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