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(a(x1))) → b(x1)
b(c(x1)) → c(c(a(a(a(a(x1))))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(x1))) → b(x1)
b(c(x1)) → c(c(a(a(a(a(x1))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(a(x1))) → b(x1)
b(c(x1)) → c(c(a(a(a(a(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(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(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(a(x1))) → b(x1)
b(c(x1)) → c(c(a(a(a(a(x1))))))
The set Q is empty.
We have obtained the following QTRS:
a(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(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(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(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(c(x1)) → A(x1)
A(a(a(x1))) → B(x1)
B(c(x1)) → A(a(a(x1)))
B(c(x1)) → A(a(x1))
B(c(x1)) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(a(a(x1))) → b(x1)
b(c(x1)) → c(c(a(a(a(a(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
B(c(x1)) → A(x1)
A(a(a(x1))) → B(x1)
B(c(x1)) → A(a(a(x1)))
B(c(x1)) → A(a(x1))
B(c(x1)) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(a(a(x1))) → b(x1)
b(c(x1)) → c(c(a(a(a(a(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B(c(x1)) → A(a(x1)) we obtained the following new rules:
B(c(a(y_0))) → A(a(a(y_0)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
Q DP problem:
The TRS P consists of the following rules:
A(a(a(x1))) → B(x1)
B(c(x1)) → A(x1)
B(c(x1)) → A(a(a(x1)))
B(c(a(y_0))) → A(a(a(y_0)))
B(c(x1)) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(a(a(x1))) → b(x1)
b(c(x1)) → c(c(a(a(a(a(x1))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B(c(x1)) → A(x1) we obtained the following new rules:
B(c(a(a(y_0)))) → A(a(a(y_0)))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
Q DP problem:
The TRS P consists of the following rules:
A(a(a(x1))) → B(x1)
B(c(x1)) → A(a(a(x1)))
B(c(a(y_0))) → A(a(a(y_0)))
B(c(a(a(y_0)))) → A(a(a(y_0)))
B(c(x1)) → A(a(a(a(x1))))
The TRS R consists of the following rules:
a(a(a(x1))) → b(x1)
b(c(x1)) → c(c(a(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
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(a(x1))) → b(x1)
b(c(x1)) → c(c(a(a(a(a(x1))))))
A(a(a(x1))) → B(x1)
B(c(x1)) → A(a(a(x1)))
B(c(a(y_0))) → A(a(a(y_0)))
B(c(a(a(y_0)))) → A(a(a(y_0)))
B(c(x1)) → A(a(a(a(x1))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(a(x1))) → b(x1)
b(c(x1)) → c(c(a(a(a(a(x1))))))
A(a(a(x1))) → B(x1)
B(c(x1)) → A(a(a(x1)))
B(c(a(y_0))) → A(a(a(y_0)))
B(c(a(a(y_0)))) → A(a(a(y_0)))
B(c(x1)) → A(a(a(a(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
a(a(A(x))) → B(x)
c(B(x)) → a(a(A(x)))
a(c(B(x))) → a(a(A(x)))
a(a(c(B(x)))) → a(a(A(x)))
c(B(x)) → a(a(a(A(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ 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(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
a(a(A(x))) → B(x)
c(B(x)) → a(a(A(x)))
a(c(B(x))) → a(a(A(x)))
a(a(c(B(x)))) → a(a(A(x)))
c(B(x)) → a(a(a(A(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
a(a(A(x))) → B(x)
c(B(x)) → a(a(A(x)))
a(c(B(x))) → a(a(A(x)))
a(a(c(B(x)))) → a(a(A(x)))
c(B(x)) → a(a(a(A(x))))
The set Q is empty.
We have obtained the following QTRS:
a(a(a(x))) → b(x)
b(c(x)) → c(c(a(a(a(a(x))))))
A(a(a(x))) → B(x)
B(c(x)) → A(a(a(x)))
B(c(a(x))) → A(a(a(x)))
B(c(a(a(x)))) → A(a(a(x)))
B(c(x)) → A(a(a(a(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ 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(a(x))) → b(x)
b(c(x)) → c(c(a(a(a(a(x))))))
A(a(a(x))) → B(x)
B(c(x)) → A(a(a(x)))
B(c(a(x))) → A(a(a(x)))
B(c(a(a(x)))) → A(a(a(x)))
B(c(x)) → A(a(a(a(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
a(a(A(x))) → B(x)
c(B(x)) → a(a(A(x)))
a(c(B(x))) → a(a(A(x)))
a(a(c(B(x)))) → a(a(A(x)))
c(B(x)) → a(a(a(A(x))))
The set Q is empty.
