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(a(x1)) → b(b(b(c(x1))))
b(c(b(x1))) → a(b(c(x1)))
Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → b(b(b(c(x1))))
b(c(b(x1))) → a(b(c(x1)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → b(b(b(c(x1))))
b(c(b(x1))) → a(b(c(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(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:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(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:
A(a(x1)) → B(c(x1))
A(a(x1)) → B(b(c(x1)))
B(c(b(x1))) → B(c(x1))
B(c(b(x1))) → A(b(c(x1)))
A(a(x1)) → B(b(b(c(x1))))
The TRS R consists of the following rules:
a(a(x1)) → b(b(b(c(x1))))
b(c(b(x1))) → a(b(c(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(x1)) → B(c(x1))
A(a(x1)) → B(b(c(x1)))
B(c(b(x1))) → B(c(x1))
B(c(b(x1))) → A(b(c(x1)))
A(a(x1)) → B(b(b(c(x1))))
The TRS R consists of the following rules:
a(a(x1)) → b(b(b(c(x1))))
b(c(b(x1))) → a(b(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(x1)) → B(c(x1))
B(c(b(x1))) → B(c(x1))
B(c(b(x1))) → A(b(c(x1)))
The TRS R consists of the following rules:
a(a(x1)) → b(b(b(c(x1))))
b(c(b(x1))) → a(b(c(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(b(x1))) → A(b(c(x1))) at position [0] we obtained the following new rules:
B(c(b(b(x0)))) → A(a(b(c(x0))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(a(x1)) → B(c(x1))
B(c(b(x1))) → B(c(x1))
B(c(b(b(x0)))) → A(a(b(c(x0))))
The TRS R consists of the following rules:
a(a(x1)) → b(b(b(c(x1))))
b(c(b(x1))) → a(b(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x1)) → b(b(b(c(x1))))
b(c(b(x1))) → a(b(c(x1)))
A(a(x1)) → B(c(x1))
B(c(b(x1))) → B(c(x1))
B(c(b(b(x0)))) → A(a(b(c(x0))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x1)) → b(b(b(c(x1))))
b(c(b(x1))) → a(b(c(x1)))
A(a(x1)) → B(c(x1))
B(c(b(x1))) → B(c(x1))
B(c(b(b(x0)))) → A(a(b(c(x0))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(a(A(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(a(A(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(a(A(x))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(b(b(c(x))))
b(c(b(x))) → a(b(c(x)))
A(a(x)) → B(c(x))
B(c(b(x))) → B(c(x))
B(c(b(b(x)))) → A(a(b(c(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(b(b(c(x))))
b(c(b(x))) → a(b(c(x)))
A(a(x)) → B(c(x))
B(c(b(x))) → B(c(x))
B(c(b(b(x)))) → A(a(b(c(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(a(A(x))))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → b(b(b(c(x))))
b(c(b(x))) → a(b(c(x)))
A(a(x)) → B(c(x))
B(c(b(x))) → B(c(x))
B(c(b(b(x)))) → A(a(b(c(x))))
The set Q is empty.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(x)) → b(b(b(c(x))))
b(c(b(x))) → a(b(c(x)))
A(a(x)) → B(c(x))
B(c(b(x))) → B(c(x))
B(c(b(b(x)))) → A(a(b(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:
A1(a(x)) → B1(x)
B1(b(c(B(x)))) → B1(a(A(x)))
B1(b(c(B(x)))) → A1(A(x))
A1(a(x)) → B1(b(b(x)))
B1(c(b(x))) → A1(x)
B1(c(b(x))) → B1(a(x))
A1(a(x)) → B1(b(x))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(x)
B1(b(c(B(x)))) → B1(a(A(x)))
B1(b(c(B(x)))) → A1(A(x))
A1(a(x)) → B1(b(b(x)))
B1(c(b(x))) → A1(x)
B1(c(b(x))) → B1(a(x))
A1(a(x)) → B1(b(x))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(x)
B1(b(c(B(x)))) → B1(a(A(x)))
A1(a(x)) → B1(b(b(x)))
B1(c(b(x))) → A1(x)
A1(a(x)) → B1(b(x))
B1(c(b(x))) → B1(a(x))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(b(c(B(x)))) → B1(a(A(x))) at position [0] we obtained the following new rules:
B1(b(c(B(x0)))) → B1(c(B(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(b(c(B(x0)))) → B1(c(B(x0)))
A1(a(x)) → B1(x)
A1(a(x)) → B1(b(b(x)))
B1(c(b(x))) → A1(x)
B1(c(b(x))) → B1(a(x))
A1(a(x)) → B1(b(x))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(x)
A1(a(x)) → B1(b(b(x)))
B1(c(b(x))) → A1(x)
A1(a(x)) → B1(b(x))
B1(c(b(x))) → B1(a(x))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(x)) → B1(b(b(x))) at position [0] we obtained the following new rules:
A1(a(c(B(x0)))) → B1(b(c(B(x0))))
A1(a(b(c(B(x0))))) → B1(b(c(b(a(A(x0))))))
