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(a(c(x1))) → c(a(c(a(a(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)
a(a(c(x1))) → c(a(c(a(a(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:
A(a(c(x1))) → C(a(a(x1)))
A(a(c(x1))) → A(c(a(a(x1))))
A(a(c(x1))) → A(x1)
A(a(c(x1))) → C(a(c(a(a(x1)))))
A(a(c(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(a(c(x1))) → c(a(c(a(a(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:
A(a(c(x1))) → C(a(a(x1)))
A(a(c(x1))) → A(c(a(a(x1))))
A(a(c(x1))) → A(x1)
A(a(c(x1))) → C(a(c(a(a(x1)))))
A(a(c(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(a(c(x1))) → c(a(c(a(a(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:
A(a(c(x1))) → A(c(a(a(x1))))
A(a(c(x1))) → A(x1)
A(a(c(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(a(c(x1))) → c(a(c(a(a(x1)))))
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(c(x1))) → A(c(a(a(x1)))) at position [0] we obtained the following new rules:
A(a(c(a(c(x0))))) → A(c(a(c(a(c(a(a(x0))))))))
A(a(c(c(x0)))) → A(c(c(a(c(a(a(x0)))))))
A(a(c(x0))) → A(c(a(b(x0))))
A(a(c(y0))) → A(c(b(a(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(c(a(c(x0))))) → A(c(a(c(a(c(a(a(x0))))))))
A(a(c(c(x0)))) → A(c(c(a(c(a(a(x0)))))))
A(a(c(x0))) → A(c(a(b(x0))))
A(a(c(y0))) → A(c(b(a(y0))))
A(a(c(x1))) → A(x1)
A(a(c(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(a(c(x1))) → c(a(c(a(a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(c(a(c(x0))))) → A(c(a(c(a(c(a(a(x0))))))))
A(a(c(c(x0)))) → A(c(c(a(c(a(a(x0)))))))
A(a(c(x0))) → A(c(a(b(x0))))
A(a(c(x1))) → A(x1)
A(a(c(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(a(c(x1))) → c(a(c(a(a(x1)))))
c(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(a(c(x0))) → A(c(a(b(x0)))) at position [0] we obtained the following new rules:
A(a(c(y0))) → A(c(b(b(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(c(a(c(x0))))) → A(c(a(c(a(c(a(a(x0))))))))
A(a(c(y0))) → A(c(b(b(y0))))
A(a(c(c(x0)))) → A(c(c(a(c(a(a(x0)))))))
A(a(c(x1))) → A(x1)
A(a(c(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(a(c(x1))) → c(a(c(a(a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(c(a(c(x0))))) → A(c(a(c(a(c(a(a(x0))))))))
A(a(c(c(x0)))) → A(c(c(a(c(a(a(x0)))))))
A(a(c(x1))) → A(x1)
A(a(c(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(x1)
a(a(c(x1))) → c(a(c(a(a(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
↳ 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(a(c(x1))) → c(a(c(a(a(x1)))))
c(c(x1)) → x1
A(a(c(a(c(x0))))) → A(c(a(c(a(c(a(a(x0))))))))
A(a(c(c(x0)))) → A(c(c(a(c(a(a(x0)))))))
A(a(c(x1))) → A(x1)
A(a(c(x1))) → A(a(x1))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(a(c(x1))) → c(a(c(a(a(x1)))))
c(c(x1)) → x1
A(a(c(a(c(x0))))) → A(c(a(c(a(c(a(a(x0))))))))
A(a(c(c(x0)))) → A(c(c(a(c(a(a(x0)))))))
A(a(c(x1))) → A(x1)
