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(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(x1) → a(c(a(x1)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(x1) → a(c(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:
A(a(b(x1))) → B(b(a(a(x1))))
A(a(b(x1))) → B(a(a(x1)))
B(x1) → A(c(a(x1)))
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
B(x1) → A(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(x1) → a(c(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(x1))) → B(b(a(a(x1))))
A(a(b(x1))) → B(a(a(x1)))
B(x1) → A(c(a(x1)))
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
B(x1) → A(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(x1) → a(c(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(b(x1))) → B(b(a(a(x1))))
A(a(b(x1))) → B(a(a(x1)))
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
B(x1) → A(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(x1) → a(c(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(x1) → a(c(a(x1)))
A(a(b(x1))) → B(b(a(a(x1))))
A(a(b(x1))) → B(a(a(x1)))
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
B(x1) → A(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(x1) → a(c(a(x1)))
A(a(b(x1))) → B(b(a(a(x1))))
A(a(b(x1))) → B(a(a(x1)))
A(a(b(x1))) → A(x1)
A(a(b(x1))) → A(a(x1))
B(x1) → A(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(x) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(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:
B1(a(A(x))) → A1(b(B(x)))
B1(a(a(x))) → A1(b(b(x)))
B1(x) → A1(c(a(x)))
B1(a(A(x))) → A1(a(B(x)))
B1(a(A(x))) → B2(x)
B1(a(a(x))) → A1(a(b(b(x))))
B1(a(a(x))) → B1(b(x))
B1(a(A(x))) → A1(a(b(B(x))))
B1(x) → A1(x)
B1(a(a(x))) → B1(x)
B1(a(A(x))) → B1(B(x))
B1(a(A(x))) → A1(B(x))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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:
B1(a(A(x))) → A1(b(B(x)))
B1(a(a(x))) → A1(b(b(x)))
B1(x) → A1(c(a(x)))
B1(a(A(x))) → A1(a(B(x)))
B1(a(A(x))) → B2(x)
B1(a(a(x))) → A1(a(b(b(x))))
B1(a(a(x))) → B1(b(x))
B1(a(A(x))) → A1(a(b(B(x))))
B1(x) → A1(x)
B1(a(a(x))) → B1(x)
B1(a(A(x))) → B1(B(x))
B1(a(A(x))) → A1(B(x))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(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 9 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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:
B1(a(a(x))) → B1(b(x))
B1(a(a(x))) → B1(x)
B1(a(A(x))) → B1(B(x))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(A(x))) → B1(B(x)) at position [0] we obtained the following new rules:
B1(a(A(x0))) → B1(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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:
B1(a(a(x))) → B1(b(x))
B1(a(A(x0))) → B1(A(x0))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(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
↳ DependencyGraphProof
↳ 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:
B1(a(a(x))) → B1(b(x))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(x))) → B1(b(x)) at position [0] we obtained the following new rules:
B1(a(a(a(A(x0))))) → B1(a(A(x0)))
B1(a(a(a(A(x0))))) → B1(A(x0))
B1(a(a(x0))) → B1(a(c(a(x0))))
B1(a(a(a(A(x0))))) → B1(a(a(B(x0))))
B1(a(a(a(a(x0))))) → B1(a(a(b(b(x0)))))
B1(a(a(a(A(x0))))) → B1(a(a(b(B(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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:
B1(a(a(a(A(x0))))) → B1(A(x0))
B1(a(a(a(a(x0))))) → B1(a(a(b(b(x0)))))
B1(a(a(x0))) → B1(a(c(a(x0))))
B1(a(a(a(A(x0))))) → B1(a(A(x0)))
B1(a(a(a(A(x0))))) → B1(a(a(B(x0))))
B1(a(a(a(A(x0))))) → B1(a(a(b(B(x0)))))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(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
