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