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

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

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

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

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
      ↳ RFCMatchBoundsTRSProof
  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

121, 122, 123, 128, 127, 126, 125, 124, 129, 134, 133, 132, 131, 130, 139, 138, 137, 136, 135, 144, 143, 142, 141, 140, 149, 148, 147, 146, 145, 154, 153, 152, 151, 150, 159, 158, 157, 156, 155, 164, 163, 162, 161, 160, 169, 168, 167, 166, 165, 171, 172, 170, 174, 173, 179, 178, 177, 176, 175, 181, 182, 180, 184, 183, 186, 187, 185, 189, 188, 191, 192, 190, 194, 193, 196, 197, 195, 199, 198, 201, 202, 200, 204, 203, 206, 207, 205, 209, 208, 211, 212, 210, 214, 213, 227, 228, 226, 230, 229, 235, 236, 234, 238, 237

Node 121 is start node and node 122 is final node.

Those nodes are connect through the following edges:



Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
QDP

Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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