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)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(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)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(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(c(x1)) → A(x1)
B(c(x1)) → B(a(x1))
A(b(x1)) → A(x1)
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(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(c(x1)) → A(x1)
B(c(x1)) → B(a(x1))
A(b(x1)) → A(x1)
A(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(x1)) → B(a(x1)) at position [0] we obtained the following new rules:
B(c(a(x0))) → B(b(x0))
B(c(b(x0))) → B(c(a(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(a(x0))) → B(b(x0))
B(c(x1)) → A(x1)
B(c(b(x0))) → B(c(a(x0)))
A(a(x1)) → B(x1)
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))
B(c(a(x0))) → B(b(x0))
B(c(x1)) → A(x1)
B(c(b(x0))) → B(c(a(x0)))
A(a(x1)) → B(x1)
A(b(x1)) → A(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))
B(c(a(x0))) → B(b(x0))
B(c(x1)) → A(x1)
B(c(b(x0))) → B(c(a(x0)))
A(a(x1)) → B(x1)
A(b(x1)) → A(x1)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → 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:
C(b(x)) → A1(b(c(x)))
A1(a(x)) → B1(x)
B1(c(B(x))) → A1(c(B(x)))
A1(c(B(x))) → B1(B(x))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(x)) → A1(c(x))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → A1(b(c(x)))
A1(a(x)) → B1(x)
B1(c(B(x))) → A1(c(B(x)))
A1(c(B(x))) → B1(B(x))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(x)) → A1(c(x))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → 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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → A1(b(c(x)))
A1(a(x)) → B1(x)
B1(c(B(x))) → A1(c(B(x)))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(x)) → A1(c(x))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(c(B(x))) → A1(c(B(x))) at position [0] we obtained the following new rules:
B1(c(B(x0))) → A1(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(x)
C(b(x)) → A1(b(c(x)))
B1(c(B(x0))) → A1(A(x0))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(x)) → A1(c(x))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → 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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x)) → A1(b(c(x)))
A1(a(x)) → B1(x)
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(x)) → A1(c(x))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(x)) → A1(c(x)) at position [0] we obtained the following new rules:
B1(a(B(x0))) → A1(A(x0))
B1(a(b(x0))) → A1(a(b(c(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(x)
C(b(x)) → A1(b(c(x)))
B1(a(B(x0))) → A1(A(x0))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → 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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(x)
C(b(x)) → A1(b(c(x)))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(x)) → A1(b(c(x))) at position [0] we obtained the following new rules:
C(b(b(x0))) → A1(b(a(b(c(x0)))))
C(b(B(x0))) → A1(b(A(x0)))
C(b(B(x0))) → A1(a(c(B(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(b(x0))) → A1(b(a(b(c(x0)))))
A1(a(x)) → B1(x)
C(b(B(x0))) → A1(b(A(x0)))
C(b(B(x0))) → A1(a(c(B(x0))))
B1(a(x)) → C(x)
C(b(x)) → B1(c(x))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(x)) → B1(c(x)) at position [0] we obtained the following new rules:
C(b(b(x0))) → B1(a(b(c(x0))))
C(b(B(x0))) → B1(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(x)
C(b(b(x0))) → A1(b(a(b(c(x0)))))
C(b(B(x0))) → A1(b(A(x0)))
C(b(B(x0))) → B1(A(x0))
B1(a(x)) → C(x)
C(b(B(x0))) → A1(a(c(B(x0))))
C(b(b(x0))) → B1(a(b(c(x0))))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → 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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(x)
C(b(b(x0))) → A1(b(a(b(c(x0)))))
C(b(B(x0))) → A1(b(A(x0)))
C(b(B(x0))) → A1(a(c(B(x0))))
B1(a(x)) → C(x)
C(b(b(x0))) → B1(a(b(c(x0))))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(B(x0))) → A1(b(A(x0))) at position [0] we obtained the following new rules:
C(b(B(x0))) → A1(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(B(x0))) → A1(A(x0))
C(b(b(x0))) → A1(b(a(b(c(x0)))))
A1(a(x)) → B1(x)
B1(a(x)) → C(x)
C(b(B(x0))) → A1(a(c(B(x0))))
C(b(b(x0))) → B1(a(b(c(x0))))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → 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
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(x)
C(b(b(x0))) → A1(b(a(b(c(x0)))))
C(b(B(x0))) → A1(a(c(B(x0))))
B1(a(x)) → C(x)
C(b(b(x0))) → B1(a(b(c(x0))))
B1(a(b(x0))) → A1(a(b(c(x0))))
C(b(x)) → C(x)
The TRS R consists of the following rules:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(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(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(x)
a(b(x)) → c(a(x))
b(c(x)) → c(b(a(x)))
B(c(a(x))) → B(b(x))
B(c(x)) → A(x)
B(c(b(x))) → B(c(a(x)))
A(a(x)) → B(x)
A(b(x)) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(x)
a(b(x)) → c(a(x))
b(c(x)) → c(b(a(x)))
B(c(a(x))) → B(b(x))
B(c(x)) → A(x)
B(c(b(x))) → B(c(a(x)))
A(a(x)) → B(x)
A(b(x)) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
a(c(B(x))) → b(B(x))
c(B(x)) → A(x)
b(c(B(x))) → a(c(B(x)))
a(A(x)) → B(x)
b(A(x)) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(x)
a(b(x)) → c(a(x))
b(c(x)) → c(b(a(x)))
B(c(a(x))) → B(b(x))
B(c(x)) → A(x)
B(c(b(x))) → B(c(a(x)))
A(a(x)) → B(x)
A(b(x)) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(x)
a(b(x)) → c(a(x))
b(c(x)) → c(b(a(x)))
B(c(a(x))) → B(b(x))
B(c(x)) → A(x)
B(c(b(x))) → B(c(a(x)))
A(a(x)) → B(x)
A(b(x)) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(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(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → b(x1)
a(b(x1)) → c(a(x1))
b(c(x1)) → c(b(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(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(x)
b(a(x)) → a(c(x))
c(b(x)) → a(b(c(x)))
Q is empty.