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)
a(b(x1)) → x1
c(c(a(x1))) → a(c(a(c(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(x1)
a(b(x1)) → x1
c(c(a(x1))) → a(c(a(c(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:
C(c(a(x1))) → A(c(c(x1)))
C(c(a(x1))) → C(a(c(c(x1))))
C(c(a(x1))) → A(c(a(c(c(x1)))))
C(c(a(x1))) → C(c(x1))
C(c(a(x1))) → C(x1)
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(x1)) → x1
c(c(a(x1))) → a(c(a(c(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:
C(c(a(x1))) → A(c(c(x1)))
C(c(a(x1))) → C(a(c(c(x1))))
C(c(a(x1))) → A(c(a(c(c(x1)))))
C(c(a(x1))) → C(c(x1))
C(c(a(x1))) → C(x1)
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(x1)) → x1
c(c(a(x1))) → a(c(a(c(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(a(x1))) → C(a(c(c(x1))))
C(c(a(x1))) → C(c(x1))
C(c(a(x1))) → C(x1)
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(x1)) → x1
c(c(a(x1))) → a(c(a(c(c(x1)))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(a(x1))) → C(a(c(c(x1)))) at position [0] we obtained the following new rules:
C(c(a(c(a(x0))))) → C(a(c(a(c(a(c(c(x0))))))))
C(c(a(a(x0)))) → C(a(a(c(a(c(c(x0)))))))
C(c(a(y0))) → C(b(c(c(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(c(a(c(a(x0))))) → C(a(c(a(c(a(c(c(x0))))))))
C(c(a(y0))) → C(b(c(c(y0))))
C(c(a(x1))) → C(c(x1))
C(c(a(x1))) → C(x1)
C(c(a(a(x0)))) → C(a(a(c(a(c(c(x0)))))))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(x1)) → x1
c(c(a(x1))) → a(c(a(c(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(c(a(c(a(x0))))) → C(a(c(a(c(a(c(c(x0))))))))
C(c(a(x1))) → C(c(x1))
C(c(a(x1))) → C(x1)
C(c(a(a(x0)))) → C(a(a(c(a(c(c(x0)))))))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(b(x1)) → x1
c(c(a(x1))) → a(c(a(c(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
↳ DependencyGraphProof
↳ 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)
a(b(x1)) → x1
c(c(a(x1))) → a(c(a(c(c(x1)))))
C(c(a(c(a(x0))))) → C(a(c(a(c(a(c(c(x0))))))))
C(c(a(x1))) → C(c(x1))
C(c(a(x1))) → C(x1)
C(c(a(a(x0)))) → C(a(a(c(a(c(c(x0)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(b(x1)) → x1
c(c(a(x1))) → a(c(a(c(c(x1)))))
C(c(a(c(a(x0))))) → C(a(c(a(c(a(c(c(x0))))))))
C(c(a(x1))) → C(c(x1))
C(c(a(x1))) → C(x1)
C(c(a(a(x0)))) → C(a(a(c(a(c(c(x0)))))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(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:
A(c(a(c(C(x))))) → A(C(x))
A(c(c(x))) → A(c(a(x)))
A(a(c(C(x)))) → A(c(a(a(C(x)))))
A(c(a(c(C(x))))) → A(c(a(c(a(C(x))))))
A(x) → B(x)
A(a(c(C(x)))) → A(C(x))
A(a(c(C(x)))) → A(a(C(x)))
A(c(a(c(C(x))))) → A(c(a(C(x))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A(c(a(c(C(x))))) → A(C(x))
A(c(c(x))) → A(c(a(x)))
A(a(c(C(x)))) → A(c(a(a(C(x)))))
A(c(a(c(C(x))))) → A(c(a(c(a(C(x))))))
A(x) → B(x)
A(a(c(C(x)))) → A(C(x))
A(a(c(C(x)))) → A(a(C(x)))
A(c(a(c(C(x))))) → A(c(a(C(x))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(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
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A(c(c(x))) → A(c(a(x)))
A(a(c(C(x)))) → A(c(a(a(C(x)))))
A(c(a(c(C(x))))) → A(c(a(c(a(C(x))))))
A(a(c(C(x)))) → A(a(C(x)))
A(c(a(c(C(x))))) → A(c(a(C(x))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(c(a(c(C(x))))) → A(c(a(C(x)))) at position [0,0] we obtained the following new rules:
A(c(a(c(C(y0))))) → A(c(b(C(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A(c(c(x))) → A(c(a(x)))
A(c(a(c(C(x))))) → A(c(a(c(a(C(x))))))
A(a(c(C(x)))) → A(c(a(a(C(x)))))
A(c(a(c(C(y0))))) → A(c(b(C(y0))))
A(a(c(C(x)))) → A(a(C(x)))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(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
↳ DependencyGraphProof
