Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
  ↳7 Instantiation (⇔)
    ↳8 QDP
  ↳9 Instantiation (⇔)
    ↳10 QDP
    ↳11 Instantiation (⇔)
    ↳12 QDP
    ↳13 Instantiation (⇔)
    ↳14 QDP
    ↳15 Instantiation (⇔)
    ↳16 QDP
    ↳17 ForwardInstantiation (⇔)
    ↳18 QDP
    ↳19 ForwardInstantiation (⇔)
    ↳20 QDP
    ↳21 ForwardInstantiation (⇔)
    ↳22 QDP
      ↳23 MNOCProof (⇔)
        ↳24 QDP
      ↳25 SemLabProof (⇐)
        ↳26 QDP
        ↳27 DependencyGraphProof (⇔)
        ↳28 QDP
        ↳29 MRRProof (⇔)
        ↳30 QDP
        ↳31 DependencyGraphProof (⇔)
        ↳32 QDP
        ↳33 MRRProof (⇔)
        ↳34 QDP
        ↳35 PisEmptyProof (⇔)
        ↳36 TRUE

(0) Obligation:

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(6) Obligation:

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.

(7) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, f(y, a)) → F(h(x), y) we obtained the following new rules [LPAR04]:

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

(8) Obligation:

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

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

(9) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, f(y, a)) → F(h(x), y) we obtained the following new rules [LPAR04]:

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

(10) Obligation:

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

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

(11) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, f(y, a)) → F(a, f(h(x), y)) we obtained the following new rules [LPAR04]:

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

(12) Obligation:

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

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

(13) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(h(z0), f(x1, a)) → F(h(h(z0)), x1) we obtained the following new rules [LPAR04]:

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

(14) Obligation:

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

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

(15) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(h(h(z0)), f(x1, a)) → F(a, f(h(h(h(z0))), x1)) we obtained the following new rules [LPAR04]:

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))

(16) Obligation:

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

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

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.

(17) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(a, f(x1, a)) → F(h(a), x1) we obtained the following new rules [LPAR04]:

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

(18) Obligation:

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

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

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.

(19) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(h(h(z0)), f(x1, a)) → F(h(h(h(z0))), x1) we obtained the following new rules [LPAR04]:

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

(20) Obligation:

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

F(a, f(x1, a)) → F(a, f(h(a), x1))
F(h(a), f(x1, a)) → F(a, f(h(h(a)), x1))
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(x0)), f(f(y_1, a), a)) → F(h(h(h(x0))), f(y_1, 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.

(21) ForwardInstantiation (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(h(a), f(x1, a)) → F(h(h(a)), x1) we obtained the following new rules [LPAR04]:

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

(22) Obligation:

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

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

(23) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [FROCOS05] to decrease Q to the empty set.

(24) Obligation:

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

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

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.

(25) SemLabProof (SOUND transformation)

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 [SEMLAB] we obtain the following labelled TRS.

(26) Obligation:

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

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

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.

(27) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(28) Obligation:

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

F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(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.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_0, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_0, 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.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.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.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.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.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.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.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.1(a.), f.0-1(f.1-1(y_0, a.), a.)) → F.0-0(h.0(h.1(a.)), f.1-1(y_0, a.))
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.0-1(f.0-1(f.0-1(y_1, a.), a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(f.0-1(y_1, a.), a.))
F.0-0(h.1(a.), f.0-1(f.0-1(f.1-1(y_1, a.), a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(f.1-1(y_1, a.), a.))

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.

(29) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, 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.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.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(a.)), f.1-1(x1, a.)) → F.1-0(a., f.0-1(h.0(h.0(h.1(a.))), 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: Polynomial interpretation [POLO]:

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   

(30) Obligation:

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

F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(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.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_0, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_0, 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.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.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.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.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.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.1(a.), f.0-1(f.1-1(y_0, a.), a.)) → F.0-0(h.0(h.1(a.)), f.1-1(y_0, a.))
F.0-0(h.1(a.), f.0-1(f.0-1(f.0-1(y_1, a.), a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(f.0-1(y_1, a.), a.))
F.0-0(h.1(a.), f.0-1(f.0-1(f.1-1(y_1, a.), a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(f.1-1(y_1, a.), a.))

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.

(31) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(32) Obligation:

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

F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(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.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_0, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_0, 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.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.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.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.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.1(a.), f.0-1(f.0-1(f.0-1(y_1, a.), a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(f.0-1(y_1, a.), a.))
F.0-0(h.1(a.), f.0-1(f.0-1(f.1-1(y_1, a.), a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(f.1-1(y_1, a.), a.))

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.

(33) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

F.1-0(a., f.0-1(x1, a.)) → F.1-0(a., f.0-0(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.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_0, a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(y_0, 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.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.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.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.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.1(a.), f.0-1(f.0-1(f.0-1(y_1, a.), a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(f.0-1(y_1, a.), a.))
F.0-0(h.1(a.), f.0-1(f.0-1(f.1-1(y_1, a.), a.), a.)) → F.0-0(h.0(h.1(a.)), f.0-1(f.1-1(y_1, a.), a.))


Used ordering: Polynomial interpretation [POLO]:

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

(34) Obligation:

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.

(35) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(36) TRUE