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(b(c(x1)))
a(c(b(x1))) → c(a(a(x1)))
c(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) → b(b(c(x1)))
a(c(b(x1))) → c(a(a(x1)))
c(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(b(x1))) → C(a(a(x1)))
A(x1) → C(x1)
A(c(b(x1))) → A(x1)
A(c(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(b(c(x1)))
a(c(b(x1))) → c(a(a(x1)))
c(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(b(x1))) → C(a(a(x1)))
A(x1) → C(x1)
A(c(b(x1))) → A(x1)
A(c(b(x1))) → A(a(x1))
The TRS R consists of the following rules:
a(x1) → b(b(c(x1)))
a(c(b(x1))) → c(a(a(x1)))
c(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
↳ QTRS Reverse
↳ QTRS Reverse
Q DP problem:
The TRS P consists of the following rules:
A(c(b(x1))) → A(a(x1))
A(c(b(x1))) → A(x1)
The TRS R consists of the following rules:
a(x1) → b(b(c(x1)))
a(c(b(x1))) → c(a(a(x1)))
c(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) → b(b(c(x1)))
a(c(b(x1))) → c(a(a(x1)))
c(b(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(b(x)))
b(c(a(x))) → a(a(c(x)))
b(c(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) → c(b(b(x)))
b(c(a(x))) → a(a(c(x)))
b(c(x)) → x
Q is empty.
We have reversed the following QTRS:
The set of rules R is
a(x1) → b(b(c(x1)))
a(c(b(x1))) → c(a(a(x1)))
c(b(x1)) → x1
The set Q is empty.
We have obtained the following QTRS:
a(x) → c(b(b(x)))
b(c(a(x))) → a(a(c(x)))
b(c(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) → c(b(b(x)))
b(c(a(x))) → a(a(c(x)))
b(c(x)) → x
Q is empty.