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VMs: Re: Numbered transcription

01/10/02 10:54:59, Gabriel Landini <G.Landini@xxxxxxxxxx> wrote:

>On Tuesday 01 October 2002 1:09 am, GC wrote:
>> EVA has already made the assumption that an <a> is not a combined
>> <c,i>, or that an <o> is not a 'c, backward-c'.  In fact EVA
>> relies on assumptions that several multiple-stroke glyphs are
>> clearly units. 

>I don't see any problem with that. Why don't you use Bitrans to convert all 
><a> into <ci>?

Let us grant that <o> is 'c,  backward-c'. I propose
that 'c' is  'upper half of c, lower half of c'.
Next that 'upper half of c' is 'upper quarter of c,
lower half of upper half of c'. Next....

But let us see <o> as 'c, backward-c' first.
<c> is <c>, for backward-c, just pull an unary
operator out of the keyboard, say... & (why not?)
So 'backward c' is '&c'. Now, to express that
<c> and <&c> are not separate, that they are
connected, just pull a binary operator out of
the keyboard, say... a period. Therefore,
EVA <o> -> SuperEVA <c.&c>

But I'm not happy about SuperEVA <c>, seeing
that I'd like to interpret it as 'upper half of
c, lower half of c'. So let's have SuperDuperEVA.
All we need an 'upper half of c' out of the 
keyboard. Well, another unary operator will
do... ~ (why not?), meaning "upper half of".
Here: <~c>. And this one for "lower half": _
So SuperEVA <c> is SuperDuperEVA... wait...
upper half and lower half are connected, yes,
but _vertically_, not horizontally. Another
binary operator. Colon (:) will do. Ready
now. SuperEVA <c> is SuperDuperEVA <~c:_c>
Now, the SuperDuperEVA of SuperEVA <&c>
(remember? see above) is clearly <&(~c:_c)>.
So that EVA <o> is SuperDuperEVA <(~c:_c).&(~c:_c)>
Of course, it become much simpler if we
agree that SuperDuperEVA <c> is actually, say,
the upper half of EVA <c>. All we need is
another unary operator to turn it upside down,
say !, eg <!c>. Then EVA <o> becomes much
more simply SuperDuperEVA <(c:!c).&(c:!c)>