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