We have obtained the following QTRS:
a(a(a(x))) → b(x)
b(c(x)) → c(c(a(a(a(a(x))))))
A(a(a(x))) → B(x)
B(c(x)) → A(a(a(x)))
B(c(a(x))) → A(a(a(x)))
B(c(a(a(x)))) → A(a(a(x)))
B(c(x)) → A(a(a(a(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ 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(a(x))) → b(x)
b(c(x)) → c(c(a(a(a(a(x))))))
A(a(a(x))) → B(x)
B(c(x)) → A(a(a(x)))
B(c(a(x))) → A(a(a(x)))
B(c(a(a(x)))) → A(a(a(x)))
B(c(x)) → A(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:
A1(c(B(x))) → A1(A(x))
A1(a(c(B(x)))) → A1(A(x))
C(b(x)) → A1(a(a(a(c(c(x))))))
C(B(x)) → A1(a(A(x)))
C(B(x)) → A1(A(x))
C(b(x)) → C(c(x))
C(B(x)) → A1(a(a(A(x))))
C(b(x)) → C(x)
C(b(x)) → A1(a(c(c(x))))
C(b(x)) → A1(a(a(c(c(x)))))
C(b(x)) → A1(c(c(x)))
A1(c(B(x))) → A1(a(A(x)))
A1(a(c(B(x)))) → A1(a(A(x)))
The TRS R consists of the following rules:
a(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
a(a(A(x))) → B(x)
c(B(x)) → a(a(A(x)))
a(c(B(x))) → a(a(A(x)))
a(a(c(B(x)))) → a(a(A(x)))
c(B(x)) → a(a(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A1(c(B(x))) → A1(A(x))
A1(a(c(B(x)))) → A1(A(x))
C(b(x)) → A1(a(a(a(c(c(x))))))
C(B(x)) → A1(a(A(x)))
C(B(x)) → A1(A(x))
C(b(x)) → C(c(x))
C(B(x)) → A1(a(a(A(x))))
C(b(x)) → C(x)
C(b(x)) → A1(a(c(c(x))))
C(b(x)) → A1(a(a(c(c(x)))))
C(b(x)) → A1(c(c(x)))
A1(c(B(x))) → A1(a(A(x)))
A1(a(c(B(x)))) → A1(a(A(x)))
The TRS R consists of the following rules:
a(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
a(a(A(x))) → B(x)
c(B(x)) → a(a(A(x)))
a(c(B(x))) → a(a(A(x)))
a(a(c(B(x)))) → a(a(A(x)))
c(B(x)) → a(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 11 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → C(c(x))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
a(a(A(x))) → B(x)
c(B(x)) → a(a(A(x)))
a(c(B(x))) → a(a(A(x)))
a(a(c(B(x)))) → a(a(A(x)))
c(B(x)) → a(a(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(x)) → C(c(x)) at position [0] we obtained the following new rules:
C(b(B(x0))) → C(a(a(a(A(x0)))))
C(b(B(x0))) → C(a(a(A(x0))))
C(b(b(x0))) → C(a(a(a(a(c(c(x0)))))))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
C(b(b(x0))) → C(a(a(a(a(c(c(x0)))))))
C(b(B(x0))) → C(a(a(A(x0))))
C(b(B(x0))) → C(a(a(a(A(x0)))))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
a(a(A(x))) → B(x)
c(B(x)) → a(a(A(x)))
a(c(B(x))) → a(a(A(x)))
a(a(c(B(x)))) → a(a(A(x)))
c(B(x)) → a(a(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(B(x0))) → C(a(a(A(x0)))) at position [0] we obtained the following new rules:
C(b(B(x0))) → C(B(x0))
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
C(b(B(x0))) → C(B(x0))
C(b(b(x0))) → C(a(a(a(a(c(c(x0)))))))
C(b(x)) → C(x)
C(b(B(x0))) → C(a(a(a(A(x0)))))
The TRS R consists of the following rules:
a(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
a(a(A(x))) → B(x)
c(B(x)) → a(a(A(x)))
a(c(B(x))) → a(a(A(x)))
a(a(c(B(x)))) → a(a(A(x)))
c(B(x)) → a(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
C(b(b(x0))) → C(a(a(a(a(c(c(x0)))))))
C(b(B(x0))) → C(a(a(a(A(x0)))))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(a(x))) → b(x)
c(b(x)) → a(a(a(a(c(c(x))))))
a(a(A(x))) → B(x)
c(B(x)) → a(a(A(x)))
a(c(B(x))) → a(a(A(x)))
a(a(c(B(x)))) → a(a(A(x)))
c(B(x)) → a(a(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.