(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) Instantiation (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), 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) Instantiation (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), a) → F(x0, h(a))
(10) 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.
(11) Instantiation (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), h(h(z1))) → F(f(x0, h(h(h(z1)))), a)
F(f(a, x0), h(a)) → F(f(x0, h(h(a))), a)
(12) 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.
(13) Instantiation (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(a)) → F(x0, h(h(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(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.
(15) Instantiation (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(a))) → F(f(x0, h(h(h(a)))), a)
(16) 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.
(17) ForwardInstantiation (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))
(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(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.
(19) ForwardInstantiation (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))))
(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(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.
(21) ForwardInstantiation (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, f(a, y_0))), h(a)) → F(f(a, f(a, y_0)), h(h(a)))
(22) 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.
(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(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))
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.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.
(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.1-1(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.
(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(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
(30) 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.
(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.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.
(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.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.)))
Used ordering: Polynomial interpretation [POLO]:
POL(F.0-0(x1, x2)) = 1 + x1 + x2
POL(F.0-1(x1, x2)) = 1 + 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
(34) Obligation:
Q DP problem:
P is empty.
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.
(35) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(36) TRUE