Step of Proof: p-fun-exp-add-sq
11,40
postcript
pdf
Inference at
*
2
2
1
2
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
(do-apply(
f
^
m
- 1;
x
))) ~ (
f
o
f
^
n
- 1 (do-apply(
f
o
f
^
m
- 1 ;
x
)))
latex
by (RW (AddrC [2;2] (UnfoldsC ``p-compose``)) 0)
CollapseTHEN ((RepUR ``do-apply`` ( 0)
)
Co
CollapseTHEN ((Fold `do-apply` 0)
CollapseTHEN (((if (0
C
) =0 then SplitOnConclITE else SplitOnHypITE (0))
)
CollapseTHENA (Auto
)
)
)
)
latex
C
1
: .....truecase..... NILNIL
C1:
13.
can-apply(
f
^
m
- 1;
x
)
C1:
(
f
o
f
^
n
(do-apply(
f
^
m
- 1;
x
))) ~ (
f
o
f
^
n
- 1 (outl(
f
(do-apply(
f
^
m
- 1;
x
)))))
C
2
: .....falsecase..... NILNIL
C2:
13.
(
can-apply(
f
^
m
- 1;
x
))
C2:
(
f
o
f
^
n
(do-apply(
f
^
m
- 1;
x
))) ~ (
f
o
f
^
n
- 1 (outl(
f
^
m
- 1(
x
))))
C
.
Definitions
f
o
g
,
do-apply(
f
;
x
)
,
left
+
right
,
Unit
,
t
T
,
P
Q
,
P
&
Q
,
x
:
A
B
(
x
)
,
A
,
b
,
s
=
t
,
,
x
:
A
.
B
(
x
)
,
P
Q
,
x
:
A
B
(
x
)
Lemmas
eqtt
to
assert
,
iff
transitivity
,
eqff
to
assert
,
assert
of
bnot
origin