A(a(c(x1))) → A(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(a(c(x))) → c(a(c(a(a(x)))))
c(c(x)) → x
A(a(c(a(c(x))))) → A(c(a(c(a(c(a(a(x))))))))
A(a(c(c(x)))) → A(c(c(a(c(a(a(x)))))))
A(a(c(x))) → A(x)
A(a(c(x))) → A(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
a(a(c(x))) → c(a(c(a(a(x)))))
c(c(x)) → x
A(a(c(a(c(x))))) → A(c(a(c(a(c(a(a(x))))))))
A(a(c(c(x)))) → A(c(c(a(c(a(a(x)))))))
A(a(c(x))) → A(x)
A(a(c(x))) → A(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(a(c(a(A(x))))) → A1(c(a(c(A(x)))))
C(a(c(a(A(x))))) → A1(a(c(a(c(a(c(A(x))))))))
C(c(a(A(x)))) → A1(c(a(c(c(A(x))))))
C(a(c(a(A(x))))) → A1(c(a(c(a(c(A(x)))))))
C(c(a(A(x)))) → A1(a(c(a(c(c(A(x)))))))
C(a(c(a(A(x))))) → C(a(c(a(c(A(x))))))
C(c(a(A(x)))) → C(a(c(c(A(x)))))
C(c(a(A(x)))) → A1(c(c(A(x))))
C(a(a(x))) → A1(a(c(a(c(x)))))
C(a(a(x))) → A1(c(a(c(x))))
C(a(a(x))) → C(a(c(x)))
C(c(a(A(x)))) → C(A(x))
C(a(a(x))) → C(x)
C(a(c(a(A(x))))) → C(a(c(A(x))))
C(a(c(a(A(x))))) → C(A(x))
C(c(a(A(x)))) → C(c(A(x)))
C(a(c(a(A(x))))) → A1(c(A(x)))
C(a(a(x))) → A1(c(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(c(a(A(x))))) → A1(c(a(c(A(x)))))
C(a(c(a(A(x))))) → A1(a(c(a(c(a(c(A(x))))))))
C(c(a(A(x)))) → A1(c(a(c(c(A(x))))))
C(a(c(a(A(x))))) → A1(c(a(c(a(c(A(x)))))))
C(c(a(A(x)))) → A1(a(c(a(c(c(A(x)))))))
C(a(c(a(A(x))))) → C(a(c(a(c(A(x))))))
C(c(a(A(x)))) → C(a(c(c(A(x)))))
C(c(a(A(x)))) → A1(c(c(A(x))))
C(a(a(x))) → A1(a(c(a(c(x)))))
C(a(a(x))) → A1(c(a(c(x))))
C(a(a(x))) → C(a(c(x)))
C(c(a(A(x)))) → C(A(x))
C(a(a(x))) → C(x)
C(a(c(a(A(x))))) → C(a(c(A(x))))
C(a(c(a(A(x))))) → C(A(x))
C(c(a(A(x)))) → C(c(A(x)))
C(a(c(a(A(x))))) → A1(c(A(x)))
C(a(a(x))) → A1(c(x))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(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 13 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(a(x))) → C(a(c(x)))
C(a(c(a(A(x))))) → C(a(c(a(c(A(x))))))
C(a(a(x))) → C(x)
C(a(c(a(A(x))))) → C(a(c(A(x))))
C(c(a(A(x)))) → C(a(c(c(A(x)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(c(a(A(x))))) → C(a(c(a(c(A(x)))))) at position [0] we obtained the following new rules:
C(a(c(a(A(y0))))) → C(a(c(b(c(A(y0))))))
C(a(c(a(A(y0))))) → C(b(c(a(c(A(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ 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(a(c(a(A(y0))))) → C(b(c(a(c(A(y0))))))
C(a(a(x))) → C(a(c(x)))
C(a(c(a(A(y0))))) → C(a(c(b(c(A(y0))))))
C(a(a(x))) → C(x)
C(a(c(a(A(x))))) → C(a(c(A(x))))
C(c(a(A(x)))) → C(a(c(c(A(x)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ 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(a(a(x))) → C(a(c(x)))
C(a(c(a(A(y0))))) → C(a(c(b(c(A(y0))))))
C(a(a(x))) → C(x)
C(a(c(a(A(x))))) → C(a(c(A(x))))
C(c(a(A(x)))) → C(a(c(c(A(x)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(c(a(A(x))))) → C(a(c(A(x)))) at position [0] we obtained the following new rules:
C(a(c(a(A(y0))))) → C(b(c(A(y0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ 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(a(a(x))) → C(a(c(x)))
C(a(c(a(A(y0))))) → C(a(c(b(c(A(y0))))))
C(a(a(x))) → C(x)
C(a(c(a(A(y0))))) → C(b(c(A(y0))))
C(c(a(A(x)))) → C(a(c(c(A(x)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(a(x))) → C(a(c(x)))
C(a(c(a(A(y0))))) → C(a(c(b(c(A(y0))))))
C(a(a(x))) → C(x)
C(c(a(A(x)))) → C(a(c(c(A(x)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(c(a(A(y0))))) → C(a(c(b(c(A(y0)))))) at position [0] we obtained the following new rules:
C(a(c(a(A(y0))))) → C(b(c(b(c(A(y0))))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(a(x))) → C(a(c(x)))
C(a(a(x))) → C(x)
C(a(c(a(A(y0))))) → C(b(c(b(c(A(y0))))))
C(c(a(A(x)))) → C(a(c(c(A(x)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(a(x))) → C(a(c(x)))
C(a(a(x))) → C(x)
C(c(a(A(x)))) → C(a(c(c(A(x)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(a(x))) → C(a(c(x))) at position [0] we obtained the following new rules:
C(a(a(y0))) → C(b(c(y0)))
C(a(a(c(x0)))) → C(a(x0))
C(a(a(a(a(x0))))) → C(a(a(a(c(a(c(x0)))))))
C(a(a(a(A(x0))))) → C(a(a(A(x0))))
C(a(a(c(a(A(x0)))))) → C(a(a(a(c(a(c(c(A(x0)))))))))
C(a(a(a(c(a(A(x0))))))) → C(a(a(a(c(a(c(a(c(A(x0))))))))))
C(a(a(a(A(x0))))) → C(a(A(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(a(c(x0)))) → C(a(x0))
C(a(a(y0))) → C(b(c(y0)))
C(a(a(a(A(x0))))) → C(a(a(A(x0))))
C(a(a(a(a(x0))))) → C(a(a(a(c(a(c(x0)))))))
C(a(a(c(a(A(x0)))))) → C(a(a(a(c(a(c(c(A(x0)))))))))
C(a(a(a(c(a(A(x0))))))) → C(a(a(a(c(a(c(a(c(A(x0))))))))))
C(a(a(x))) → C(x)
C(a(a(a(A(x0))))) → C(a(A(x0)))
C(c(a(A(x)))) → C(a(c(c(A(x)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(a(c(x0)))) → C(a(x0))
C(a(a(a(A(x0))))) → C(a(a(A(x0))))
C(a(a(a(a(x0))))) → C(a(a(a(c(a(c(x0)))))))
C(a(a(c(a(A(x0)))))) → C(a(a(a(c(a(c(c(A(x0)))))))))
C(a(a(a(c(a(A(x0))))))) → C(a(a(a(c(a(c(a(c(A(x0))))))))))
C(a(a(x))) → C(x)
C(a(a(a(A(x0))))) → C(a(A(x0)))
C(c(a(A(x)))) → C(a(c(c(A(x)))))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(a(A(x)))) → C(a(c(c(A(x))))) at position [0] we obtained the following new rules:
C(c(a(A(y0)))) → C(b(c(c(A(y0)))))
C(c(a(A(y0)))) → C(a(A(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(a(c(x0)))) → C(a(x0))
C(a(a(a(a(x0))))) → C(a(a(a(c(a(c(x0)))))))
C(a(a(a(A(x0))))) → C(a(a(A(x0))))
C(a(a(c(a(A(x0)))))) → C(a(a(a(c(a(c(c(A(x0)))))))))
C(c(a(A(y0)))) → C(b(c(c(A(y0)))))
C(a(a(x))) → C(x)
C(a(a(a(c(a(A(x0))))))) → C(a(a(a(c(a(c(a(c(A(x0))))))))))
C(c(a(A(y0)))) → C(a(A(y0)))
C(a(a(a(A(x0))))) → C(a(A(x0)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(a(c(x0)))) → C(a(x0))
C(a(a(a(A(x0))))) → C(a(a(A(x0))))
C(a(a(a(a(x0))))) → C(a(a(a(c(a(c(x0)))))))
C(a(a(c(a(A(x0)))))) → C(a(a(a(c(a(c(c(A(x0)))))))))
C(a(a(a(c(a(A(x0))))))) → C(a(a(a(c(a(c(a(c(A(x0))))))))))
C(a(a(x))) → C(x)
C(c(a(A(y0)))) → C(a(A(y0)))
