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(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(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(b(x1)))) → C(x1)
A(c(b(b(x1)))) → B(b(c(x1)))
A(c(b(b(x1)))) → B(c(x1))
B(b(b(x1))) → C(x1)
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(c(x1))) → A(b(b(x1)))
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(c(x1))) → B(a(b(b(x1))))
B(b(c(x1))) → B(b(x1))
B(b(b(x1))) → C(b(a(c(x1))))
C(a(a(a(x1)))) → C(x1)
B(b(b(x1))) → B(a(c(x1)))
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A(c(b(b(x1)))) → C(x1)
A(c(b(b(x1)))) → B(b(c(x1)))
A(c(b(b(x1)))) → B(c(x1))
B(b(b(x1))) → C(x1)
A(c(b(b(x1)))) → B(b(b(c(x1))))
B(b(c(x1))) → A(b(b(x1)))
B(b(b(x1))) → A(c(x1))
C(a(a(a(x1)))) → A(c(x1))
B(b(c(x1))) → B(x1)
B(b(c(x1))) → B(a(b(b(x1))))
B(b(c(x1))) → B(b(x1))
B(b(b(x1))) → C(b(a(c(x1))))
C(a(a(a(x1)))) → C(x1)
B(b(b(x1))) → B(a(c(x1)))
The TRS R consists of the following rules:
a(c(b(b(x1)))) → b(b(b(c(x1))))
b(b(c(x1))) → b(a(b(b(x1))))
c(a(a(a(x1)))) → a(c(x1))
b(b(b(x1))) → c(b(a(c(x1))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.