↳ 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:
A(c(c(x))) → A(c(a(x)))
A(a(c(C(x)))) → A(c(a(a(C(x)))))
A(c(a(c(C(x))))) → A(c(a(c(a(C(x))))))
A(a(c(C(x)))) → A(a(C(x)))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(c(C(x)))) → A(a(C(x))) at position [0] we obtained the following new rules:
A(a(c(C(y0)))) → A(b(C(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A(c(c(x))) → A(c(a(x)))
A(c(a(c(C(x))))) → A(c(a(c(a(C(x))))))
A(a(c(C(x)))) → A(c(a(a(C(x)))))
A(a(c(C(y0)))) → A(b(C(y0)))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(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
↳ DependencyGraphProof
↳ 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:
A(c(c(x))) → A(c(a(x)))
A(a(c(C(x)))) → A(c(a(a(C(x)))))
A(c(a(c(C(x))))) → A(c(a(c(a(C(x))))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(c(a(c(C(x))))) → A(c(a(c(a(C(x)))))) at position [0,0] we obtained the following new rules:
A(c(a(c(C(y0))))) → A(c(a(c(b(C(y0))))))
A(c(a(c(C(y0))))) → A(c(b(c(a(C(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A(c(a(c(C(y0))))) → A(c(b(c(a(C(y0))))))
A(c(c(x))) → A(c(a(x)))
A(c(a(c(C(y0))))) → A(c(a(c(b(C(y0))))))
A(a(c(C(x)))) → A(c(a(a(C(x)))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A(c(c(x))) → A(c(a(x)))
A(c(a(c(C(y0))))) → A(c(a(c(b(C(y0))))))
A(a(c(C(x)))) → A(c(a(a(C(x)))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(c(C(x)))) → A(c(a(a(C(x))))) at position [0,0] we obtained the following new rules:
A(a(c(C(y0)))) → A(c(a(b(C(y0)))))
A(a(c(C(y0)))) → A(c(b(a(C(y0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A(c(c(x))) → A(c(a(x)))
A(c(a(c(C(y0))))) → A(c(a(c(b(C(y0))))))
A(a(c(C(y0)))) → A(c(a(b(C(y0)))))
A(a(c(C(y0)))) → A(c(b(a(C(y0)))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(c(a(c(C(y0))))) → A(c(a(c(b(C(y0)))))) at position [0,0] we obtained the following new rules:
A(c(a(c(C(y0))))) → A(c(b(c(b(C(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A(c(c(x))) → A(c(a(x)))
A(c(a(c(C(y0))))) → A(c(b(c(b(C(y0))))))
A(a(c(C(y0)))) → A(c(a(b(C(y0)))))
A(a(c(C(y0)))) → A(c(b(a(C(y0)))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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:
A(c(c(x))) → A(c(a(x)))
A(a(c(C(y0)))) → A(c(a(b(C(y0)))))
A(a(c(C(y0)))) → A(c(b(a(C(y0)))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(c(c(x))) → A(c(a(x))) at position [0,0] we obtained the following new rules:
A(c(c(c(c(x0))))) → A(c(c(c(a(c(a(x0)))))))
A(c(c(a(c(C(x0)))))) → A(c(c(c(a(c(a(a(C(x0)))))))))
A(c(c(x0))) → A(c(b(x0)))
A(c(c(c(C(x0))))) → A(c(C(x0)))
A(c(c(c(C(x0))))) → A(c(c(C(x0))))
A(c(c(c(a(c(C(x0))))))) → A(c(c(c(a(c(a(c(a(C(x0))))))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A(c(c(a(c(C(x0)))))) → A(c(c(c(a(c(a(a(C(x0)))))))))
A(c(c(c(C(x0))))) → A(c(C(x0)))
A(c(c(c(c(x0))))) → A(c(c(c(a(c(a(x0)))))))
A(c(c(x0))) → A(c(b(x0)))
A(c(c(c(C(x0))))) → A(c(c(C(x0))))
A(a(c(C(y0)))) → A(c(a(b(C(y0)))))
A(c(c(c(a(c(C(x0))))))) → A(c(c(c(a(c(a(c(a(C(x0))))))))))
A(a(c(C(y0)))) → A(c(b(a(C(y0)))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A(c(c(a(c(C(x0)))))) → A(c(c(c(a(c(a(a(C(x0)))))))))
A(c(c(c(c(x0))))) → A(c(c(c(a(c(a(x0)))))))
A(c(c(x0))) → A(c(b(x0)))
A(c(c(c(C(x0))))) → A(c(c(C(x0))))
A(a(c(C(y0)))) → A(c(a(b(C(y0)))))
A(c(c(c(a(c(C(x0))))))) → A(c(c(c(a(c(a(c(a(C(x0))))))))))
A(a(c(C(y0)))) → A(c(b(a(C(y0)))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(c(C(y0)))) → A(c(a(b(C(y0))))) at position [0,0] we obtained the following new rules:
A(a(c(C(y0)))) → A(c(b(b(C(y0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A(c(c(a(c(C(x0)))))) → A(c(c(c(a(c(a(a(C(x0)))))))))
A(c(c(c(c(x0))))) → A(c(c(c(a(c(a(x0)))))))
A(c(c(x0))) → A(c(b(x0)))
A(c(c(c(C(x0))))) → A(c(c(C(x0))))
A(c(c(c(a(c(C(x0))))))) → A(c(c(c(a(c(a(c(a(C(x0))))))))))
A(a(c(C(y0)))) → A(c(b(a(C(y0)))))
A(a(c(C(y0)))) → A(c(b(b(C(y0)))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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:
A(c(c(a(c(C(x0)))))) → A(c(c(c(a(c(a(a(C(x0)))))))))
A(c(c(c(c(x0))))) → A(c(c(c(a(c(a(x0)))))))
A(c(c(x0))) → A(c(b(x0)))
A(c(c(c(C(x0))))) → A(c(c(C(x0))))
A(c(c(c(a(c(C(x0))))))) → A(c(c(c(a(c(a(c(a(C(x0))))))))))
A(a(c(C(y0)))) → A(c(b(a(C(y0)))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(c(C(y0)))) → A(c(b(a(C(y0))))) at position [0,0] we obtained the following new rules:
A(a(c(C(y0)))) → A(c(C(y0)))
A(a(c(C(y0)))) → A(c(b(b(C(y0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ 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:
A(c(c(a(c(C(x0)))))) → A(c(c(c(a(c(a(a(C(x0)))))))))
A(c(c(c(c(x0))))) → A(c(c(c(a(c(a(x0)))))))
A(c(c(x0))) → A(c(b(x0)))
A(c(c(c(C(x0))))) → A(c(c(C(x0))))
A(c(c(c(a(c(C(x0))))))) → A(c(c(c(a(c(a(c(a(C(x0))))))))))
A(a(c(C(y0)))) → A(c(C(y0)))
A(a(c(C(y0)))) → A(c(b(b(C(y0)))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(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
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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:
A(c(c(a(c(C(x0)))))) → A(c(c(c(a(c(a(a(C(x0)))))))))
A(c(c(c(c(x0))))) → A(c(c(c(a(c(a(x0)))))))
A(c(c(x0))) → A(c(b(x0)))
A(c(c(c(C(x0))))) → A(c(c(C(x0))))
A(c(c(c(a(c(C(x0))))))) → A(c(c(c(a(c(a(c(a(C(x0))))))))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(c(c(x0))) → A(c(b(x0))) at position [0,0] we obtained the following new rules:
A(c(c(a(x0)))) → A(c(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(c(a(x0)))) → A(c(x0))
A(c(c(a(c(C(x0)))))) → A(c(c(c(a(c(a(a(C(x0)))))))))
A(c(c(c(c(x0))))) → A(c(c(c(a(c(a(x0)))))))
A(c(c(c(C(x0))))) → A(c(c(C(x0))))
A(c(c(c(a(c(C(x0))))))) → A(c(c(c(a(c(a(c(a(C(x0))))))))))
A(c(c(x))) → A(x)
The TRS R consists of the following rules:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(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) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(b(x)) → x
c(c(a(x))) → a(c(a(c(c(x)))))
C(c(a(c(a(x))))) → C(a(c(a(c(a(c(c(x))))))))
C(c(a(x))) → C(c(x))
C(c(a(x))) → C(x)
C(c(a(a(x)))) → C(a(a(c(a(c(c(x)))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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)
a(b(x)) → x
c(c(a(x))) → a(c(a(c(c(x)))))
C(c(a(c(a(x))))) → C(a(c(a(c(a(c(c(x))))))))
C(c(a(x))) → C(c(x))
C(c(a(x))) → C(x)
C(c(a(a(x)))) → C(a(a(c(a(c(c(x)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
a(c(a(c(C(x))))) → c(c(a(c(a(c(a(C(x))))))))
a(c(C(x))) → c(C(x))
a(c(C(x))) → C(x)
a(a(c(C(x)))) → c(c(a(c(a(a(C(x)))))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(b(x)) → x
c(c(a(x))) → a(c(a(c(c(x)))))
C(c(a(c(a(x))))) → C(a(c(a(c(a(c(c(x))))))))
C(c(a(x))) → C(c(x))
C(c(a(x))) → C(x)
C(c(a(a(x)))) → C(a(a(c(a(c(c(x)))))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ 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)
a(b(x)) → x
c(c(a(x))) → a(c(a(c(c(x)))))
C(c(a(c(a(x))))) → C(a(c(a(c(a(c(c(x))))))))
C(c(a(x))) → C(c(x))
C(c(a(x))) → C(x)
C(c(a(a(x)))) → C(a(a(c(a(c(c(x)))))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(b(x1)) → x1
c(c(a(x1))) → a(c(a(c(c(x1)))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(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) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(b(x1)) → x1
c(c(a(x1))) → a(c(a(c(c(x1)))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(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) → b(x)
b(a(x)) → x
a(c(c(x))) → c(c(a(c(a(x)))))
Q is empty.