Termination proof
1: switching to dependency pairs
The following set of initial dependency pairs has been identified.
|
f#(
empty
,
cons(
a
,
k
)
)
|
→ |
f#(
cons(
a
,
k
)
,
k
)
|
|
f#(
cons(
a
,
k
)
,
y
)
|
→ |
f#(
y
,
k
)
|
1.1: reduction pair processor
Using the following reduction pair
Linear polynomial
interpretation over
the naturals
|
[f
(x1, x2)
]
|
= |
x1 +
2
x2
|
|
[f#
(x1, x2)
]
|
= |
x1 +
2
x2
|
|
[empty]
|
= |
1
|
|
[cons
(x1, x2)
]
|
= |
3
x1 +
3
x2
|
|
[f(x1, ..., xn)]
|
= |
x1 + ... + xn + 1
|
for all other symbols f of arity n
|
one remains with the following pair(s).
|
f#(
cons(
a
,
k
)
,
y
)
|
→ |
f#(
y
,
k
)
|
1.1.1: reduction pair processor
Using the following reduction pair
Linear polynomial
interpretation over
the naturals
|
[f
(x1, x2)
]
|
= |
x1 +
2
x2
|
|
[f#
(x1, x2)
]
|
= |
x1 +
2
x2
|
|
[empty]
|
= |
2
|
|
[cons
(x1, x2)
]
|
= |
3
x1 +
3
x2
+
3
|
|
[f(x1, ..., xn)]
|
= |
x1 + ... + xn + 1
|
for all other symbols f of arity n
|
one remains with the following pair(s).
1.1.1.1: P is empty
All dependency pairs have been removed.