Solution via decomposition into basic predicates several problems 18-th currently posted to Informatics_100 News Wire

Solution via decomposition into basic predicates
E(77)=E(64) v E(8) v E(4) v E(1) ;
E(12)=E(8) v E(4)
Now original equation has been converted per De Morgan rules:-
¬E(77) v E(12) v E(A) ≡ 1;
¬E(64)^¬E(8)^¬E(4)^¬E(1) v E(8) v E(4) v E(A) ≡ 1
Suppress E(8) and E(4) in conjunction
¬E(64)^¬E(1) v E(12) v E(A) ≡ 1 ;
¬E(65) v E(12) v E(A) ≡ 1
Hence A(min) = 65

¬E(33)=>(E(45)=>E(A)) ≡ 1
E(33) v ¬E(45) v E(A) ≡ 1
E(33)=E(32) v E(1) ;
E(45)=E(32) v E(8) v E(4) v E(1)
E(32) v E(1) v ¬E(32)^¬E(8)^¬E(4)^¬E(1) v E(A) ≡ 1
Suppress E(32) and E(1) in conjunction
E(32) v E(1) v ¬E(8)^¬E(4) v E(A) ≡ 1;
E(33) v ¬(E(8) v E(4)) v E(A) ≡ 1
E(33) v ¬E(12) v E(A) ≡ 1
Hence
A(min)=12