C(a(a(a(A(x0))))) → C(a(A(x0)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(a(a(A(x0))))) → C(a(A(x0))) at position [0] we obtained the following new rules:
C(a(a(a(A(y0))))) → C(b(A(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(a(c(x0)))) → C(a(x0))
C(a(a(a(a(x0))))) → C(a(a(a(c(a(c(x0)))))))
C(a(a(a(A(x0))))) → C(a(a(A(x0))))
C(a(a(c(a(A(x0)))))) → C(a(a(a(c(a(c(c(A(x0)))))))))
C(a(a(x))) → C(x)
C(a(a(a(c(a(A(x0))))))) → C(a(a(a(c(a(c(a(c(A(x0))))))))))
C(a(a(a(A(y0))))) → C(b(A(y0)))
C(c(a(A(y0)))) → C(a(A(y0)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(a(c(x0)))) → C(a(x0))
C(a(a(a(A(x0))))) → C(a(a(A(x0))))
C(a(a(a(a(x0))))) → C(a(a(a(c(a(c(x0)))))))
C(a(a(c(a(A(x0)))))) → C(a(a(a(c(a(c(c(A(x0)))))))))
C(a(a(a(c(a(A(x0))))))) → C(a(a(a(c(a(c(a(c(A(x0))))))))))
C(a(a(x))) → C(x)
C(c(a(A(y0)))) → C(a(A(y0)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(c(a(A(y0)))) → C(a(A(y0))) at position [0] we obtained the following new rules:
C(c(a(A(y0)))) → C(b(A(y0)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(a(c(x0)))) → C(a(x0))
C(a(a(a(a(x0))))) → C(a(a(a(c(a(c(x0)))))))
C(a(a(a(A(x0))))) → C(a(a(A(x0))))
C(a(a(c(a(A(x0)))))) → C(a(a(a(c(a(c(c(A(x0)))))))))
C(a(a(x))) → C(x)
C(a(a(a(c(a(A(x0))))))) → C(a(a(a(c(a(c(a(c(A(x0))))))))))
C(c(a(A(y0)))) → C(b(A(y0)))
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ 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(a(a(c(x0)))) → C(a(x0))
C(a(a(a(A(x0))))) → C(a(a(A(x0))))
C(a(a(a(a(x0))))) → C(a(a(a(c(a(c(x0)))))))
C(a(a(c(a(A(x0)))))) → C(a(a(a(c(a(c(c(A(x0)))))))))
C(a(a(a(c(a(A(x0))))))) → C(a(a(a(c(a(c(a(c(A(x0))))))))))
C(a(a(x))) → C(x)
The TRS R consists of the following rules:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(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
a(x) → b(x)
c(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
c(a(c(a(A(x))))) → a(a(c(a(c(a(c(A(x))))))))
c(c(a(A(x)))) → a(a(c(a(c(c(A(x)))))))
c(a(A(x))) → A(x)
c(a(A(x))) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
a(a(c(x))) → c(a(c(a(a(x)))))
c(c(x)) → x
A(a(c(a(c(x))))) → A(c(a(c(a(c(a(a(x))))))))
A(a(c(c(x)))) → A(c(c(a(c(a(a(x)))))))
A(a(c(x))) → A(x)
A(a(c(x))) → A(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → b(x)
a(a(c(x))) → c(a(c(a(a(x)))))
c(c(x)) → x
A(a(c(a(c(x))))) → A(c(a(c(a(c(a(a(x))))))))
A(a(c(c(x)))) → A(c(c(a(c(a(a(x)))))))
A(a(c(x))) → A(x)
A(a(c(x))) → A(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(x1)
a(a(c(x1))) → c(a(c(a(a(x1)))))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(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(a(a(x))) → a(a(c(a(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)
a(a(c(x1))) → c(a(c(a(a(x1)))))
c(c(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → b(x)
c(a(a(x))) → a(a(c(a(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(a(a(x))) → a(a(c(a(c(x)))))
c(c(x)) → x
Q is empty.