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:

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

Q is empty.


QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
      ↳ RFCMatchBoundsTRSProof

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

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

Q is empty.

Termination of the TRS R could be shown with a Match Bound [6,7] of 2. This implies Q-termination of R.
The following rules were used to construct the certificate:

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

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

128, 129, 133, 134, 132, 130, 135, 131, 139, 140, 138, 136, 141, 137, 145, 146, 144, 142, 147, 143, 151, 152, 150, 148, 153, 149

Node 128 is start node and node 129 is final node.

Those nodes are connect through the following edges: