Original challenging Question proposed by MEA

Updated original

((D(35)⊕D(56)) => ¬D(a)^D(14)) v E(3) ≡ 1

Replace implication by disjunction

(D(35) ≡ D(56)) v ¬D(a)^D(2)^D(7) v E(3) ≡ 1

First notice

E(3)=E(1)+E(2) = ¬(¬E(1)^¬E(2)) = ¬D(4)

So, converting  to following below

(D(35) ≡ D(56)) v ¬D(a)^D(2)^D(7) v  ¬D(4) ≡ 1
D(35)^D(56) v ¬D(35)^¬D(56) v ¬D(A)^D(7) v ¬D(4) ≡ 1

D(35)^D(56) = D(7)^D(5)^D(8)

¬D(35)^¬D(56) = (¬D(5) v ¬D(7))^(¬D(8) v ¬D(7) =
  = ¬D(5)^¬D(8) v ¬D(7)
D(7)^D(5)^D(8) v ¬D(5)^¬D(8) v ¬D(7) v
                     v ¬D(A)^D(7) v ¬D(4) ≡ 1
D(5)^D(8) v ¬D(5)^¬D(8) v
                  v ¬D(7) v ¬D(A) v ¬D(4) ≡ 1
D(40) v ¬D(5)^¬D(8) v ¬D(28) v ¬D(A) ≡ 1

Thus A(min) = 40