Step of Proof: p-fun-exp-add-sq
11,40
postcript
pdf
Inference at
*
2
2
1
I
of proof for Lemma
p-fun-exp-add-sq
:
1.
A
: Type
2.
f
:
A
(
A
+ Top)
3.
x
:
A
4.
m
:
5. 0 <
m
6.
n
:
. (
can-apply(
f
^
m
- 1;
x
))
((
f
^
n
+(
m
- 1)(
x
)) ~ (
f
^
n
(do-apply(
f
^
m
- 1;
x
))))
7.
n
:
8.
can-apply(
f
^
m
;
x
)
9.
(
n
= 0)
10.
(
n
+
m
= 0)
11.
(
n
= 0)
12.
(
m
= 0)
(
f
o
f
^(
n
+
m
) - 1 (
x
)) ~ (
f
o
f
^
n
- 1 (do-apply(
f
o
f
^
m
- 1 ;
x
)))
latex
by Subst' (
f
o
f
^(
n
+
m
) - 1 (
x
)) ~ (
f
o
f
^
n
(do-apply(
f
^
m
- 1;
x
))) ( 0)
latex
1
: .....equality..... NILNIL
1:
(
f
o
f
^(
n
+
m
) - 1 (
x
)) ~ (
f
o
f
^
n
(do-apply(
f
^
m
- 1;
x
)))
2
:
2:
(
f
o
f
^
n
(do-apply(
f
^
m
- 1;
x
))) ~ (
f
o
f
^
n
- 1 (do-apply(
f
o
f
^
m
- 1 ;
x
)))
.
Definitions
s
~
t
,
n
+
m
,
f
(
a
)
,
f
o
g
,
do-apply(
f
;
x
)
,
f
^
n
,
n
-
m
,
#$n
origin