I've taken the 12360 equations and turned them into table in the following code which should reproduce what it looks like in the addres space of the CPU (except with every byte duplicated since A0 isn't connected). Each row represents a bit, each column represents an address at which that bit becomes 0.
#include <stdio.h>
unsigned short btab[][7] = {
{ 0x023E, 0x047C, 0x05C1, 0x091B, 0x098B, 0x0BF1, 0x0E61 },
{ 0x0263, 0x04B2, 0x05B3, 0x0627, 0x0899, 0x090B, 0x0B80 },
{ 0x0078, 0x03CD, 0x04E3, 0x0645, 0x06BE, 0x0732, 0x0991 },
{ 0x0113, 0x0355, 0x0373, 0x0398, 0x03B9, 0x0627, 0x068D },
{ 0x0162, 0x02AF, 0x02C4, 0x053C, 0x07AD, 0x081D, 0x0A86 },
{ 0x0567, 0x0764, 0x087E, 0x0AE4, 0x0D4E, 0x0FB9, 0x1022 },
{ 0x0207, 0x0470, 0x0683, 0x06F4, 0x096B, 0x0B88, 0x0DF2 },
{ 0x0057, 0x00A8, 0x00A9, 0x0116, 0x0180, 0x01EB, 0x025A },
};
int main() {
int i;
for(i=0;i<8192;i++) {
int j;
unsigned char c = 255;
for(j=0;j<8;j++) {
int k;
for(k=0;k<7;k++)
c &= ~((i == btab[j][k]) << j);
}
putchar(c);
}
return 0;
}Attached is the output of the above code, it is indeed very sparse.