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(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → b(c(a(c(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(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → b(c(a(c(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(x1) → B(c(x1))
C(c(b(x1))) → A(c(x1))
C(c(b(x1))) → B(c(a(c(x1))))
A(x1) → C(x1)
C(c(b(x1))) → C(x1)
C(c(b(x1))) → C(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → b(c(a(c(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(x1) → B(c(x1))
C(c(b(x1))) → A(c(x1))
C(c(b(x1))) → B(c(a(c(x1))))
A(x1) → C(x1)
C(c(b(x1))) → C(x1)
C(c(b(x1))) → C(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → b(c(a(c(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:
C(c(b(x1))) → A(c(x1))
A(x1) → C(x1)
C(c(b(x1))) → C(x1)
C(c(b(x1))) → C(a(c(x1)))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → b(c(a(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(b(x1))) → C(a(c(x1))) at position [0] we obtained the following new rules:
C(c(b(y0))) → C(b(c(c(y0))))
C(c(b(c(b(x0))))) → C(a(b(c(a(c(x0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(c(b(y0))) → C(b(c(c(y0))))
C(c(b(c(b(x0))))) → C(a(b(c(a(c(x0))))))
C(c(b(x1))) → A(c(x1))
A(x1) → C(x1)
C(c(b(x1))) → C(x1)
The TRS R consists of the following rules:
a(x1) → b(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → b(c(a(c(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(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → b(c(a(c(x1))))
C(c(b(y0))) → C(b(c(c(y0))))
C(c(b(c(b(x0))))) → C(a(b(c(a(c(x0))))))
C(c(b(x1))) → A(c(x1))
A(x1) → C(x1)
C(c(b(x1))) → C(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → b(c(a(c(x1))))
C(c(b(y0))) → C(b(c(c(y0))))
C(c(b(c(b(x0))))) → C(a(b(c(a(c(x0))))))
C(c(b(x1))) → A(c(x1))
A(x1) → C(x1)
C(c(b(x1))) → C(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
b(c(C(x))) → c(c(b(C(x))))
b(c(b(c(C(x))))) → c(a(c(b(a(C(x))))))
b(c(C(x))) → c(A(x))
A(x) → C(x)
b(c(C(x))) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
b(c(C(x))) → c(c(b(C(x))))
b(c(b(c(C(x))))) → c(a(c(b(a(C(x))))))
b(c(C(x))) → c(A(x))
A(x) → C(x)
b(c(C(x))) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
b(c(C(x))) → c(c(b(C(x))))
b(c(b(c(C(x))))) → c(a(c(b(a(C(x))))))
b(c(C(x))) → c(A(x))
A(x) → C(x)
b(c(C(x))) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(c(x))
b(b(x)) → x
c(c(b(x))) → b(c(a(c(x))))
C(c(b(x))) → C(b(c(c(x))))
C(c(b(c(b(x))))) → C(a(b(c(a(c(x))))))
C(c(b(x))) → A(c(x))
A(x) → C(x)
C(c(b(x))) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(c(x))
b(b(x)) → x
c(c(b(x))) → b(c(a(c(x))))
C(c(b(x))) → C(b(c(c(x))))
C(c(b(c(b(x))))) → C(a(b(c(a(c(x))))))
C(c(b(x))) → A(c(x))
A(x) → C(x)
C(c(b(x))) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
b(c(C(x))) → c(c(b(C(x))))
b(c(b(c(C(x))))) → c(a(c(b(a(C(x))))))
b(c(C(x))) → c(A(x))
A(x) → C(x)
b(c(C(x))) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(c(x))
b(b(x)) → x
c(c(b(x))) → b(c(a(c(x))))
C(c(b(x))) → C(b(c(c(x))))
C(c(b(c(b(x))))) → C(a(b(c(a(c(x))))))
C(c(b(x))) → A(c(x))
A(x) → C(x)
C(c(b(x))) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(c(x))
b(b(x)) → x
c(c(b(x))) → b(c(a(c(x))))
C(c(b(x))) → C(b(c(c(x))))
C(c(b(c(b(x))))) → C(a(b(c(a(c(x))))))
C(c(b(x))) → A(c(x))
A(x) → C(x)
C(c(b(x))) → 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:
B(c(C(x))) → B(C(x))
B(c(b(c(C(x))))) → B(a(C(x)))
A1(x) → B(x)
B(c(c(x))) → B(x)
B(c(b(c(C(x))))) → A1(c(b(a(C(x)))))
