Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.


↳ QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

We have reversed the following QTRS:
The set of rules R is

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

The set Q is empty.
We have obtained the following QTRS:

b(a(x)) → a(a(b(x)))
c(b(x)) → b(c(x))
a(a(x)) → a(c(a(x)))

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
      ↳ RFCMatchBoundsTRSProof

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

b(a(x)) → a(a(b(x)))
c(b(x)) → b(c(x))
a(a(x)) → a(c(a(x)))

Q is empty.

Termination of the TRS R could be shown with a Match Bound [6,7] of 3. This implies Q-termination of R.
The following rules were used to construct the certificate:

b(a(x)) → a(a(b(x)))
c(b(x)) → b(c(x))
a(a(x)) → a(c(a(x)))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

147, 148, 149, 151, 150, 152, 153, 154, 156, 155, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170

Node 147 is start node and node 148 is final node.

Those nodes are connect through the following edges: