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(d(x1)) → d(b(c(b(d(x1)))))
a(x1) → b(b(f(b(b(x1)))))
b(d(b(x1))) → a(d(x1))
d(f(x1)) → b(d(x1))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(d(x1)) → d(b(c(b(d(x1)))))
a(x1) → b(b(f(b(b(x1)))))
b(d(b(x1))) → a(d(x1))
d(f(x1)) → b(d(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(x1) → B(f(b(b(x1))))
A(x1) → B(b(x1))
B(d(b(x1))) → A(d(x1))
B(d(b(x1))) → D(x1)
A(d(x1)) → B(d(x1))
A(x1) → B(x1)
A(x1) → B(b(f(b(b(x1)))))
D(f(x1)) → D(x1)
D(f(x1)) → B(d(x1))
A(d(x1)) → B(c(b(d(x1))))
A(d(x1)) → D(b(c(b(d(x1)))))
The TRS R consists of the following rules:
a(d(x1)) → d(b(c(b(d(x1)))))
a(x1) → b(b(f(b(b(x1)))))
b(d(b(x1))) → a(d(x1))
d(f(x1)) → b(d(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A(x1) → B(f(b(b(x1))))
A(x1) → B(b(x1))
B(d(b(x1))) → A(d(x1))
B(d(b(x1))) → D(x1)
A(d(x1)) → B(d(x1))
A(x1) → B(x1)
A(x1) → B(b(f(b(b(x1)))))
D(f(x1)) → D(x1)
D(f(x1)) → B(d(x1))
A(d(x1)) → B(c(b(d(x1))))
A(d(x1)) → D(b(c(b(d(x1)))))
The TRS R consists of the following rules:
a(d(x1)) → d(b(c(b(d(x1)))))
a(x1) → b(b(f(b(b(x1)))))
b(d(b(x1))) → a(d(x1))
d(f(x1)) → b(d(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 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A(x1) → B(b(x1))
B(d(b(x1))) → A(d(x1))
B(d(b(x1))) → D(x1)
A(d(x1)) → B(d(x1))
A(x1) → B(x1)
D(f(x1)) → D(x1)
D(f(x1)) → B(d(x1))
The TRS R consists of the following rules:
a(d(x1)) → d(b(c(b(d(x1)))))
a(x1) → b(b(f(b(b(x1)))))
b(d(b(x1))) → a(d(x1))
d(f(x1)) → b(d(x1))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.