A1(a(c(B(x0)))) → B1(c(b(a(A(x0)))))
A1(a(c(b(x0)))) → B1(b(c(b(a(x0)))))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(a(x)) → B1(x)
A1(a(c(B(x0)))) → B1(b(c(B(x0))))
A1(a(b(c(B(x0))))) → B1(b(c(b(a(A(x0))))))
A1(a(c(B(x0)))) → B1(c(b(a(A(x0)))))
B1(c(b(x))) → A1(x)
A1(a(c(b(x0)))) → B1(b(c(b(a(x0)))))
B1(c(b(x))) → B1(a(x))
A1(a(x)) → B1(b(x))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B1(c(b(x))) → B1(a(x)) at position [0] we obtained the following new rules:
B1(c(b(a(x0)))) → B1(c(b(b(b(x0)))))
B1(c(b(A(x0)))) → B1(c(B(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(c(b(a(x0)))) → B1(c(b(b(b(x0)))))
A1(a(x)) → B1(x)
B1(c(b(A(x0)))) → B1(c(B(x0)))
A1(a(c(B(x0)))) → B1(b(c(B(x0))))
A1(a(b(c(B(x0))))) → B1(b(c(b(a(A(x0))))))
B1(c(b(x))) → A1(x)
A1(a(c(B(x0)))) → B1(c(b(a(A(x0)))))
A1(a(c(b(x0)))) → B1(b(c(b(a(x0)))))
A1(a(x)) → B1(b(x))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(c(b(a(x0)))) → B1(c(b(b(b(x0)))))
A1(a(x)) → B1(x)
A1(a(c(B(x0)))) → B1(b(c(B(x0))))
A1(a(b(c(B(x0))))) → B1(b(c(b(a(A(x0))))))
A1(a(c(B(x0)))) → B1(c(b(a(A(x0)))))
B1(c(b(x))) → A1(x)
A1(a(c(b(x0)))) → B1(b(c(b(a(x0)))))
A1(a(x)) → B1(b(x))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(x)) → B1(b(x)) at position [0] we obtained the following new rules:
A1(a(c(b(x0)))) → B1(c(b(a(x0))))
A1(a(b(c(B(x0))))) → B1(c(b(a(A(x0)))))
A1(a(c(B(x0)))) → B1(c(B(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(c(b(a(x0)))) → B1(c(b(b(b(x0)))))
A1(a(x)) → B1(x)
A1(a(c(B(x0)))) → B1(c(B(x0)))
A1(a(c(B(x0)))) → B1(b(c(B(x0))))
A1(a(c(b(x0)))) → B1(c(b(a(x0))))
A1(a(b(c(B(x0))))) → B1(b(c(b(a(A(x0))))))
A1(a(b(c(B(x0))))) → B1(c(b(a(A(x0)))))
B1(c(b(x))) → A1(x)
A1(a(c(B(x0)))) → B1(c(b(a(A(x0)))))
A1(a(c(b(x0)))) → B1(b(c(b(a(x0)))))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(c(b(a(x0)))) → B1(c(b(b(b(x0)))))
A1(a(x)) → B1(x)
A1(a(c(B(x0)))) → B1(b(c(B(x0))))
A1(a(c(b(x0)))) → B1(c(b(a(x0))))
A1(a(b(c(B(x0))))) → B1(b(c(b(a(A(x0))))))
A1(a(b(c(B(x0))))) → B1(c(b(a(A(x0)))))
A1(a(c(B(x0)))) → B1(c(b(a(A(x0)))))
B1(c(b(x))) → A1(x)
A1(a(c(b(x0)))) → B1(b(c(b(a(x0)))))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(c(B(x0)))) → B1(b(c(B(x0)))) at position [0] we obtained the following new rules:
A1(a(c(B(x0)))) → B1(c(B(x0)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(c(b(a(x0)))) → B1(c(b(b(b(x0)))))
A1(a(x)) → B1(x)
A1(a(c(B(x0)))) → B1(c(B(x0)))
A1(a(c(b(x0)))) → B1(c(b(a(x0))))
A1(a(b(c(B(x0))))) → B1(b(c(b(a(A(x0))))))
B1(c(b(x))) → A1(x)
A1(a(c(B(x0)))) → B1(c(b(a(A(x0)))))
A1(a(b(c(B(x0))))) → B1(c(b(a(A(x0)))))
A1(a(c(b(x0)))) → B1(b(c(b(a(x0)))))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(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
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(c(b(a(x0)))) → B1(c(b(b(b(x0)))))
A1(a(x)) → B1(x)
A1(a(c(b(x0)))) → B1(c(b(a(x0))))
A1(a(b(c(B(x0))))) → B1(b(c(b(a(A(x0))))))
A1(a(b(c(B(x0))))) → B1(c(b(a(A(x0)))))
A1(a(c(B(x0)))) → B1(c(b(a(A(x0)))))
B1(c(b(x))) → A1(x)
A1(a(c(b(x0)))) → B1(b(c(b(a(x0)))))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule B1(c(b(x))) → A1(x) we obtained the following new rules:
B1(c(b(a(y_0)))) → A1(a(y_0))
B1(c(b(a(b(y_1))))) → A1(a(b(y_1)))
↳ QTRS
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B1(c(b(a(x0)))) → B1(c(b(b(b(x0)))))
A1(a(x)) → B1(x)
B1(c(b(a(y_0)))) → A1(a(y_0))
B1(c(b(a(b(y_1))))) → A1(a(b(y_1)))
A1(a(c(b(x0)))) → B1(c(b(a(x0))))
A1(a(b(c(B(x0))))) → B1(b(c(b(a(A(x0))))))
A1(a(c(B(x0)))) → B1(c(b(a(A(x0)))))
A1(a(b(c(B(x0))))) → B1(c(b(a(A(x0)))))
A1(a(c(b(x0)))) → B1(b(c(b(a(x0)))))
The TRS R consists of the following rules:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
a(A(x)) → c(B(x))
b(c(B(x))) → c(B(x))
b(b(c(B(x)))) → c(b(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(a(x1)) → b(b(b(c(x1))))
b(c(b(x1))) → a(b(c(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(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:
a(a(x)) → c(b(b(b(x))))
b(c(b(x))) → c(b(a(x)))
Q is empty.