Termination w.r.t. Q proof of /tmp/tmp

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(x1)) → b(b(b(x1)))
b(b(x1)) → c(c(c(x1)))
c(c(x1)) → d(d(d(x1)))
b(x1) → d(d(x1))
c(d(d(x1))) → a(x1)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

a(a(x1)) → b(b(b(x1)))
b(b(x1)) → c(c(c(x1)))
c(c(x1)) → d(d(d(x1)))
b(x1) → d(d(x1))
c(d(d(x1))) → 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:

C(d(d(x1))) → A(x1)
B(b(x1)) → C(c(x1))
A(a(x1)) → B(b(x1))
B(b(x1)) → C(c(c(x1)))
A(a(x1)) → B(b(b(x1)))
B(b(x1)) → C(x1)
A(a(x1)) → B(x1)

The TRS R consists of the following rules:

a(a(x1)) → b(b(b(x1)))
b(b(x1)) → c(c(c(x1)))
c(c(x1)) → d(d(d(x1)))
b(x1) → d(d(x1))
c(d(d(x1))) → a(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:

C(d(d(x1))) → A(x1)
B(b(x1)) → C(c(x1))
A(a(x1)) → B(b(x1))
B(b(x1)) → C(c(c(x1)))
A(a(x1)) → B(b(b(x1)))
B(b(x1)) → C(x1)
A(a(x1)) → B(x1)

The TRS R consists of the following rules:

a(a(x1)) → b(b(b(x1)))
b(b(x1)) → c(c(c(x1)))
c(c(x1)) → d(d(d(x1)))
b(x1) → d(d(x1))
c(d(d(x1))) → a(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.