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(a(x1)) → b(b(c(x1)))
b(x1) → x1
c(b(x1)) → b(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) → x1
a(a(x1)) → b(b(c(x1)))
b(x1) → x1
c(b(x1)) → b(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(a(x1)) → B(c(x1))
A(a(x1)) → B(b(c(x1)))
C(b(x1)) → B(a(c(x1)))
C(b(x1)) → A(c(x1))
A(a(x1)) → C(x1)
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(a(x1)) → b(b(c(x1)))
b(x1) → x1
c(b(x1)) → b(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(a(x1)) → B(c(x1))
A(a(x1)) → B(b(c(x1)))
C(b(x1)) → B(a(c(x1)))
C(b(x1)) → A(c(x1))
A(a(x1)) → C(x1)
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(a(x1)) → b(b(c(x1)))
b(x1) → x1
c(b(x1)) → b(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 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(b(x1)) → A(c(x1))
A(a(x1)) → C(x1)
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(a(x1)) → b(b(c(x1)))
b(x1) → x1
c(b(x1)) → b(a(c(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C(b(x1)) → A(c(x1)) at position [0] we obtained the following new rules:
C(b(b(x0))) → A(b(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(b(b(x0))) → A(b(a(c(x0))))
A(a(x1)) → C(x1)
C(b(x1)) → C(x1)
The TRS R consists of the following rules:
a(x1) → x1
a(a(x1)) → b(b(c(x1)))
b(x1) → x1
c(b(x1)) → b(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) → x1
a(a(x1)) → b(b(c(x1)))
b(x1) → x1
c(b(x1)) → b(a(c(x1)))
C(b(b(x0))) → A(b(a(c(x0))))
A(a(x1)) → C(x1)
C(b(x1)) → C(x1)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(x1)) → b(b(c(x1)))
b(x1) → x1
c(b(x1)) → b(a(c(x1)))
C(b(b(x0))) → A(b(a(c(x0))))
A(a(x1)) → C(x1)
C(b(x1)) → C(x1)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(C(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(C(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(b(C(x))) → A1(b(A(x)))
A1(a(x)) → B(x)
B(b(C(x))) → B(A(x))
A1(a(x)) → B(b(x))
B(c(x)) → A1(b(x))
B(c(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
B(b(C(x))) → A1(b(A(x)))
A1(a(x)) → B(x)
B(b(C(x))) → B(A(x))
A1(a(x)) → B(b(x))
B(c(x)) → A1(b(x))
B(c(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(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
↳ 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(b(C(x))) → A1(b(A(x)))
A1(a(x)) → B(x)
A1(a(x)) → B(b(x))
B(c(x)) → A1(b(x))
B(c(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(b(C(x))) → A1(b(A(x))) at position [0] we obtained the following new rules:
B(b(C(y0))) → A1(A(y0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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:
A1(a(x)) → B(x)
B(b(C(y0))) → A1(A(y0))
A1(a(x)) → B(b(x))
B(c(x)) → A1(b(x))
B(c(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(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
↳ 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:
A1(a(x)) → B(x)
A1(a(x)) → B(b(x))
B(c(x)) → A1(b(x))
B(c(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(a(x)) → B(b(x)) at position [0] we obtained the following new rules:
A1(a(c(x0))) → B(c(a(b(x0))))
A1(a(x0)) → B(x0)
A1(a(C(x0))) → B(C(x0))
A1(a(b(C(x0)))) → B(c(a(b(A(x0)))))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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:
A1(a(c(x0))) → B(c(a(b(x0))))
A1(a(x)) → B(x)
A1(a(C(x0))) → B(C(x0))
B(c(x)) → A1(b(x))
A1(a(b(C(x0)))) → B(c(a(b(A(x0)))))
B(c(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(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
↳ 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:
A1(a(c(x0))) → B(c(a(b(x0))))
A1(a(x)) → B(x)
B(c(x)) → A1(b(x))
A1(a(b(C(x0)))) → B(c(a(b(A(x0)))))
B(c(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(C(x)) → C(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(c(x)) → A1(b(x)) at position [0] we obtained the following new rules:
B(c(b(C(x0)))) → A1(c(a(b(A(x0)))))
B(c(C(x0))) → A1(C(x0))
B(c(c(x0))) → A1(c(a(b(x0))))
B(c(x0)) → A1(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(c(b(C(x0)))) → A1(c(a(b(A(x0)))))
A1(a(x)) → B(x)
A1(a(c(x0))) → B(c(a(b(x0))))
B(c(c(x0))) → A1(c(a(b(x0))))
B(c(x0)) → A1(x0)
B(c(C(x0))) → A1(C(x0))
A1(a(b(C(x0)))) → B(c(a(b(A(x0)))))
B(c(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(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 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ 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:
A1(a(c(x0))) → B(c(a(b(x0))))
A1(a(x)) → B(x)
B(c(x0)) → A1(x0)
A1(a(b(C(x0)))) → B(c(a(b(A(x0)))))
B(c(x)) → B(x)
The TRS R consists of the following rules:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(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(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(C(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(a(x)) → b(b(c(x)))
b(x) → x
c(b(x)) → b(a(c(x)))
C(b(b(x))) → A(b(a(c(x))))
A(a(x)) → C(x)
C(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(a(x)) → b(b(c(x)))
b(x) → x
c(b(x)) → b(a(c(x)))
C(b(b(x))) → A(b(a(c(x))))
A(a(x)) → C(x)
C(b(x)) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
b(b(C(x))) → c(a(b(A(x))))
a(A(x)) → C(x)
b(C(x)) → C(x)
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(a(x)) → b(b(c(x)))
b(x) → x
c(b(x)) → b(a(c(x)))
C(b(b(x))) → A(b(a(c(x))))
A(a(x)) → C(x)
C(b(x)) → C(x)
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ 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(a(x)) → b(b(c(x)))
b(x) → x
c(b(x)) → b(a(c(x)))
C(b(b(x))) → A(b(a(c(x))))
A(a(x)) → C(x)
C(b(x)) → C(x)
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(x1)) → b(b(c(x1)))
b(x1) → x1
c(b(x1)) → b(a(c(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(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
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(x1)) → b(b(c(x1)))
b(x1) → x1
c(b(x1)) → b(a(c(x1)))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(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
a(a(x)) → c(b(b(x)))
b(x) → x
b(c(x)) → c(a(b(x)))
Q is empty.