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

Q is empty.


QTRS
  ↳ QTRS Reverse

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

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

Q is empty.

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

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

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

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

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

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

141, 142, 148, 149, 147, 144, 145, 143, 146, 155, 156, 154, 151, 152, 150, 153, 162, 163, 161, 158, 159, 157, 160, 169, 170, 168, 165, 166, 164, 167

Node 141 is start node and node 142 is final node.

Those nodes are connect through the following edges: