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) → x1
a(b(x1)) → b(c(a(x1)))
c(a(c(x1))) → b(a(a(x1)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
c(a(c(x1))) → b(a(a(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(a(c(x1))) → A(x1)
A(b(x1)) → C(a(x1))
C(a(c(x1))) → A(a(x1))
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
c(a(c(x1))) → b(a(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(a(c(x1))) → A(x1)
A(b(x1)) → C(a(x1))
C(a(c(x1))) → A(a(x1))
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
c(a(c(x1))) → b(a(a(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(a(c(x1))) → A(a(x1)) at position [0] we obtained the following new rules:
C(a(c(x0))) → A(x0)
C(a(c(b(x0)))) → A(b(c(a(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → C(a(x1))
C(a(c(x1))) → A(x1)
C(a(c(b(x0)))) → A(b(c(a(x0))))
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
c(a(c(x1))) → b(a(a(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
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
c(a(c(x1))) → b(a(a(x1)))
A(b(x1)) → C(a(x1))
C(a(c(x1))) → A(x1)
C(a(c(b(x0)))) → A(b(c(a(x0))))
A(b(x1)) → A(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
c(a(c(x1))) → b(a(a(x1)))
A(b(x1)) → C(a(x1))
C(a(c(x1))) → A(x1)
C(a(c(b(x0)))) → A(b(c(a(x0))))
A(b(x1)) → A(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → 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:
B(c(a(C(x)))) → A1(c(b(A(x))))
B(a(x)) → A1(c(b(x)))
B(c(a(C(x)))) → C1(b(A(x)))
B(c(a(C(x)))) → B(A(x))
B(A(x)) → A1(C(x))
C1(a(c(x))) → A1(a(b(x)))
C1(a(c(x))) → B(x)
B(a(x)) → B(x)
C1(a(c(x))) → A1(b(x))
B(a(x)) → C1(b(x))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ 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:
B(c(a(C(x)))) → A1(c(b(A(x))))
B(a(x)) → A1(c(b(x)))
B(c(a(C(x)))) → C1(b(A(x)))
B(c(a(C(x)))) → B(A(x))
B(A(x)) → A1(C(x))
C1(a(c(x))) → A1(a(b(x)))
C1(a(c(x))) → B(x)
B(a(x)) → B(x)
C1(a(c(x))) → A1(b(x))
B(a(x)) → C1(b(x))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → 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 6 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ 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:
B(c(a(C(x)))) → C1(b(A(x)))
C1(a(c(x))) → B(x)
B(a(x)) → B(x)
B(a(x)) → C1(b(x))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(x)) → C1(b(x)) at position [0] we obtained the following new rules:
B(a(c(a(C(x0))))) → C1(a(c(b(A(x0)))))
B(a(A(x0))) → C1(A(x0))
B(a(A(x0))) → C1(a(C(x0)))
B(a(a(x0))) → C1(a(c(b(x0))))
↳ QTRS
↳ DependencyPairsProof
↳ 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:
B(a(c(a(C(x0))))) → C1(a(c(b(A(x0)))))
B(c(a(C(x)))) → C1(b(A(x)))
B(a(a(x0))) → C1(a(c(b(x0))))
C1(a(c(x))) → B(x)
B(a(A(x0))) → C1(a(C(x0)))
B(a(A(x0))) → C1(A(x0))
B(a(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → 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
↳ 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:
B(a(c(a(C(x0))))) → C1(a(c(b(A(x0)))))
B(c(a(C(x)))) → C1(b(A(x)))
B(a(a(x0))) → C1(a(c(b(x0))))
C1(a(c(x))) → B(x)
B(a(A(x0))) → C1(a(C(x0)))
B(a(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(a(C(x)))) → C1(b(A(x))) at position [0] we obtained the following new rules:
B(c(a(C(x0)))) → C1(A(x0))
B(c(a(C(x0)))) → C1(a(C(x0)))
↳ QTRS
↳ DependencyPairsProof
↳ 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:
B(a(c(a(C(x0))))) → C1(a(c(b(A(x0)))))
B(a(a(x0))) → C1(a(c(b(x0))))
B(c(a(C(x0)))) → C1(A(x0))
C1(a(c(x))) → B(x)
B(c(a(C(x0)))) → C1(a(C(x0)))
B(a(x)) → B(x)
B(a(A(x0))) → C1(a(C(x0)))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → 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
↳ Narrowing
↳ 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:
B(a(c(a(C(x0))))) → C1(a(c(b(A(x0)))))
B(a(a(x0))) → C1(a(c(b(x0))))
C1(a(c(x))) → B(x)
B(c(a(C(x0)))) → C1(a(C(x0)))
B(a(A(x0))) → C1(a(C(x0)))
B(a(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(a(A(x0))) → C1(a(C(x0))) at position [0] we obtained the following new rules:
B(a(A(y0))) → C1(C(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ 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:
B(a(c(a(C(x0))))) → C1(a(c(b(A(x0)))))
B(a(a(x0))) → C1(a(c(b(x0))))
B(a(A(y0))) → C1(C(y0))
C1(a(c(x))) → B(x)
B(a(x)) → B(x)
B(c(a(C(x0)))) → C1(a(C(x0)))
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → 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
↳ Narrowing
↳ 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:
B(a(c(a(C(x0))))) → C1(a(c(b(A(x0)))))
B(a(a(x0))) → C1(a(c(b(x0))))
C1(a(c(x))) → B(x)
B(c(a(C(x0)))) → C1(a(C(x0)))
B(a(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → A(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(a(C(x0)))) → C1(a(C(x0))) at position [0] we obtained the following new rules:
B(c(a(C(y0)))) → C1(C(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ 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
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(c(a(C(x0))))) → C1(a(c(b(A(x0)))))
B(a(a(x0))) → C1(a(c(b(x0))))
B(c(a(C(y0)))) → C1(C(y0))
C1(a(c(x))) → B(x)
B(a(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → 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
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ 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
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(a(c(a(C(x0))))) → C1(a(c(b(A(x0)))))
B(a(a(x0))) → C1(a(c(b(x0))))
C1(a(c(x))) → B(x)
B(a(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → 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) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(x)) → b(c(a(x)))
c(a(c(x))) → b(a(a(x)))
A(b(x)) → C(a(x))
C(a(c(x))) → A(x)
C(a(c(b(x)))) → A(b(c(a(x))))
A(b(x)) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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) → x
a(b(x)) → b(c(a(x)))
c(a(c(x))) → b(a(a(x)))
A(b(x)) → C(a(x))
C(a(c(x))) → A(x)
C(a(c(b(x)))) → A(b(c(a(x))))
A(b(x)) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
b(A(x)) → a(C(x))
c(a(C(x))) → A(x)
b(c(a(C(x)))) → a(c(b(A(x))))
b(A(x)) → A(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(b(x)) → b(c(a(x)))
c(a(c(x))) → b(a(a(x)))
A(b(x)) → C(a(x))
C(a(c(x))) → A(x)
C(a(c(b(x)))) → A(b(c(a(x))))
A(b(x)) → A(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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) → x
a(b(x)) → b(c(a(x)))
c(a(c(x))) → b(a(a(x)))
A(b(x)) → C(a(x))
C(a(c(x))) → A(x)
C(a(c(b(x)))) → A(b(c(a(x))))
A(b(x)) → A(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
c(a(c(x1))) → b(a(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(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) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(b(x1)) → b(c(a(x1)))
c(a(c(x1))) → b(a(a(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(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) → x
b(a(x)) → a(c(b(x)))
c(a(c(x))) → a(a(b(x)))
Q is empty.