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