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, f(y, a)) → f(f(a, f(h(x), y)), a)

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

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

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

f(x0, f(x1, a))


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, f(y, a)) → F(h(x), y)
F(x, f(y, a)) → F(a, f(h(x), y))
F(x, f(y, a)) → F(f(a, f(h(x), y)), a)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(x1, a))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

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

F(x, f(y, a)) → F(h(x), y)
F(x, f(y, a)) → F(a, f(h(x), y))
F(x, f(y, a)) → F(f(a, f(h(x), y)), a)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(x1, a))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP
              ↳ Instantiation

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

F(x, f(y, a)) → F(h(x), y)
F(x, f(y, a)) → F(a, f(h(x), y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(x1, a))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(x, f(y, a)) → F(h(x), y) we obtained the following new rules:

F(a, f(x1, a)) → F(h(a), x1)
F(h(z0), f(x1, a)) → F(h(h(z0)), x1)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
QDP
                  ↳ Instantiation

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

F(a, f(x1, a)) → F(h(a), x1)
F(h(z0), f(x1, a)) → F(h(h(z0)), x1)
F(x, f(y, a)) → F(a, f(h(x), y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(x1, a))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(x, f(y, a)) → F(a, f(h(x), y)) we obtained the following new rules:

F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(h(h(z0)), f(x1, a)) → F(a, f(h(h(h(z0))), x1))
F(a, f(x1, a)) → F(a, f(h(a), x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
QDP
                      ↳ Instantiation

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

F(a, f(x1, a)) → F(h(a), x1)
F(h(z0), f(x1, a)) → F(h(h(z0)), x1)
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(a, f(x1, a)) → F(a, f(h(a), x1))
F(h(h(z0)), f(x1, a)) → F(a, f(h(h(h(z0))), x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(x1, a))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(h(z0), f(x1, a)) → F(h(h(z0)), x1) we obtained the following new rules:

F(h(a), f(x1, a)) → F(h(h(a)), x1)
F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation
QDP
                          ↳ Instantiation

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

F(h(a), f(x1, a)) → F(h(h(a)), x1)
F(a, f(x1, a)) → F(h(a), x1)
F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1)
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(h(h(z0)), f(x1, a)) → F(a, f(h(h(h(z0))), x1))
F(a, f(x1, a)) → F(a, f(h(a), x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(x1, a))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule F(h(h(z0)), f(x1, a)) → F(a, f(h(h(h(z0))), x1)) we obtained the following new rules:

F(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ Instantiation
QDP
                              ↳ ForwardInstantiation

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

F(h(a), f(x1, a)) → F(h(h(a)), x1)
F(a, f(x1, a)) → F(h(a), x1)
F(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))
F(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1)
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(a, f(x1, a)) → F(a, f(h(a), x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(x1, a))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(a, f(x1, a)) → F(h(a), x1) we obtained the following new rules:

F(a, f(f(y_0, a), a)) → F(h(a), f(y_0, a))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ Instantiation
                            ↳ QDP
                              ↳ ForwardInstantiation
QDP
                                  ↳ ForwardInstantiation

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

F(h(a), f(x1, a)) → F(h(h(a)), x1)
F(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))
F(a, f(f(y_0, a), a)) → F(h(a), f(y_0, a))
F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1)
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(a, f(x1, a)) → F(a, f(h(a), x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(x1, a))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(h(a), f(x1, a)) → F(h(h(a)), x1) we obtained the following new rules:

F(h(a), f(f(y_1, a), a)) → F(h(h(a)), f(y_1, a))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ Instantiation
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ ForwardInstantiation
QDP
                                      ↳ ForwardInstantiation

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

F(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))
F(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1)
F(a, f(f(y_0, a), a)) → F(h(a), f(y_0, a))
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(h(a), f(f(y_1, a), a)) → F(h(h(a)), f(y_1, a))
F(a, f(x1, a)) → F(a, f(h(a), x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(x1, a))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1) we obtained the following new rules:

F(h(h(x0)), f(f(y_1, a), a)) → F(h(h(h(x0))), f(y_1, a))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ Instantiation
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ ForwardInstantiation
                                    ↳ QDP
                                      ↳ ForwardInstantiation
QDP
                                          ↳ MNOCProof
                                          ↳ SemLabProof

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

F(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))
F(a, f(f(y_0, a), a)) → F(h(a), f(y_0, a))
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(h(h(x0)), f(f(y_1, a), a)) → F(h(h(h(x0))), f(y_1, a))
F(h(a), f(f(y_1, a), a)) → F(h(h(a)), f(y_1, a))
F(a, f(x1, a)) → F(a, f(h(a), x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, f(x1, a))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ Instantiation
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ ForwardInstantiation
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ QDP
                                          ↳ MNOCProof
QDP
                                          ↳ SemLabProof

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

F(h(h(a)), f(x1, a)) → F(a, f(h(h(h(a))), x1))
F(h(h(h(z0))), f(x1, a)) → F(a, f(h(h(h(h(z0)))), x1))
F(a, f(f(y_0, a), a)) → F(h(a), f(y_0, a))
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
F(h(a), f(f(y_1, a), a)) → F(h(h(a)), f(y_1, a))
F(h(h(x0)), f(f(y_1, a), a)) → F(h(h(h(x0))), f(y_1, a))
F(a, f(x1, a)) → F(a, f(h(a), x1))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.a: 1
f: 0
h: 0
F: 0
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

F.0-0(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.0-0(h.1(a.), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.1-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.0(h.0(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.1(a.)), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), x1))
F.0-0(h.1(a.), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.1(a.)), x1))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.1(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.0(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.0(z0)))), x1))
F.1-0(a., f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.1(a.), x1))
F.1-0(a., f.0-1(f.1-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.1-1(y_0, a.))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))