↳ DependencyGraphProof
↳ 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:
B1(a(a(a(a(x0))))) → B1(a(a(b(b(x0)))))
B1(a(a(x0))) → B1(a(c(a(x0))))
B1(a(a(a(A(x0))))) → B1(a(A(x0)))
B1(a(a(a(A(x0))))) → B1(a(a(B(x0))))
B1(a(a(a(A(x0))))) → B1(a(a(b(B(x0)))))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(a(A(x0))))) → B1(a(A(x0))) at position [0] we obtained the following new rules:
B1(a(a(a(A(y0))))) → B1(A(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
B1(a(a(a(A(y0))))) → B1(A(y0))
B1(a(a(a(a(x0))))) → B1(a(a(b(b(x0)))))
B1(a(a(x0))) → B1(a(c(a(x0))))
B1(a(a(a(A(x0))))) → B1(a(a(B(x0))))
B1(a(a(a(A(x0))))) → B1(a(a(b(B(x0)))))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
B1(a(a(a(a(x0))))) → B1(a(a(b(b(x0)))))
B1(a(a(x0))) → B1(a(c(a(x0))))
B1(a(a(a(A(x0))))) → B1(a(a(B(x0))))
B1(a(a(a(A(x0))))) → B1(a(a(b(B(x0)))))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(x0))) → B1(a(c(a(x0)))) at position [0] we obtained the following new rules:
B1(a(a(x0))) → B1(a(c(x0)))
B1(a(a(y0))) → B1(c(a(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
B1(a(a(a(a(x0))))) → B1(a(a(b(b(x0)))))
B1(a(a(x0))) → B1(a(c(x0)))
B1(a(a(y0))) → B1(c(a(y0)))
B1(a(a(a(A(x0))))) → B1(a(a(B(x0))))
B1(a(a(a(A(x0))))) → B1(a(a(b(B(x0)))))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
B1(a(a(a(a(x0))))) → B1(a(a(b(b(x0)))))
B1(a(a(x0))) → B1(a(c(x0)))
B1(a(a(a(A(x0))))) → B1(a(a(B(x0))))
B1(a(a(a(A(x0))))) → B1(a(a(b(B(x0)))))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(x) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(a(a(x0))) → B1(a(c(x0))) at position [0] we obtained the following new rules:
B1(a(a(y0))) → B1(c(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
B1(a(a(a(a(x0))))) → B1(a(a(b(b(x0)))))
B1(a(a(y0))) → B1(c(y0))
B1(a(a(a(A(x0))))) → B1(a(a(B(x0))))
B1(a(a(a(A(x0))))) → B1(a(a(b(B(x0)))))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
B1(a(a(a(a(x0))))) → B1(a(a(b(b(x0)))))
B1(a(a(a(A(x0))))) → B1(a(a(B(x0))))
B1(a(a(a(A(x0))))) → B1(a(a(b(B(x0)))))
B1(a(a(x))) → B1(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(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(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(x) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(a(b(x))) → b(b(a(a(x))))
b(x) → a(c(a(x)))
A(a(b(x))) → B(b(a(a(x))))
A(a(b(x))) → B(a(a(x)))
A(a(b(x))) → A(x)
A(a(b(x))) → A(a(x))
B(x) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(x) → x
a(a(b(x))) → b(b(a(a(x))))
b(x) → a(c(a(x)))
A(a(b(x))) → B(b(a(a(x))))
A(a(b(x))) → B(a(a(x)))
A(a(b(x))) → A(x)
A(a(b(x))) → A(a(x))
B(x) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
b(a(A(x))) → a(a(b(B(x))))
b(a(A(x))) → a(a(B(x)))
b(a(A(x))) → A(x)
b(a(A(x))) → a(A(x))
B(x) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(a(b(x))) → b(b(a(a(x))))
b(x) → a(c(a(x)))
A(a(b(x))) → B(b(a(a(x))))
A(a(b(x))) → B(a(a(x)))
A(a(b(x))) → A(x)
A(a(b(x))) → A(a(x))
B(x) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(x) → x
a(a(b(x))) → b(b(a(a(x))))
b(x) → a(c(a(x)))
A(a(b(x))) → B(b(a(a(x))))
A(a(b(x))) → B(a(a(x)))
A(a(b(x))) → A(x)
A(a(b(x))) → A(a(x))
B(x) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(x1) → a(c(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(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(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(b(x1))) → b(b(a(a(x1))))
b(x1) → a(c(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(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(x) → x
b(a(a(x))) → a(a(b(b(x))))
b(x) → a(c(a(x)))
Q is empty.