Termination w.r.t. Q proof of Waldmann_06

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

0 QTRS

↳1 Overlay + Local Confluence (⇔, 0 ms)

↳2 QTRS

↳3 DependencyPairsProof (⇔, 0 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 QDP

↳7 TransformationProof (⇔, 0 ms)

↳8 QDP

↳9 TransformationProof (⇔, 0 ms)

↳10 QDP

↳11 TransformationProof (⇔, 0 ms)

↳12 QDP

↳13 TransformationProof (⇔, 0 ms)

↳14 QDP

↳15 TransformationProof (⇔, 0 ms)

↳16 QDP

↳17 TransformationProof (⇔, 0 ms)

↳18 QDP

↳19 TransformationProof (⇔, 0 ms)

↳20 QDP

↳21 SemLabProof (⇒, 149 ms)

↳22 QDP

↳23 DependencyGraphProof (⇔, 0 ms)

↳24 QDP

↳25 MRRProof (⇔, 5 ms)

↳26 QDP

↳27 DependencyGraphProof (⇔, 0 ms)

↳28 QDP

↳29 MRRProof (⇔, 0 ms)

↳30 QDP

↳31 DependencyGraphProof (⇔, 0 ms)

↳32 TRUE

(0) Obligation:

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

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

The set Q consists of the following terms:

f(f(a, x0), x1)

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(a, x0), x1)

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(a, x0), x1)

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

(7) TransformationProof (EQUIVALENT transformation)

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

F(f(a, x0), h(z1)) → F(x0, h(h(z1))) → F(f(a, x0), h(z1)) → F(x0, h(h(z1)))
F(f(a, x0), a) → F(x0, h(a)) → F(f(a, x0), a) → F(x0, h(a))

(8) Obligation:

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

F(f(a, x), y) → F(f(x, h(y)), a)
F(f(a, x0), h(z1)) → F(x0, h(h(z1)))
F(f(a, x0), a) → F(x0, h(a))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(a, x0), x1)

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

(9) TransformationProof (EQUIVALENT transformation)

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

F(f(a, x0), a) → F(f(x0, h(a)), a) → F(f(a, x0), a) → F(f(x0, h(a)), a)
F(f(a, x0), h(h(z1))) → F(f(x0, h(h(h(z1)))), a) → F(f(a, x0), h(h(z1))) → F(f(x0, h(h(h(z1)))), a)
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a) → F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)

(10) Obligation:

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

F(f(a, x0), h(z1)) → F(x0, h(h(z1)))
F(f(a, x0), a) → F(x0, h(a))
F(f(a, x0), a) → F(f(x0, h(a)), a)
F(f(a, x0), h(h(z1))) → F(f(x0, h(h(h(z1)))), a)
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(a, x0), x1)

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

(11) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(f(a, x0), h(z1)) → F(x0, h(h(z1))) we obtained the following new rules [LPAR04]:

F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1)))) → F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1))))
F(f(a, x0), h(a)) → F(x0, h(h(a))) → F(f(a, x0), h(a)) → F(x0, h(h(a)))

(12) Obligation:

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

F(f(a, x0), a) → F(x0, h(a))
F(f(a, x0), a) → F(f(x0, h(a)), a)
F(f(a, x0), h(h(z1))) → F(f(x0, h(h(h(z1)))), a)
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1))))
F(f(a, x0), h(a)) → F(x0, h(h(a)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(a, x0), x1)

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

(13) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(f(a, x0), h(h(z1))) → F(f(x0, h(h(h(z1)))), a) we obtained the following new rules [LPAR04]:

F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a) → F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a)
F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a) → F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a)

(14) Obligation:

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

F(f(a, x0), a) → F(x0, h(a))
F(f(a, x0), a) → F(f(x0, h(a)), a)
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1))))
F(f(a, x0), h(a)) → F(x0, h(h(a)))
F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a)
F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(a, x0), x1)

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

(15) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(f(a, x0), a) → F(x0, h(a)) we obtained the following new rules [LPAR04]:

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

(16) Obligation:

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

F(f(a, x0), a) → F(f(x0, h(a)), a)
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1))))
F(f(a, x0), h(a)) → F(x0, h(h(a)))
F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a)
F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a)
F(f(a, f(a, y_0)), a) → F(f(a, y_0), h(a))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(a, x0), x1)

