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(b(x1))) → c(a(x1))
a(c(x1)) → c(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(a(b(x1))) → c(a(x1))
a(c(x1)) → c(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:
A(c(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(c(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → c(a(x1))
a(c(x1)) → c(b(a(a(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(c(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → x1
a(a(b(x1))) → c(a(x1))
a(c(x1)) → c(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
↳ 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(b(x1))) → c(a(x1))
a(c(x1)) → c(b(a(a(x1))))
A(c(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(c(x1)) → A(a(x1))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(b(x1))) → c(a(x1))
a(c(x1)) → c(b(a(a(x1))))
A(c(x1)) → A(x1)
A(a(b(x1))) → A(x1)
A(c(x1)) → A(a(x1))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
c(A(x)) → A(x)
b(a(A(x))) → A(x)
c(A(x)) → a(A(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
c(A(x)) → A(x)
b(a(A(x))) → A(x)
c(A(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(x)) → A1(A(x))
C(a(x)) → B(c(x))
C(a(x)) → C(x)
B(a(a(x))) → A1(c(x))
B(a(a(x))) → C(x)
C(a(x)) → A1(b(c(x)))
C(a(x)) → A1(a(b(c(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
c(A(x)) → A(x)
b(a(A(x))) → A(x)
c(A(x)) → a(A(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ 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:
C(A(x)) → A1(A(x))
C(a(x)) → B(c(x))
C(a(x)) → C(x)
B(a(a(x))) → A1(c(x))
B(a(a(x))) → C(x)
C(a(x)) → A1(b(c(x)))
C(a(x)) → A1(a(b(c(x))))
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
c(A(x)) → A(x)
b(a(A(x))) → A(x)
c(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 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ 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:
C(a(x)) → B(c(x))
C(a(x)) → C(x)
B(a(a(x))) → C(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
c(A(x)) → A(x)
b(a(A(x))) → A(x)
c(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(x)) → B(c(x)) at position [0] we obtained the following new rules:
C(a(a(x0))) → B(a(a(b(c(x0)))))
C(a(A(x0))) → B(a(A(x0)))
C(a(A(x0))) → B(A(x0))
↳ QTRS
↳ DependencyPairsProof
↳ 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:
C(a(a(x0))) → B(a(a(b(c(x0)))))
C(a(x)) → C(x)
C(a(A(x0))) → B(a(A(x0)))
C(a(A(x0))) → B(A(x0))
B(a(a(x))) → C(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
c(A(x)) → A(x)
b(a(A(x))) → A(x)
c(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
↳ 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:
C(a(a(x0))) → B(a(a(b(c(x0)))))
C(a(x)) → C(x)
C(a(A(x0))) → B(a(A(x0)))
B(a(a(x))) → C(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
c(A(x)) → A(x)
b(a(A(x))) → A(x)
c(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(x0))) → B(a(A(x0))) at position [0] we obtained the following new rules:
C(a(A(y0))) → B(A(y0))
↳ QTRS
↳ DependencyPairsProof
↳ 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:
C(a(a(x0))) → B(a(a(b(c(x0)))))
C(a(x)) → C(x)
C(a(A(y0))) → B(A(y0))
B(a(a(x))) → C(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
c(A(x)) → A(x)
b(a(A(x))) → A(x)
c(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ 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:
C(a(a(x0))) → B(a(a(b(c(x0)))))
C(a(x)) → C(x)
B(a(a(x))) → C(x)
The TRS R consists of the following rules:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
c(A(x)) → A(x)
b(a(A(x))) → A(x)
c(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) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
c(A(x)) → A(x)
b(a(A(x))) → A(x)
c(A(x)) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(a(b(x))) → c(a(x))
a(c(x)) → c(b(a(a(x))))
A(c(x)) → A(x)
A(a(b(x))) → A(x)
A(c(x)) → A(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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(b(x))) → c(a(x))
a(c(x)) → c(b(a(a(x))))
A(c(x)) → A(x)
A(a(b(x))) → A(x)
A(c(x)) → A(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
c(A(x)) → A(x)
b(a(A(x))) → A(x)
c(A(x)) → a(A(x))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
a(a(b(x))) → c(a(x))
a(c(x)) → c(b(a(a(x))))
A(c(x)) → A(x)
A(a(b(x))) → A(x)
A(c(x)) → A(a(x))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ 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(b(x))) → c(a(x))
a(c(x)) → c(b(a(a(x))))
A(c(x)) → A(x)
A(a(b(x))) → A(x)
A(c(x)) → A(a(x))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(b(x1))) → c(a(x1))
a(c(x1)) → c(b(a(a(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(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(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → x1
a(a(b(x1))) → c(a(x1))
a(c(x1)) → c(b(a(a(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → x
b(a(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(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(a(x))) → a(c(x))
c(a(x)) → a(a(b(c(x))))
Q is empty.