → C(c(c(x)))
C(c(a(a(x)))) → A1(c(c(c(c(x)))))
C(c(a(a(x)))) → C(x)
A1(a(b(b(b(b(a(a(x)))))))) → C(c(a(a(x))))
Used ordering: POLO with Polynomial interpretation [25]:
POL(A(x1)) = 2·x1
POL(A1(x1)) = 2·x1
POL(C(x1)) = 2 + 2·x1
POL(a(x1)) = 2 + x1
POL(b(x1)) = x1
POL(c(x1)) = x1
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(b(b(a(a(x)))))))) → A1(a(c(c(a(a(x))))))
A1(a(b(b(b(b(a(A(x)))))))) → A1(a(c(c(a(A(x))))))
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(a(A(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(b(b(b(b(a(A(x)))))))) → A1(a(c(c(a(A(x)))))) at position [0] we obtained the following new rules:
A1(a(b(b(b(b(a(A(x0)))))))) → A1(a(a(A(x0))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(b(b(a(a(x)))))))) → A1(a(c(c(a(a(x))))))
A1(a(b(b(b(b(a(A(x0)))))))) → A1(a(a(A(x0))))
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(b(b(a(a(x)))))))) → A1(a(c(c(a(a(x))))))
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(a(A(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(b(b(b(b(a(a(x)))))))) → A1(a(c(c(a(a(x)))))) at position [0] we obtained the following new rules:
A1(a(b(b(b(b(a(a(a(b(b(b(b(a(a(x0))))))))))))))) → A1(a(c(c(a(b(b(a(a(c(c(a(a(x0)))))))))))))
A1(a(b(b(b(b(a(a(a(b(b(b(b(a(A(x0))))))))))))))) → A1(a(c(c(a(b(b(a(a(c(c(a(A(x0)))))))))))))
A1(a(b(b(b(b(a(a(x0)))))))) → A1(a(a(a(c(c(c(c(x0))))))))
A1(a(b(b(b(b(a(a(b(b(b(b(a(A(x0)))))))))))))) → A1(a(c(c(b(b(a(a(c(c(a(A(x0))))))))))))
A1(a(b(b(b(b(a(a(b(b(b(b(a(a(x0)))))))))))))) → A1(a(c(c(b(b(a(a(c(c(a(a(x0))))))))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(b(b(a(a(a(b(b(b(b(a(a(x0))))))))))))))) → A1(a(c(c(a(b(b(a(a(c(c(a(a(x0)))))))))))))
A1(a(b(b(b(b(a(a(x0)))))))) → A1(a(a(a(c(c(c(c(x0))))))))
A1(a(b(b(b(b(a(a(b(b(b(b(a(a(x0)))))))))))))) → A1(a(c(c(b(b(a(a(c(c(a(a(x0))))))))))))
A1(a(b(b(b(b(a(a(a(b(b(b(b(a(A(x0))))))))))))))) → A1(a(c(c(a(b(b(a(a(c(c(a(A(x0)))))))))))))
A1(a(b(b(b(b(a(a(b(b(b(b(a(A(x0)))))))))))))) → A1(a(c(c(b(b(a(a(c(c(a(A(x0))))))))))))
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A1(a(b(b(b(b(a(a(x0)))))))) → A1(a(a(a(c(c(c(c(x0))))))))
The TRS R consists of the following rules:
a(a(b(b(b(b(a(a(x)))))))) → b(b(a(a(c(c(a(a(x))))))))
c(c(a(a(x)))) → a(a(c(c(c(c(x))))))
c(c(c(c(c(c(x)))))) → b(b(c(c(b(b(x))))))
c(c(a(A(x)))) → a(A(x))
a(a(b(b(b(b(a(A(x)))))))) → b(b(a(a(c(c(a(A(x))))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.