B(c(C(x))) → A2(x)
B(c(c(x))) → A1(c(b(x)))
B(c(b(c(C(x))))) → A1(C(x))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
b(c(C(x))) → c(c(b(C(x))))
b(c(b(c(C(x))))) → c(a(c(b(a(C(x))))))
b(c(C(x))) → c(A(x))
A(x) → C(x)
b(c(C(x))) → C(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
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(C(x))) → B(C(x))
B(c(b(c(C(x))))) → B(a(C(x)))
A1(x) → B(x)
B(c(c(x))) → B(x)
B(c(b(c(C(x))))) → A1(c(b(a(C(x)))))
B(c(C(x))) → A2(x)
B(c(c(x))) → A1(c(b(x)))
B(c(b(c(C(x))))) → A1(C(x))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
b(c(C(x))) → c(c(b(C(x))))
b(c(b(c(C(x))))) → c(a(c(b(a(C(x))))))
b(c(C(x))) → c(A(x))
A(x) → C(x)
b(c(C(x))) → 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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(b(c(C(x))))) → B(a(C(x)))
A1(x) → B(x)
B(c(c(x))) → B(x)
B(c(b(c(C(x))))) → A1(c(b(a(C(x)))))
B(c(c(x))) → A1(c(b(x)))
B(c(b(c(C(x))))) → A1(C(x))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
b(c(C(x))) → c(c(b(C(x))))
b(c(b(c(C(x))))) → c(a(c(b(a(C(x))))))
b(c(C(x))) → c(A(x))
A(x) → C(x)
b(c(C(x))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(b(c(C(x))))) → B(a(C(x))) at position [0] we obtained the following new rules:
B(c(b(c(C(y0))))) → B(c(b(C(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B(x)
B(c(b(c(C(y0))))) → B(c(b(C(y0))))
B(c(c(x))) → B(x)
B(c(b(c(C(x))))) → A1(c(b(a(C(x)))))
B(c(c(x))) → A1(c(b(x)))
B(c(b(c(C(x))))) → A1(C(x))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
b(c(C(x))) → c(c(b(C(x))))
b(c(b(c(C(x))))) → c(a(c(b(a(C(x))))))
b(c(C(x))) → c(A(x))
A(x) → C(x)
b(c(C(x))) → 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
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → B(x)
B(c(c(x))) → B(x)
B(c(b(c(C(x))))) → A1(c(b(a(C(x)))))
B(c(c(x))) → A1(c(b(x)))
B(c(b(c(C(x))))) → A1(C(x))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
b(c(C(x))) → c(c(b(C(x))))
b(c(b(c(C(x))))) → c(a(c(b(a(C(x))))))
b(c(C(x))) → c(A(x))
A(x) → C(x)
b(c(C(x))) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule A1(x) → B(x) we obtained the following new rules:
A1(C(y_0)) → B(C(y_0))
A1(c(y_1)) → B(c(y_1))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(c(x))) → B(x)
B(c(b(c(C(x))))) → A1(c(b(a(C(x)))))
A1(C(y_0)) → B(C(y_0))
B(c(c(x))) → A1(c(b(x)))
B(c(b(c(C(x))))) → A1(C(x))
A1(c(y_1)) → B(c(y_1))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
b(c(C(x))) → c(c(b(C(x))))
b(c(b(c(C(x))))) → c(a(c(b(a(C(x))))))
b(c(C(x))) → c(A(x))
A(x) → C(x)
b(c(C(x))) → 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 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Instantiation
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(c(c(x))) → B(x)
B(c(b(c(C(x))))) → A1(c(b(a(C(x)))))
B(c(c(x))) → A1(c(b(x)))
A1(c(y_1)) → B(c(y_1))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
b(c(C(x))) → c(c(b(C(x))))
b(c(b(c(C(x))))) → c(a(c(b(a(C(x))))))
b(c(C(x))) → c(A(x))
A(x) → C(x)
b(c(C(x))) → 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(x1) → b(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → b(c(a(c(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(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) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(c(x1))
b(b(x1)) → x1
c(c(b(x1))) → b(c(a(c(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(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) → c(b(x))
b(b(x)) → x
b(c(c(x))) → c(a(c(b(x))))
Q is empty.