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

(17) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(f(a, x0), h(h(z1))) → F(x0, h(h(h(z1)))) we obtained the following new rules [LPAR04]:

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

(18) Obligation:

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

F(f(a, x0), a) → F(f(x0, h(a)), a)
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, x0), h(a)) → F(x0, h(h(a)))
F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a)
F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a)
F(f(a, f(a, y_0)), a) → F(f(a, y_0), h(a))
F(f(a, f(a, y_0)), h(h(x1))) → F(f(a, y_0), h(h(h(x1))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(a, x0), x1)

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

(19) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(f(a, x0), h(a)) → F(x0, h(h(a))) we obtained the following new rules [LPAR04]:

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

(20) Obligation:

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

F(f(a, x0), a) → F(f(x0, h(a)), a)
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
F(f(a, x0), h(h(h(z1)))) → F(f(x0, h(h(h(h(z1))))), a)
F(f(a, x0), h(h(a))) → F(f(x0, h(h(h(a)))), a)
F(f(a, f(a, y_0)), a) → F(f(a, y_0), h(a))
F(f(a, f(a, y_0)), h(h(x1))) → F(f(a, y_0), h(h(h(x1))))
F(f(a, f(a, y_0)), h(a)) → F(f(a, y_0), h(h(a)))
F(f(a, f(a, f(a, y_0))), h(a)) → F(f(a, f(a, y_0)), h(h(a)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(f(a, x0), x1)

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

(21) SemLabProof (SOUND transformation)

We found the following model for the rules of the TRSs R and P. 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 QDP problem.

(22) Obligation:

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

F.0-1(f.1-0(a., x0), a.) → F.0-1(f.0-0(x0, h.1(a.)), a.)
F.0-1(f.1-1(a., x0), a.) → F.0-1(f.1-0(x0, h.1(a.)), a.)
F.0-1(f.1-0(a., f.1-0(a., y_0)), a.) → F.0-0(f.1-0(a., y_0), h.1(a.))
F.0-1(f.1-0(a., f.1-1(a., y_0)), a.) → F.0-0(f.1-1(a., y_0), h.1(a.))
F.0-0(f.1-0(a., x0), h.1(a.)) → F.0-1(f.0-0(x0, h.0(h.1(a.))), a.)
F.0-0(f.1-1(a., x0), h.1(a.)) → F.0-1(f.1-0(x0, h.0(h.1(a.))), a.)
F.0-0(f.1-0(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-0(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-1(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-1(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-0(a., x0), h.0(h.1(a.))) → F.0-1(f.0-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-1(a., x0), h.0(h.1(a.))) → F.0-1(f.1-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.1(a.)) → F.0-0(f.1-0(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.1(a.)) → F.0-0(f.1-1(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-0(a., f.1-0(a., y_0))), h.1(a.)) → F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-0(a., f.1-1(a., y_0))), h.1(a.)) → F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.1(x1))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(24) Obligation:

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

F.0-1(f.1-0(a., x0), a.) → F.0-1(f.0-0(x0, h.1(a.)), a.)
F.0-1(f.1-0(a., f.1-0(a., y_0)), a.) → F.0-0(f.1-0(a., y_0), h.1(a.))
F.0-0(f.1-0(a., x0), h.1(a.)) → F.0-1(f.0-0(x0, h.0(h.1(a.))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.1(a.)) → F.0-0(f.1-0(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., x0), h.0(h.1(a.))) → F.0-1(f.0-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-1(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-1(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.1(a.)) → F.0-0(f.1-1(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-1(a., x0), h.0(h.1(a.))) → F.0-1(f.1-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., f.1-0(a., f.1-0(a., y_0))), h.1(a.)) → F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-0(a., f.1-1(a., y_0))), h.1(a.)) → F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(a.)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(25) 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(f.1-1(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-1(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.1-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-1(a., x0), h.0(h.1(a.))) → F.0-1(f.1-0(x0, h.0(h.0(h.1(a.)))), a.)

