Nuprl Definition : abs-S
11,40
postcript
pdf
abs-S (
j
,
i
,
e
) == (
(
e
In
)) c
((
In
(
e
).1) = <
j
,
i
>)
latex
clarification:
abs-S(
C
;
In
)(
j
,
i
,
e
) == (
(
e
In
)) c
((
In
(
e
).1) = <
j
,
i
>
(
:
C
C
))
latex
Definitions
x
.
A
(
x
)
,
A
c
B
,
b
,
e
X
,
s
=
t
,
x
:
A
B
(
x
)
,
t
.1
,
X
(
e
)
,
<
a
,
b
>
FDL editor aliases
abs-S
origin