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