The TRS R consists of the following rules:

f.1-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.1(x), y)), a.)
f.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.0(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)

The set Q consists of the following terms:

f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ Instantiation
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ ForwardInstantiation
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ QDP
                                          ↳ MNOCProof
                                          ↳ SemLabProof
QDP
                                              ↳ DependencyGraphProof

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

F.0-0(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.0-0(h.1(a.), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.1-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.0(h.0(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.1(a.)), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), x1))
F.0-0(h.1(a.), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.1(a.)), x1))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.1(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.0(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.0(z0)))), x1))
F.1-0(a., f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.1(a.), x1))
F.1-0(a., f.0-1(f.1-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.1-1(y_0, a.))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))

The TRS R consists of the following rules:

f.1-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.1(x), y)), a.)
f.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.0(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)

The set Q consists of the following terms:

f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ Instantiation
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ ForwardInstantiation
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ QDP
                                          ↳ MNOCProof
                                          ↳ SemLabProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
QDP
                                                  ↳ RuleRemovalProof

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

F.0-0(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.0-0(h.1(a.), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.1-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.0(h.1(a.)), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), x1))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.0(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.1-1(y_1, a.))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.1(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.0(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.0(z0)))), x1))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))

The TRS R consists of the following rules:

f.1-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.1(x), y)), a.)
f.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.0(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)

The set Q consists of the following terms:

f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))

We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F.0-0(h.0(h.1(a.)), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), x1))
F.0-0(h.0(h.0(h.1(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.1(z0)))), x1))
F.0-0(h.0(h.0(h.0(z0))), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.0(h.0(z0)))), x1))

Strictly oriented rules of the TRS R:

f.1-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.1(x), y)), a.)
f.0-0(x, f.1-1(y, a.)) → f.0-1(f.1-0(a., f.0-1(h.0(x), y)), a.)

Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-0(x1, x2)) = x1 + x2   
POL(F.1-0(x1, x2)) = x1 + x2   
POL(a.) = 0   
POL(f.0-0(x1, x2)) = x1 + x2   
POL(f.0-1(x1, x2)) = x1 + x2   
POL(f.1-0(x1, x2)) = x1 + x2   
POL(f.1-1(x1, x2)) = 1 + x1 + x2   
POL(h.0(x1)) = x1   
POL(h.1(x1)) = x1   



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ Instantiation
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ ForwardInstantiation
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ QDP
                                          ↳ MNOCProof
                                          ↳ SemLabProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
QDP
                                                      ↳ DependencyGraphProof

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

F.0-0(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.1(a.), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
F.0-0(h.0(h.0(x0)), f.0-1(f.1-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.1-1(y_1, a.))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))

The TRS R consists of the following rules:

f.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)

The set Q consists of the following terms:

f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ Instantiation
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ ForwardInstantiation
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ QDP
                                          ↳ MNOCProof
                                          ↳ SemLabProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ DependencyGraphProof
QDP
                                                          ↳ RuleRemovalProof

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

F.0-0(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))

The TRS R consists of the following rules:

f.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)

The set Q consists of the following terms:

f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))

We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F.0-0(h.1(a.), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.1(a.)), x1))
F.0-0(h.0(h.1(a.)), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.1(a.))), x1))
F.0-0(h.1(a.), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_1, a.))
F.0-0(h.0(h.0(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.0(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(f.0-1(y_0, a.), a.)) → F.0-0(h.1(a.), f.0-1(y_0, a.))
F.0-0(h.0(h.1(x0)), f.0-1(f.0-1(y_1, a.), a.)) → F.0-0(h.0(h.0(h.1(x0))), f.0-1(y_1, a.))
F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.1(a.), x1))
F.0-0(h.0(h.0(h.0(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.0(z0)))), x1))
F.0-0(h.0(h.0(h.1(z0))), f.0-1(x1, a.)) → F.1-0(a., f.0-0(h.0(h.0(h.0(h.1(z0)))), x1))


Used ordering: POLO with Polynomial interpretation [25]:

POL(F.0-0(x1, x2)) = 1 + x1 + x2   
POL(F.1-0(x1, x2)) = 1 + x1 + x2   
POL(a.) = 0   
POL(f.0-0(x1, x2)) = x1 + x2   
POL(f.0-1(x1, x2)) = 1 + x1 + x2   
POL(f.1-0(x1, x2)) = x1 + x2   
POL(h.0(x1)) = x1   
POL(h.1(x1)) = x1   



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ Instantiation
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ Instantiation
                            ↳ QDP
                              ↳ ForwardInstantiation
                                ↳ QDP
                                  ↳ ForwardInstantiation
                                    ↳ QDP
                                      ↳ ForwardInstantiation
                                        ↳ QDP
                                          ↳ MNOCProof
                                          ↳ SemLabProof
                                            ↳ QDP
                                              ↳ DependencyGraphProof
                                                ↳ QDP
                                                  ↳ RuleRemovalProof
                                                    ↳ QDP
                                                      ↳ DependencyGraphProof
                                                        ↳ QDP
                                                          ↳ RuleRemovalProof
QDP
                                                              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f.1-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.1(x), y)), a.)
f.0-0(x, f.0-1(y, a.)) → f.0-1(f.1-0(a., f.0-0(h.0(x), y)), a.)

The set Q consists of the following terms:

f.0-0(x0, f.0-1(x1, a.))
f.0-0(x0, f.1-1(x1, a.))
f.1-0(x0, f.0-1(x1, a.))
f.1-0(x0, f.1-1(x1, a.))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.