YES Termination w.r.t. Q proof of ../tpdb/TRS/Zantema/z30.trs
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:

f(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → f(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → f(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))

Q is empty.

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

F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, f(a, f(a, f(a, f(b, x))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, x)
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, f(a, f(a, f(a, f(b, x)))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, x))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(a, f(b, x))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x))))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, x)))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))

The TRS R consists of the following rules:

f(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → f(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, f(a, f(a, f(a, f(b, x))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, x)
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, f(a, f(a, f(a, f(b, x)))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, x))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(a, f(b, x))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x))))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, x)))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))

The TRS R consists of the following rules:

f(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → f(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ UsableRulesReductionPairsProof

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

F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(a, f(b, x))))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, x)))
F(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → F(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))

The TRS R consists of the following rules:

f(a, f(a, f(b, f(a, f(a, f(b, f(a, x))))))) → f(a, f(b, f(a, f(a, f(b, f(a, f(a, f(a, f(b, x)))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
First, we A-transformed [17] the QDP-Problem. Then we obtain the following A-transformed DP problem.
The pairs P are:

a1(a(b(a(a(b(a(x))))))) → a1(a(a(b(x))))
a1(a(b(a(a(b(a(x))))))) → a1(a(b(x)))
a1(a(b(a(a(b(a(x))))))) → a1(a(b(a(a(a(b(x)))))))

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

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

Q is empty.

By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(a(x1)) = x1   
POL(a1(x1)) = x1   
POL(b(x1)) = x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesReductionPairsProof
QDP
              ↳ RFCMatchBoundsDPProof

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

a1(a(b(a(a(b(a(x))))))) → a1(a(b(a(a(a(b(x)))))))
a1(a(b(a(a(b(a(x))))))) → a1(a(a(b(x))))
a1(a(b(a(a(b(a(x))))))) → a1(a(b(x)))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
Finiteness of the DP problem can be shown by a matchbound of 3.
As the DP problem is minimal we only have to initialize the certificate graph by the rules of P:

a1(a(b(a(a(b(a(x))))))) → a1(a(b(a(a(a(b(x)))))))
a1(a(b(a(a(b(a(x))))))) → a1(a(a(b(x))))
a1(a(b(a(a(b(a(x))))))) → a1(a(b(x)))

To find matches we regarded all rules of R and P:

a(a(b(a(a(b(a(x))))))) → a(b(a(a(b(a(a(a(b(x)))))))))
a1(a(b(a(a(b(a(x))))))) → a1(a(b(a(a(a(b(x)))))))
a1(a(b(a(a(b(a(x))))))) → a1(a(a(b(x))))
a1(a(b(a(a(b(a(x))))))) → a1(a(b(x)))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

44, 45, 51, 47, 48, 50, 49, 46, 59, 55, 56, 52, 53, 58, 57, 54, 67, 63, 64, 60, 61, 66, 65, 62, 68, 72, 73, 71, 70, 69, 80, 81, 79, 78, 76, 77, 75, 74, 88, 89, 87, 86, 84, 85, 83, 82, 96, 97, 95, 94, 92, 93, 91, 90, 104, 105, 103, 102, 100, 101, 99, 98, 112, 113, 111, 110, 108, 109, 107, 106, 121, 117, 118, 114, 115, 120, 119, 116, 129, 125, 126, 122, 123, 128, 127, 124, 136, 137, 135, 134, 132, 133, 131, 130

Node 44 is start node and node 45 is final node.

Those nodes are connect through the following edges: