Visibility is controlled by the first bit of the group code 70.
In attached LISP file this group code is stored in xx variable: (setq xx (cdr (assoc 70 ed))).
This is description of all four bits, but I think you are not interested in other three (2, 4 and 8):
1 = Attribute is invisible (does not appear)
2 = This is a constant attribute
4 = Verification is required on input of this attribute
8 = Attribute is preset (no prompt during insertion)
Some examples (bit 1 which controls visibility is marked red):
Group code=9. Binary it is 1001
Attribute is invisible (bit1=1), Not constant (bit2=0), Verification not required (bit3=0), Preset(bit4=1)
Group code=12. Binary it is 1100
Attribute is visible (bit1=0), Not constant (bit2=0), Verification is required (bit3=1), Preset(bit4=1)
Group code=3. Binary it is 0011
Attribute is invisible (bit1=1), Constant (bit2=1), Verification not required (bit3=0), Not preset(bit4=0)
Group code=1. Binary it is 0001
Attribute is invisible (bit1=1), Not constant (bit2=0), Verification not required (bit3=0), Not preset(bit4=0)
If (= (logand xx 1) 0) expression IS TRUE than the first bit is eqaul to 0 (that means that attribute is visible).
To make it invisible use (1+ xx) expression. This expression only increments first bit without changing other bits.
If you want to make it visible than do the opposite:
If (= (logand xx 1) 0) expression IS NOT TRUE than the first bit is eqaul to 1 (that means that attribute is invisible)
To make it visible use (1- xx) expression. This expression only decrements first bit without changing other bits.
Regards,
Vladimir