Solution three tasks from Eugene Dzhobs recent Video (06/01/20) per bit conjunction related via Algebra of predicates {E(k)}

https://vk.com/inform_web?z=video-184870282_456239124%2F85fc69eef8cd9338ce%2Fpl_wall_-184870282


Below we follow technique proposed in 
http://kpolyakov.spb.ru/download/mea18bit.pdf

 (1)
¬E(21) V (¬E(11)=>E(A)) =1
¬E(16)^¬E(4)^¬E(1) V E(8) V E(2) V E(1) + E(A) =1
Suppress ¬E(1) in conjunction
¬E(16)^¬E(4) V E(11) V E(A) ≡1
¬E(20) V E(11) V E(A) ≡1
Thus A(min) = 20
(2)
E(58)=E(32) V E(16) V E(8) V E(2)
E(22)=E(16) V E(4) V E(2)
¬E58)=¬E(32)^¬E(16)^¬E(8)^¬E(2)
¬E(58) V E(22) V ¬E(A) =1
Suppress ¬E(16) and ¬E(2) in conjunction
¬E(32)^¬E(16)^¬E(8)^¬E(2) V E(16) V E(4) V E(2) V E(A) ≡1
¬E(40) V E(22) V E(A) ≡1
Thus A(min)=40
(3)
(x&A=0) V ((x&69 = 4) => (x&118 = 6)) ≡1
(x&69 = 4)=¬E(64)^E(4)^¬E(1)
(x&69 ! = 4)=E(64) V ¬E(4) V E(1)
(x&118 = 6)=¬E(64)^¬E(32)^¬E(16)^E(4)^E(2)
E(64)  V ¬E(4) V E(1) V ¬E(32)^¬E(16)^E(2) V ¬E(A) ≡1
Suppress ¬E(64) and E(4) in conjunction
E(65) V ¬E(4) V ¬E(32)^¬E(16)^E(2) V ¬E(A) ≡1
Thus A(max)=65

Original idea proposed by Helen Mironchick

Solution equation kind of
(x&67 != 3) v ((x&55 = 7) => (x&A =0))
via Algebra of predicates {E(k)}
(x&67 = 3) = ¬E(64)^E(2)^E(1)
(x&67 != 3) = E(64) V ¬E(2) V ¬E(1)
(x&55 = 7) = ¬E(32)^¬E(16)^E(4)^E(2)^E(1)
(x&55 != 7) = E(32) V E(16) V ¬E(4) V ¬E(2) V ¬E(1)
E(64) V ¬E(2) V ¬E(1) V E(32) V E(16) V
V ¬E(4) V ¬E(2) V ¬E(1) V ¬E(A) ≡1
E(112) V ¬(E(4)^E(2)^E(1)) V ¬E(A) ≡1
Thus A(max) = 112