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