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(x1)) → a(b(x1))
b(a(c(x1))) → c(c(a(a(a(x1)))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → a(b(x1))
b(a(c(x1))) → c(c(a(a(a(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:
B(a(c(x1))) → A(a(a(x1)))
A(a(x1)) → A(b(x1))
B(a(c(x1))) → A(a(x1))
B(a(c(x1))) → A(x1)
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(a(x1)) → a(b(x1))
b(a(c(x1))) → c(c(a(a(a(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:
B(a(c(x1))) → A(a(a(x1)))
A(a(x1)) → A(b(x1))
B(a(c(x1))) → A(a(x1))
B(a(c(x1))) → A(x1)
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(a(x1)) → a(b(x1))
b(a(c(x1))) → c(c(a(a(a(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(x1)) → A(b(x1)) at position [0] we obtained the following new rules:
A(a(a(c(x0)))) → A(c(c(a(a(a(x0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(c(x1))) → A(a(a(x1)))
A(a(a(c(x0)))) → A(c(c(a(a(a(x0))))))
B(a(c(x1))) → A(a(x1))
B(a(c(x1))) → A(x1)
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(a(x1)) → a(b(x1))
b(a(c(x1))) → c(c(a(a(a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(c(x1))) → A(a(a(x1)))
B(a(c(x1))) → A(a(x1))
B(a(c(x1))) → A(x1)
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(a(x1)) → a(b(x1))
b(a(c(x1))) → c(c(a(a(a(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → a(b(x1))
b(a(c(x1))) → c(c(a(a(a(x1)))))
B(a(c(x1))) → A(a(a(x1)))
B(a(c(x1))) → A(a(x1))
B(a(c(x1))) → A(x1)
A(a(x1)) → B(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → a(b(x1))
b(a(c(x1))) → c(c(a(a(a(x1)))))
B(a(c(x1))) → A(a(a(x1)))
B(a(c(x1))) → A(a(x1))
B(a(c(x1))) → A(x1)
A(a(x1)) → B(x1)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → B(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → B(x)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → a(b(x))
b(a(c(x))) → c(c(a(a(a(x)))))
B(a(c(x))) → A(a(a(x)))
B(a(c(x))) → A(a(x))
B(a(c(x))) → A(x)
A(a(x)) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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(a(x)) → a(b(x))
b(a(c(x))) → c(c(a(a(a(x)))))
B(a(c(x))) → A(a(a(x)))
B(a(c(x))) → A(a(x))
B(a(c(x))) → A(x)
A(a(x)) → B(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → B(x)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → a(b(x))
b(a(c(x))) → c(c(a(a(a(x)))))
B(a(c(x))) → A(a(a(x)))
B(a(c(x))) → A(a(x))
B(a(c(x))) → A(x)
A(a(x)) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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(a(x)) → a(b(x))
b(a(c(x))) → c(c(a(a(a(x)))))
B(a(c(x))) → A(a(a(x)))
B(a(c(x))) → A(a(x))
B(a(c(x))) → A(x)
A(a(x)) → 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(a(b(x))) → C(c(x))
C(a(b(x))) → A1(a(c(c(x))))
C(a(b(x))) → A1(a(a(c(c(x)))))
C(a(B(x))) → A1(A(x))
C(a(B(x))) → A1(a(A(x)))
C(a(b(x))) → C(x)
C(a(b(x))) → A1(c(c(x)))
The TRS R consists of the following rules:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C(a(b(x))) → C(c(x))
C(a(b(x))) → A1(a(c(c(x))))
C(a(b(x))) → A1(a(a(c(c(x)))))
C(a(B(x))) → A1(A(x))
C(a(B(x))) → A1(a(A(x)))
C(a(b(x))) → C(x)
C(a(b(x))) → A1(c(c(x)))
The TRS R consists of the following rules:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → 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 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C(a(b(x))) → C(c(x))
C(a(b(x))) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(b(x))) → C(c(x)) at position [0] we obtained the following new rules:
C(a(b(a(B(x0))))) → C(a(A(x0)))
C(a(b(a(B(x0))))) → C(A(x0))
C(a(b(a(b(x0))))) → C(a(a(a(c(c(x0))))))
C(a(b(a(B(x0))))) → C(a(a(A(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C(a(b(a(B(x0))))) → C(A(x0))
C(a(b(a(B(x0))))) → C(a(A(x0)))
C(a(b(a(b(x0))))) → C(a(a(a(c(c(x0))))))
C(a(b(a(B(x0))))) → C(a(a(A(x0))))
C(a(b(x))) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → 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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C(a(b(a(B(x0))))) → C(a(A(x0)))
C(a(b(a(b(x0))))) → C(a(a(a(c(c(x0))))))
C(a(b(a(B(x0))))) → C(a(a(A(x0))))
C(a(b(x))) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(b(a(B(x0))))) → C(a(A(x0))) at position [0] we obtained the following new rules:
C(a(b(a(B(x0))))) → C(B(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
C(a(b(a(B(x0))))) → C(B(x0))
C(a(b(a(b(x0))))) → C(a(a(a(c(c(x0))))))
C(a(b(x))) → C(x)
C(a(b(a(B(x0))))) → C(a(a(A(x0))))
The TRS R consists of the following rules:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → 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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(b(a(b(x0))))) → C(a(a(a(c(c(x0))))))
C(a(b(a(B(x0))))) → C(a(a(A(x0))))
C(a(b(x))) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(b(a(B(x0))))) → C(a(a(A(x0)))) at position [0] we obtained the following new rules:
C(a(b(a(B(y0))))) → C(b(a(A(y0))))
C(a(b(a(B(x0))))) → C(a(B(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
C(a(b(a(B(y0))))) → C(b(a(A(y0))))
C(a(b(a(B(x0))))) → C(a(B(x0)))
C(a(b(a(b(x0))))) → C(a(a(a(c(c(x0))))))
C(a(b(x))) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → 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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(b(a(b(x0))))) → C(a(a(a(c(c(x0))))))
C(a(b(x))) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
c(a(B(x))) → a(a(A(x)))
c(a(B(x))) → a(A(x))
c(a(B(x))) → A(x)
a(A(x)) → B(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(a(x1)) → a(b(x1))
b(a(c(x1))) → c(c(a(a(a(x1)))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(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:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → a(b(x1))
b(a(c(x1))) → c(c(a(a(a(x1)))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(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:
a(a(x)) → b(a(x))
c(a(b(x))) → a(a(a(c(c(x)))))
Q is empty.