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(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → b(b(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(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → b(b(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(b(x1)) → C(a(a(c(x1))))
C(c(x1)) → B(x1)
C(c(x1)) → B(b(x1))
A(b(x1)) → A(c(x1))
A(b(x1)) → C(x1)
A(b(x1)) → A(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → b(b(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(b(x1)) → C(a(a(c(x1))))
C(c(x1)) → B(x1)
C(c(x1)) → B(b(x1))
A(b(x1)) → A(c(x1))
A(b(x1)) → C(x1)
A(b(x1)) → A(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → 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 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → A(c(x1))
A(b(x1)) → A(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → b(b(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → A(c(x1)) at position [0] we obtained the following new rules:
A(b(c(x0))) → A(b(b(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(c(x0))) → A(b(b(x0)))
A(b(x1)) → A(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → b(b(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(x1)) → A(a(c(x1))) at position [0] we obtained the following new rules:
A(b(c(x0))) → A(a(b(b(x0))))
A(b(y0)) → A(c(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(c(x0))) → A(b(b(x0)))
A(b(c(x0))) → A(a(b(b(x0))))
A(b(y0)) → A(c(y0))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → b(b(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(y0)) → A(c(y0)) at position [0] we obtained the following new rules:
A(b(c(x0))) → A(b(b(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(c(x0))) → A(b(b(x0)))
A(b(c(x0))) → A(a(b(b(x0))))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → b(b(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(c(x0))) → A(b(b(x0))) at position [0] we obtained the following new rules:
A(b(c(y0))) → A(b(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(c(x0))) → A(a(b(b(x0))))
A(b(c(y0))) → A(b(y0))
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → 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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → b(b(x1))
A(b(c(x0))) → A(a(b(b(x0))))
A(b(c(y0))) → A(b(y0))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → b(b(x1))
A(b(c(x0))) → A(a(b(b(x0))))
A(b(c(y0))) → A(b(y0))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(A(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(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(A(x))) → B(a(A(x)))
B(a(x)) → A1(a(c(x)))
B(a(x)) → C(x)
C(c(x)) → B(x)
B(a(x)) → A1(c(x))
C(b(A(x))) → A1(A(x))
C(c(x)) → B(b(x))
B(a(x)) → C(a(a(c(x))))
C(b(A(x))) → B(b(a(A(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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(A(x))) → B(a(A(x)))
B(a(x)) → A1(a(c(x)))
B(a(x)) → C(x)
C(c(x)) → B(x)
B(a(x)) → A1(c(x))
C(b(A(x))) → A1(A(x))
C(c(x)) → B(b(x))
B(a(x)) → C(a(a(c(x))))
C(b(A(x))) → B(b(a(A(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(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 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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(A(x))) → B(a(A(x)))
B(a(x)) → C(x)
C(c(x)) → B(x)
C(c(x)) → B(b(x))
B(a(x)) → C(a(a(c(x))))
C(b(A(x))) → B(b(a(A(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(x)) → B(b(x)) at position [0] we obtained the following new rules:
C(c(x0)) → B(x0)
C(c(a(x0))) → B(c(a(a(c(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ 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(A(x))) → B(a(A(x)))
B(a(x)) → C(x)
C(c(x)) → B(x)
B(a(x)) → C(a(a(c(x))))
C(c(a(x0))) → B(c(a(a(c(x0)))))
C(b(A(x))) → B(b(a(A(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(x)) → C(a(a(c(x)))) at position [0] we obtained the following new rules:
B(a(b(A(x0)))) → C(a(a(b(A(x0)))))
B(a(c(x0))) → C(a(a(b(b(x0)))))
B(a(b(A(x0)))) → C(a(a(b(b(a(A(x0)))))))
