Bitwise2 extended functionality

Down here we sequentially apply

¬A + A*B = (¬A + A)*(¬A + B) = ¬A + B
 A + ¬A*B = (A + ¬A)*(A + B) = A + B

********************************************************************
(x&248 ¬= 152)v(x&252 = 156)v(x&250 = 154)v(x&A = 0) = 1
********************************************************************
252 = 11111100
156 = 10011100
248 = 11111000
152 = 10011000
250 = 11111010
154 = 10011010

(x&248 ¬= 152) = ¬Z(96) + Z(128) + Z(16) + Z(8)
(x&252 =  156)  =  Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(4)
(x&250 =  154)  =  Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(2)

¬Z(96) + Z(128) + Z(16) + Z(8) + Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(4)
       + Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(2) + A = 1

Converting Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(4) and get
¬Z(96) + Z(128) + Z(16) + Z(8) entries like before run (1)
¬Z(96) + Z(128) + Z(16) + Z(8) + ¬Z(4)
       + Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(2)  + A =  1

Converting Z(96)*¬Z(128)*¬Z(16)*¬Z(8)*¬Z(2) and get
¬Z(96) + Z(128) + Z(16) + Z(8) entries like before run (2)
¬Z(96) + Z(128) + Z(16) + Z(8) + ¬Z(4) + ¬Z(2)  + A =  1

Now we are ready to apply De Morgan rules :-
     ¬(Z(96)*Z(4)*Z(2)) + Z(128) + Z(16) + Z(8) + A = 1

Finally getting done
Z(102) => Z(128) + Z(16) + Z(8) + A = 1
(Z(102)=>Z(128)) + (Z(102)=>Z(16)) + (Z(102)=>Z(8)) + (Z(102)=>A) = 1
Z(102) => A = 1