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(c(x1))
b(a(b(x1))) → x1
c(c(x1)) → a(a(a(b(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(c(x1))
b(a(b(x1))) → x1
c(c(x1)) → a(a(a(b(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(c(x1)) → A(a(b(x1)))
A(x1) → B(c(x1))
C(c(x1)) → A(b(x1))
A(x1) → C(x1)
C(c(x1)) → B(x1)
C(c(x1)) → A(a(a(b(x1))))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
b(a(b(x1))) → x1
c(c(x1)) → a(a(a(b(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:
C(c(x1)) → A(a(b(x1)))
A(x1) → B(c(x1))
C(c(x1)) → A(b(x1))
A(x1) → C(x1)
C(c(x1)) → B(x1)
C(c(x1)) → A(a(a(b(x1))))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
b(a(b(x1))) → x1
c(c(x1)) → a(a(a(b(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
↳ QDPToSRSProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C(c(x1)) → A(a(b(x1)))
C(c(x1)) → A(b(x1))
A(x1) → C(x1)
C(c(x1)) → A(a(a(b(x1))))
The TRS R consists of the following rules:
a(x1) → b(c(x1))
b(a(b(x1))) → x1
c(c(x1)) → a(a(a(b(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(c(x1))
b(a(b(x1))) → x1
c(c(x1)) → a(a(a(b(x1))))
C(c(x1)) → A(a(b(x1)))
C(c(x1)) → A(b(x1))
A(x1) → C(x1)
C(c(x1)) → A(a(a(b(x1))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(c(x1))
b(a(b(x1))) → x1
c(c(x1)) → a(a(a(b(x1))))
C(c(x1)) → A(a(b(x1)))
C(c(x1)) → A(b(x1))
A(x1) → C(x1)
C(c(x1)) → A(a(a(b(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(x))))
c(C(x)) → b(a(A(x)))
c(C(x)) → b(A(x))
A(x) → C(x)
c(C(x)) → b(a(a(A(x))))
The set Q is empty.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QTRS Reverse
↳ QTRS Reverse
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(x))))
c(C(x)) → b(a(A(x)))
c(C(x)) → b(A(x))
A(x) → C(x)
c(C(x)) → b(a(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:
A1(x) → C1(b(x))
C1(c(x)) → B(a(a(a(x))))
A1(x) → B(x)
C1(c(x)) → A1(a(x))
C1(c(x)) → A1(x)
C1(C(x)) → B(a(A(x)))
C1(C(x)) → A1(A(x))
C1(C(x)) → A1(a(A(x)))
C1(C(x)) → B(a(a(A(x))))
C1(C(x)) → B(A(x))
C1(C(x)) → A2(x)
C1(c(x)) → A1(a(a(x)))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(x))))
c(C(x)) → b(a(A(x)))
c(C(x)) → b(A(x))
A(x) → C(x)
c(C(x)) → b(a(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → C1(b(x))
C1(c(x)) → B(a(a(a(x))))
A1(x) → B(x)
C1(c(x)) → A1(a(x))
C1(c(x)) → A1(x)
C1(C(x)) → B(a(A(x)))
C1(C(x)) → A1(A(x))
C1(C(x)) → A1(a(A(x)))
C1(C(x)) → B(a(a(A(x))))
C1(C(x)) → B(A(x))
C1(C(x)) → A2(x)
C1(c(x)) → A1(a(a(x)))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(x))))
c(C(x)) → b(a(A(x)))
c(C(x)) → b(A(x))
A(x) → C(x)
c(C(x)) → b(a(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 6 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A1(x) → C1(b(x))
C1(c(x)) → A1(a(x))
C1(c(x)) → A1(x)
C1(C(x)) → A1(A(x))
C1(C(x)) → A1(a(A(x)))
C1(c(x)) → A1(a(a(x)))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(x))))
c(C(x)) → b(a(A(x)))
c(C(x)) → b(A(x))
A(x) → C(x)
c(C(x)) → b(a(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A1(x) → C1(b(x)) at position [0] we obtained the following new rules:
A1(a(b(x0))) → C1(x0)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(c(x)) → A1(a(x))
C1(c(x)) → A1(x)
C1(C(x)) → A1(A(x))
A1(a(b(x0))) → C1(x0)
C1(C(x)) → A1(a(A(x)))
C1(c(x)) → A1(a(a(x)))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(x))))
c(C(x)) → b(a(A(x)))
c(C(x)) → b(A(x))
A(x) → C(x)
c(C(x)) → b(a(a(A(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule C1(C(x)) → A1(A(x)) at position [0] we obtained the following new rules:
C1(C(x0)) → A1(C(x0))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(c(x)) → A1(a(x))
C1(c(x)) → A1(x)
A1(a(b(x0))) → C1(x0)
C1(C(x0)) → A1(C(x0))
C1(C(x)) → A1(a(A(x)))
C1(c(x)) → A1(a(a(x)))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(x))))
c(C(x)) → b(a(A(x)))
c(C(x)) → b(A(x))
A(x) → C(x)
c(C(x)) → b(a(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
↳ QDPToSRSProof
↳ QTRS
↳ QTRS Reverse
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
C1(c(x)) → A1(a(x))
C1(c(x)) → A1(x)
A1(a(b(x0))) → C1(x0)
C1(C(x)) → A1(a(A(x)))
C1(c(x)) → A1(a(a(x)))
The TRS R consists of the following rules:
a(x) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(x))))
c(C(x)) → b(a(A(x)))
c(C(x)) → b(A(x))
A(x) → C(x)
c(C(x)) → b(a(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(x1) → b(c(x1))
b(a(b(x1))) → x1
c(c(x1)) → a(a(a(b(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(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) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(x))))
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(c(x1))
b(a(b(x1))) → x1
c(c(x1)) → a(a(a(b(x1))))
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(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) → c(b(x))
b(a(b(x))) → x
c(c(x)) → b(a(a(a(x))))
Q is empty.