B(a(y0)) → C(a(c(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(c(x0))) → C(a(a(b(b(x0)))))
B(a(b(A(x0)))) → C(a(a(b(A(x0)))))
C(b(A(x))) → B(a(A(x)))
B(a(y0)) → C(a(c(y0)))
B(a(b(A(x0)))) → C(a(a(b(b(a(A(x0)))))))
B(a(x)) → C(x)
C(c(x)) → B(x)
C(b(A(x))) → B(b(a(A(x))))
C(c(a(x0))) → B(c(a(a(c(x0)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(A(x))) → B(b(a(A(x)))) at position [0] we obtained the following new rules:
C(b(A(y0))) → B(a(A(y0)))
C(b(A(y0))) → B(b(A(y0)))
C(b(A(y0))) → B(c(a(a(c(A(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(b(A(x0)))) → C(a(a(b(A(x0)))))
B(a(c(x0))) → C(a(a(b(b(x0)))))
C(b(A(x))) → B(a(A(x)))
B(a(b(A(x0)))) → C(a(a(b(b(a(A(x0)))))))
B(a(y0)) → C(a(c(y0)))
B(a(x)) → C(x)
C(c(x)) → B(x)
C(b(A(y0))) → B(b(A(y0)))
C(b(A(y0))) → B(c(a(a(c(A(y0))))))
C(c(a(x0))) → B(c(a(a(c(x0)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(A(y0))) → B(b(A(y0))) at position [0] we obtained the following new rules:
C(b(A(y0))) → B(A(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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:
B(a(c(x0))) → C(a(a(b(b(x0)))))
B(a(b(A(x0)))) → C(a(a(b(A(x0)))))
C(b(A(x))) → B(a(A(x)))
B(a(y0)) → C(a(c(y0)))
B(a(b(A(x0)))) → C(a(a(b(b(a(A(x0)))))))
B(a(x)) → C(x)
C(c(x)) → B(x)
C(b(A(y0))) → B(c(a(a(c(A(y0))))))
C(b(A(y0))) → B(A(y0))
C(c(a(x0))) → B(c(a(a(c(x0)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(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
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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:
B(a(c(x0))) → C(a(a(b(b(x0)))))
B(a(b(A(x0)))) → C(a(a(b(A(x0)))))
C(b(A(x))) → B(a(A(x)))
B(a(y0)) → C(a(c(y0)))
B(a(b(A(x0)))) → C(a(a(b(b(a(A(x0)))))))
B(a(x)) → C(x)
C(c(x)) → B(x)
C(b(A(y0))) → B(c(a(a(c(A(y0))))))
C(c(a(x0))) → B(c(a(a(c(x0)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(A(y0))) → B(c(a(a(c(A(y0)))))) at position [0] we obtained the following new rules:
C(b(A(y0))) → B(c(a(c(A(y0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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:
B(a(b(A(x0)))) → C(a(a(b(A(x0)))))
B(a(c(x0))) → C(a(a(b(b(x0)))))
C(b(A(x))) → B(a(A(x)))
C(b(A(y0))) → B(c(a(c(A(y0)))))
B(a(b(A(x0)))) → C(a(a(b(b(a(A(x0)))))))
B(a(y0)) → C(a(c(y0)))
B(a(x)) → C(x)
C(c(x)) → B(x)
C(c(a(x0))) → B(c(a(a(c(x0)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(A(y0))) → B(c(a(c(A(y0))))) at position [0] we obtained the following new rules:
C(b(A(y0))) → B(c(c(A(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(c(x0))) → C(a(a(b(b(x0)))))
B(a(b(A(x0)))) → C(a(a(b(A(x0)))))
C(b(A(x))) → B(a(A(x)))
B(a(y0)) → C(a(c(y0)))
B(a(b(A(x0)))) → C(a(a(b(b(a(A(x0)))))))
B(a(x)) → C(x)
C(c(x)) → B(x)
C(b(A(y0))) → B(c(c(A(y0))))
C(c(a(x0))) → B(c(a(a(c(x0)))))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(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(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
A(b(c(x))) → A(a(b(b(x))))
A(b(c(x))) → A(b(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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(x) → x
a(b(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
A(b(c(x))) → A(a(b(b(x))))
A(b(c(x))) → A(b(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
c(b(A(x))) → b(b(a(A(x))))
c(b(A(x))) → b(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
A(b(c(x))) → A(a(b(b(x))))
A(b(c(x))) → A(b(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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(x) → x
a(b(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
A(b(c(x))) → A(a(b(b(x))))
A(b(c(x))) → A(b(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → b(b(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(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(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → c(a(a(c(x1))))
b(x1) → x1
c(c(x1)) → b(b(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(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(x)) → c(a(a(c(x))))
b(x) → x
c(c(x)) → b(b(x))
Q is empty.