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

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

(26) Obligation:

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

F.0-1(f.1-0(a., x0), a.) → F.0-1(f.0-0(x0, h.1(a.)), a.)
F.0-1(f.1-0(a., f.1-0(a., y_0)), a.) → F.0-0(f.1-0(a., y_0), h.1(a.))
F.0-0(f.1-0(a., x0), h.1(a.)) → F.0-1(f.0-0(x0, h.0(h.1(a.))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.1(a.)) → F.0-0(f.1-0(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., x0), h.0(h.1(a.))) → F.0-1(f.0-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-1(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., f.1-1(a., y_0)), h.1(a.)) → F.0-0(f.1-1(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-0(a., f.1-0(a., y_0))), h.1(a.)) → F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-0(a., f.1-1(a., y_0))), h.1(a.)) → F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(a.)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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.0-1(f.1-0(a., x0), a.) → F.0-1(f.0-0(x0, h.1(a.)), a.)
F.0-1(f.1-0(a., f.1-0(a., y_0)), a.) → F.0-0(f.1-0(a., y_0), h.1(a.))
F.0-0(f.1-0(a., x0), h.1(a.)) → F.0-1(f.0-0(x0, h.0(h.1(a.))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.1(a.)) → F.0-0(f.1-0(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., x0), h.0(h.1(a.))) → F.0-1(f.0-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-0(a., f.1-0(a., f.1-0(a., y_0))), h.1(a.)) → F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-0(a., f.1-1(a., y_0))), h.1(a.)) → F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(a.)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(29) MRRProof (EQUIVALENT transformation)

F.0-1(f.1-0(a., x0), a.) → F.0-1(f.0-0(x0, h.1(a.)), a.)
F.0-0(f.1-0(a., x0), h.1(a.)) → F.0-1(f.0-0(x0, h.0(h.1(a.))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.1(a.)) → F.0-0(f.1-0(a., y_0), h.0(h.1(a.)))
F.0-0(f.1-0(a., x0), h.0(h.1(a.))) → F.0-1(f.0-0(x0, h.0(h.0(h.1(a.)))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.1(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.1(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.1(z1))))), a.)
F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.0(x1))) → F.0-0(f.1-0(a., y_0), h.0(h.0(h.0(x1))))
F.0-0(f.1-0(a., x0), h.0(h.0(h.0(z1)))) → F.0-1(f.0-0(x0, h.0(h.0(h.0(h.0(z1))))), a.)
F.0-0(f.1-0(a., f.1-0(a., f.1-0(a., y_0))), h.1(a.)) → F.0-0(f.1-0(a., f.1-0(a., y_0)), h.0(h.1(a.)))
F.0-0(f.1-0(a., f.1-0(a., f.1-1(a., y_0))), h.1(a.)) → F.0-0(f.1-0(a., f.1-1(a., y_0)), h.0(h.1(a.)))

Used ordering: Polynomial interpretation [POLO]:

POL(F.0-0(x₁, x₂)) = 1 + x₁ + x₂
POL(F.0-1(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)) = 1 + x₁ + x₂
POL(f.1-1(x₁, x₂)) = 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.0-1(f.1-0(a., f.1-0(a., y_0)), a.) → F.0-0(f.1-0(a., y_0), h.1(a.))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(31) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(0) Obligation:

(1) Overlay + Local Confluence (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Obligation:

(7) TransformationProof (EQUIVALENT transformation)

(8) Obligation:

(9) TransformationProof (EQUIVALENT transformation)

(10) Obligation:

(11) TransformationProof (EQUIVALENT transformation)

(12) Obligation:

(13) TransformationProof (EQUIVALENT transformation)

(14) Obligation:

(15) TransformationProof (EQUIVALENT transformation)

(16) Obligation:

(17) TransformationProof (EQUIVALENT transformation)

(18) Obligation:

(19) TransformationProof (EQUIVALENT transformation)

(20) Obligation:

(21) SemLabProof (SOUND transformation)

(22) Obligation:

(23) DependencyGraphProof (EQUIVALENT transformation)

(24) Obligation:

(25) MRRProof (EQUIVALENT transformation)

(26) Obligation:

(27) DependencyGraphProof (EQUIVALENT transformation)

(28) Obligation:

(29) MRRProof (EQUIVALENT transformation)

(30) Obligation:

(31) DependencyGraphProof (EQUIVALENT transformation)

(32) TRUE