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) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → x1
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → 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(b(c(x1))) → B(x1)
B(b(c(x1))) → C(a(b(x1)))
B(b(c(x1))) → A(b(x1))
A(x1) → B(x1)
B(b(c(x1))) → B(c(a(b(x1))))
B(b(c(x1))) → C(b(c(a(b(x1)))))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → 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:
B(b(c(x1))) → B(x1)
B(b(c(x1))) → C(a(b(x1)))
B(b(c(x1))) → A(b(x1))
A(x1) → B(x1)
B(b(c(x1))) → B(c(a(b(x1))))
B(b(c(x1))) → C(b(c(a(b(x1)))))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → 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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(c(x1))) → B(x1)
A(x1) → B(x1)
B(b(c(x1))) → A(b(x1))
B(b(c(x1))) → B(c(a(b(x1))))
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(c(x1))) → B(c(a(b(x1)))) at position [0] we obtained the following new rules:
B(b(c(b(c(x0))))) → B(c(a(c(b(c(a(b(x0))))))))
B(b(c(y0))) → B(c(b(b(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(c(x1))) → B(x1)
B(b(c(y0))) → B(c(b(b(y0))))
B(b(c(b(c(x0))))) → B(c(a(c(b(c(a(b(x0))))))))
B(b(c(x1))) → A(b(x1))
A(x1) → B(x1)
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → 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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → x1
B(b(c(x1))) → B(x1)
B(b(c(y0))) → B(c(b(b(y0))))
B(b(c(b(c(x0))))) → B(c(a(c(b(c(a(b(x0))))))))
B(b(c(x1))) → A(b(x1))
A(x1) → B(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → x1
B(b(c(x1))) → B(x1)
B(b(c(y0))) → B(c(b(b(y0))))
B(b(c(b(c(x0))))) → B(c(a(c(b(c(a(b(x0))))))))
B(b(c(x1))) → A(b(x1))
A(x1) → B(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → 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:
C(b(b(x))) → C(b(c(x)))
C(b(c(b(B(x))))) → A1(c(B(x)))
C(b(c(b(B(x))))) → C(b(c(a(c(B(x))))))
C(b(c(b(B(x))))) → C(B(x))
C(b(b(x))) → C(x)
C(b(B(x))) → A2(x)
C(b(c(b(B(x))))) → C(a(c(B(x))))
C(b(B(x))) → C(B(x))
C(b(c(b(B(x))))) → A1(c(b(c(a(c(B(x)))))))
C(b(b(x))) → A1(c(b(c(x))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(b(x))) → C(b(c(x)))
C(b(c(b(B(x))))) → A1(c(B(x)))
C(b(c(b(B(x))))) → C(b(c(a(c(B(x))))))
C(b(c(b(B(x))))) → C(B(x))
C(b(b(x))) → C(x)
C(b(B(x))) → A2(x)
C(b(c(b(B(x))))) → C(a(c(B(x))))
C(b(B(x))) → C(B(x))
C(b(c(b(B(x))))) → A1(c(b(c(a(c(B(x)))))))
C(b(b(x))) → A1(c(b(c(x))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(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 6 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(b(x))) → C(b(c(x)))
C(b(c(b(B(x))))) → C(b(c(a(c(B(x))))))
C(b(b(x))) → C(x)
C(b(c(b(B(x))))) → C(a(c(B(x))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(c(b(B(x))))) → C(b(c(a(c(B(x)))))) at position [0,0] we obtained the following new rules:
C(b(c(b(B(y0))))) → C(b(c(b(c(B(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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:
C(b(c(b(B(y0))))) → C(b(c(b(c(B(y0))))))
C(b(b(x))) → C(b(c(x)))
C(b(b(x))) → C(x)
C(b(c(b(B(x))))) → C(a(c(B(x))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(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
↳ 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(b(x))) → C(b(c(x)))
C(b(b(x))) → C(x)
C(b(c(b(B(x))))) → C(a(c(B(x))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(c(b(B(x))))) → C(a(c(B(x)))) at position [0] we obtained the following new rules:
C(b(c(b(B(y0))))) → C(b(c(B(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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:
C(b(b(x))) → C(b(c(x)))
C(b(b(x))) → C(x)
C(b(c(b(B(y0))))) → C(b(c(B(y0))))
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ 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:
C(b(b(x))) → C(b(c(x)))
C(b(b(x))) → C(x)
The TRS R consists of the following rules:
a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(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) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(b(c(x))) → c(b(c(a(b(x)))))
c(c(x)) → x
B(b(c(x))) → B(x)
B(b(c(x))) → B(c(b(b(x))))
B(b(c(b(c(x))))) → B(c(a(c(b(c(a(b(x))))))))
B(b(c(x))) → A(b(x))
A(x) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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) → b(x)
b(b(c(x))) → c(b(c(a(b(x)))))
c(c(x)) → x
B(b(c(x))) → B(x)
B(b(c(x))) → B(c(b(b(x))))
B(b(c(b(c(x))))) → B(c(a(c(b(c(a(b(x))))))))
B(b(c(x))) → A(b(x))
A(x) → B(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
c(b(B(x))) → B(x)
c(b(B(x))) → b(b(c(B(x))))
c(b(c(b(B(x))))) → b(a(c(b(c(a(c(B(x))))))))
c(b(B(x))) → b(A(x))
A(x) → B(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(b(c(x))) → c(b(c(a(b(x)))))
c(c(x)) → x
B(b(c(x))) → B(x)
B(b(c(x))) → B(c(b(b(x))))
B(b(c(b(c(x))))) → B(c(a(c(b(c(a(b(x))))))))
B(b(c(x))) → A(b(x))
A(x) → B(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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) → b(x)
b(b(c(x))) → c(b(c(a(b(x)))))
c(c(x)) → x
B(b(c(x))) → B(x)
B(b(c(x))) → B(c(b(b(x))))
B(b(c(b(c(x))))) → B(c(a(c(b(c(a(b(x))))))))
B(b(c(x))) → A(b(x))
A(x) → B(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → 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) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
b(b(c(x1))) → c(b(c(a(b(x1)))))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → 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) → b(x)
c(b(b(x))) → b(a(c(b(c(x)))))
c(c(x)) → x
Q is empty.