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

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

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

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

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

The signature Sigma is {f}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(x0, a(a(b(b(x1)))))
f(a(x0), x1)
f(b(x0), x1)


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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, a(a(b(b(x1)))))
f(a(x0), x1)
f(b(x0), x1)

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, a(a(b(b(x1)))))
f(a(x0), x1)
f(b(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

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

R is empty.
The set Q consists of the following terms:

f(x0, a(a(b(b(x1)))))
f(a(x0), x1)
f(b(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(x0, a(a(b(b(x1)))))
f(a(x0), x1)
f(b(x0), x1)



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The finiteness of this DP problem is implied by strong termination of a SRS due to [12].


↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof

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

a(a(b(b(x)))) → b(b(b(a(a(a(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:

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

The TRS R consists of the following rules:

a(a(b(b(x)))) → b(b(b(a(a(a(x))))))

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ MNOCProof

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

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

The TRS R consists of the following rules:

a(a(b(b(x)))) → b(b(b(a(a(a(x))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ QDPToSRSProof
                    ↳ QTRS
                      ↳ DependencyPairsProof
                        ↳ QDP
                          ↳ MNOCProof
                            ↳ QDP
                              ↳ RFCMatchBoundsDPProof

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

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

The TRS R consists of the following rules:

a(a(b(b(x)))) → b(b(b(a(a(a(x))))))

The set Q consists of the following terms:

a(a(b(b(x0))))

We have to consider all minimal (P,Q,R)-chains.
Finiteness of the DP problem can be shown by a matchbound of 8.
As the DP problem is minimal we only have to initialize the certificate graph by the rules of P:

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

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

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

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

44, 45, 46, 48, 49, 47, 51, 50, 53, 54, 52, 56, 55, 58, 59, 57, 61, 60, 63, 64, 62, 66, 65, 71, 70, 69, 68, 67, 73, 74, 72, 76, 75, 81, 80, 79, 78, 77, 86, 85, 84, 83, 82, 91, 90, 89, 88, 87, 96, 95, 94, 93, 92, 101, 100, 99, 98, 97, 106, 105, 104, 103, 102, 111, 110, 109, 108, 107, 116, 115, 114, 113, 112, 121, 120, 119, 118, 117, 126, 125, 124, 123, 122, 131, 130, 129, 128, 127, 133, 134, 132, 136, 135, 141, 140, 139, 138, 137, 146, 145, 144, 143, 142, 151, 150, 149, 148, 147, 153, 154, 152, 156, 155

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

Those nodes are connect through the following edges: