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(b(x1)) → x1
a(b(c(x1))) → b(c(b(c(a(a(b(x1)))))))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x1)) → x1
a(b(c(x1))) → b(c(b(c(a(a(b(x1)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → x1
a(b(c(x1))) → b(c(b(c(a(a(b(x1)))))))
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(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:
A(b(c(x1))) → A(a(b(x1)))
A(b(c(x1))) → A(b(x1))
The TRS R consists of the following rules:
a(b(x1)) → x1
a(b(c(x1))) → b(c(b(c(a(a(b(x1)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(c(x1))) → A(a(b(x1)))
A(b(c(x1))) → A(b(x1))
The TRS R consists of the following rules:
a(b(x1)) → x1
a(b(c(x1))) → b(c(b(c(a(a(b(x1)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(b(c(x1))) → A(a(b(x1))) at position [0] we obtained the following new rules:
A(b(c(c(x0)))) → A(b(c(b(c(a(a(b(x0))))))))
A(b(c(x0))) → A(x0)
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(c(c(x0)))) → A(b(c(b(c(a(a(b(x0))))))))
A(b(c(x1))) → A(b(x1))
A(b(c(x0))) → A(x0)
The TRS R consists of the following rules:
a(b(x1)) → x1
a(b(c(x1))) → b(c(b(c(a(a(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x1)) → x1
a(b(c(x1))) → b(c(b(c(a(a(b(x1)))))))
A(b(c(c(x0)))) → A(b(c(b(c(a(a(b(x0))))))))
A(b(c(x1))) → A(b(x1))
A(b(c(x0))) → A(x0)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → x1
a(b(c(x1))) → b(c(b(c(a(a(b(x1)))))))
A(b(c(c(x0)))) → A(b(c(b(c(a(a(b(x0))))))))
A(b(c(x1))) → A(b(x1))
A(b(c(x0))) → A(x0)
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(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(c(b(A(x)))) → C(b(c(b(A(x)))))
C(c(b(A(x)))) → B(a(a(c(b(c(b(A(x))))))))
C(c(b(A(x)))) → B(c(b(A(x))))
C(b(a(x))) → B(x)
C(b(a(x))) → C(b(c(b(x))))
C(b(a(x))) → B(c(b(x)))
C(b(a(x))) → C(b(x))
C(b(a(x))) → B(a(a(c(b(c(b(x)))))))
The TRS R consists of the following rules:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(c(b(A(x)))) → C(b(c(b(A(x)))))
C(c(b(A(x)))) → B(a(a(c(b(c(b(A(x))))))))
C(c(b(A(x)))) → B(c(b(A(x))))
C(b(a(x))) → B(x)
C(b(a(x))) → C(b(c(b(x))))
C(b(a(x))) → B(c(b(x)))
C(b(a(x))) → C(b(x))
C(b(a(x))) → B(a(a(c(b(c(b(x)))))))
The TRS R consists of the following rules:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(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 5 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(c(b(A(x)))) → C(b(c(b(A(x)))))
C(b(a(x))) → C(b(c(b(x))))
C(b(a(x))) → C(b(x))
The TRS R consists of the following rules:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(b(A(x)))) → C(b(c(b(A(x))))) at position [0] we obtained the following new rules:
C(c(b(A(x0)))) → C(b(b(A(x0))))
C(c(b(A(x0)))) → C(b(A(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(c(b(A(x0)))) → C(b(A(x0)))
C(b(a(x))) → C(b(c(b(x))))
C(c(b(A(x0)))) → C(b(b(A(x0))))
C(b(a(x))) → C(b(x))
The TRS R consists of the following rules:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(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 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(a(x))) → C(b(c(b(x))))
C(b(a(x))) → C(b(x))
The TRS R consists of the following rules:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(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(a(x))) → C(b(c(b(x)))) at position [0] we obtained the following new rules:
C(b(a(A(x0)))) → C(b(A(x0)))
C(b(a(a(x0)))) → C(b(b(a(a(c(b(c(b(x0)))))))))
C(b(a(A(x0)))) → C(b(b(A(x0))))
C(b(a(a(x0)))) → C(b(c(x0)))
↳ QTRS
↳ QTRS Reverse
↳ 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
Q DP problem:
The TRS P consists of the following rules:
C(b(a(A(x0)))) → C(b(A(x0)))
C(b(a(a(x0)))) → C(b(b(a(a(c(b(c(b(x0)))))))))
C(b(a(A(x0)))) → C(b(b(A(x0))))
C(b(a(a(x0)))) → C(b(c(x0)))
C(b(a(x))) → C(b(x))
The TRS R consists of the following rules:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(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 2 less nodes.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(a(a(x0)))) → C(b(b(a(a(c(b(c(b(x0)))))))))
C(b(a(a(x0)))) → C(b(c(x0)))
C(b(a(x))) → C(b(x))
The TRS R consists of the following rules:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(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
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → x
a(b(c(x))) → b(c(b(c(a(a(b(x)))))))
A(b(c(c(x)))) → A(b(c(b(c(a(a(b(x))))))))
A(b(c(x))) → A(b(x))
A(b(c(x))) → A(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → x
a(b(c(x))) → b(c(b(c(a(a(b(x)))))))
A(b(c(c(x)))) → A(b(c(b(c(a(a(b(x))))))))
A(b(c(x))) → A(b(x))
A(b(c(x))) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
c(c(b(A(x)))) → b(a(a(c(b(c(b(A(x))))))))
c(b(A(x))) → b(A(x))
c(b(A(x))) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(b(x)) → x
a(b(c(x))) → b(c(b(c(a(a(b(x)))))))
A(b(c(c(x)))) → A(b(c(b(c(a(a(b(x))))))))
A(b(c(x))) → A(b(x))
A(b(c(x))) → A(x)
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → x
a(b(c(x))) → b(c(b(c(a(a(b(x)))))))
A(b(c(c(x)))) → A(b(c(b(c(a(a(b(x))))))))
A(b(c(x))) → A(b(x))
A(b(c(x))) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(b(x1)) → x1
a(b(c(x1))) → b(c(b(c(a(a(b(x1)))))))
The set Q is empty.
We have obtained the following QTRS:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
Q restricted rewrite system:
The TRS R consists of the following rules:
b(a(x)) → x
c(b(a(x))) → b(a(a(c(b(c(b(x)))))))
Q is empty.