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