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(x1) → c(a(c(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(x1) → c(a(c(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(x1) → c(a(c(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)))
b(x) → c(a(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)))
b(x) → c(a(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)))
b(x) → c(a(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, 150, 149, 152, 151, 153, 154, 156, 155, 158, 157, 159, 160, 161, 162, 164, 163, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174

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

Those nodes are connect through the following edges: