Step of Proof: append_overlapping_sublists
11,40
postcript
pdf
Inference at
*
1
2
3
I
of proof for Lemma
append
overlapping
sublists
:
.....wf..... NILNIL
1.
T
: Type
2.
L1
:
T
List
3.
L2
:
T
List
4.
L
:
T
List
5.
x
:
T
6.
i
,
j
:
. (
i
< ||
L
||)
(
j
< ||
L
||)
(
(
i
=
j
))
(
(
L
[
i
] =
L
[
j
]))
7.
f1
: {0..||
L1
@ [
x
]||
}
{0..||
L
||
}
8. increasing(
f1
;||
L1
@ [
x
]||)
9.
j
:{0..||
L1
@ [
x
]||
}. (
L1
@ [
x
])[
j
] =
L
[(
f1
(
j
))]
10.
f
: {0..(||
L2
||+1)
}
{0..||
L
||
}
11. increasing(
f
;||
L2
||+1)
12.
j
:{0..(||
L2
||+1)
}. [
x
/
L2
][
j
] =
L
[(
f
(
j
))]
13. ||
L1
@ [
x
/
L2
]|| = ||
L1
||+||
L2
||+1
14. ||[]||
0
15.
f2
: {0..||
L1
@ [
x
/
L2
]||
}
{0..||
L
||
}
(increasing(
f2
;||
L1
@ [
x
/
L2
]||)
(
& (
j
:{0..||
L1
@ [
x
/
L2
]||
}. (
L1
@ [
x
/
L2
])[
j
] =
L
[(
f2
(
j
))]))
latex
by (Auto_aux (first_nat 1:n) ((first_nat 2:n),(first_nat 3:n)) (first_tok SupInf:t) inil_term)
b
latex
.
Definitions
suptype(
S
;
T
)
,
S
T
,
False
,
A
B
,
Y
,
t
...$L
,
P
&
Q
,
,
t
T
,
i
j
,
A
,
||
as
||
,
P
Q
,
x
:
A
.
B
(
x
)
,
{
i
..
j
}
Lemmas
select
wf
,
length
wf1
,
int
seg
wf
,
length
append
,
append
wf
,
length
nil
,
non
neg
length
,
length
cons
,
increasing
wf
origin