Solution of the equation E(M)⊕E(N) => E(A)*¬E(M & N) ≡ 1 via the calculus of basic predicates according to E.A.Mironchick

In general we follow guidelines of technique developed in
http://kpolyakov.spb.ru/download/mea18bit.pdf

Per link mentioned above (quoting Helen A. Mironchick)
Let Et (x) be a predicate whose truth set is all x for which x & t ≠ 0.
If t is a power of two, then such a predicate will be called basic.
The basic predicate describes (fixes) a single unit in the binary notation.
Further, for brevity, the predicate Et (x) will be denoted by E(t);
we will also denote the truth set of this predicate.
(quoting ends)


Denote by {X} the binary representation of a natural number X.
The core statement of the post below  is :-

Let R, M, N be natural numbers. R is the minimum
satisfying the condition {M OR N} = {R OR {M & N}},
where "OR" is a bitwise disjunction, and "&" is a bitwise conjunction
Then the smallest A satisfying the equation
E(M)⊕E(N)=>E(A)*¬E (M & N) ≡ 1 would be equal R.

First we intend to show that, E(M) v E(N) = E(R) v E(M&N).
Notice also that everywhere below, "*" is "^".

Consider expansions in the logical sum of  basic predicates.
E (M) and E(N). All pairs of equal basic predicates will be
collapsed into one and the logical sum of such predicate pairs
will obviously give E(M&N). The logical sum of all those
remaining is exactly E (R). It remains to apply the formulas
of De Morgan.

               ¬(E(M) v E(N)) = ¬(E(R) v E(M & N))

and get the required equality below

               ¬E(M)*¬E(N) = ¬E(R)*¬E(M&N)  (1)

Bitwise2 has a familiar formula. Due to the fact that ¬E(N) = Z(N)

  Z(M)*Z(N) = Z(M OR N) = Z(R)*Z(M & N)

See :-   Solving the equation ¬Z(M)⊕¬Z(N) => ¬A*Z(M&N) ≡ 1 in the Bitwise2 technique  https://informatics-ege.blogspot.com/2018/12/zmn-am-1-bitwise2.html

Thus, ¬E(M)*¬E(N) = ¬E(R)*¬E (M & N) can be obtained
as a result of Statement 3 of http://kpolyakov.spb.ru/download/bitwise2.pdf

   Convert the original equation as follows

   E(M)⊕E(N) => E(A)*¬E(M&N) ≡ 1
   (E(M)≡E(N)) v E(A)*¬E(M&N) ≡ 1
 ¬E(M)*¬E(N) v E(M)*E(N) v E(A)*¬E(M&N) ≡ 1

    From the decomposition of M and N into basic predicates
    define the numbers REST-M and REST-N such that
    each of them has no common unit bits with M&N and in doing so
    obtain

     {REST-M} + {M & N} = {M}
     {REST-N} + {M & N} = {N}

     Consequently

     E(M) = E(REST-M) v E(M&N)
     E(N) = E(REST-N) v E(M&N)

Apply formula (1) to ¬E(M)*¬E(N):-

   ¬E(R)*¬E(M&N) v (E(REST-M) v E(M & N))*(E(REST-N) v E(M&N)) v
       v E(A)*¬E(M&N) ≡ 1
   ¬E(R)*¬E(M&N) v E(REST-M)*E(REST-N) v E (M&N) v
       v E(A)*¬E(M&N) ≡ 1
   ¬E(R) v E(REST-M)*E(REST-N) v E(M&N) v E(A)≡ 1
  
    Thus A (min) = R

The rest is not critical for the core result
already obtained, but makes things quite clear
Here we rely the duality between the calculus
of basic predicates and Bitwise2 to make sure
that ∃ x : E(REST-M)(x)*E(REST-N)(x) = False

E(REST-M)*E(REST-N) = ¬Z(REST-M)*¬Z(REST-N)
¬Z(REST-M)*¬Z(REST-N) = ¬(Z(REST-M) v Z(REST-N))
∃ x : Z(REST-M)(x) v Z(REST-N)(x) =True
∃ x : ¬(Z(REST-M)(x) v Z(REST-N)(x)) = ¬Z((REST-M)(x)&(REST-M)(x)) = ¬Z(0)
¬Z(0) = ¬True = False

Finally we get ∃ x : E(REST-M)(x)*E(REST-N)(x)=False
So, in fact:-

 ∀ х :  ¬E(R)(x)*¬E(M&N)(x) v (E(REST-M)(x) v E(M & N))(x)*(E(REST-N)(x) v E(M&N)(x)) v E(A)(x)*¬E(M&N)(x) = 1

means that
 
   ¬E(R)*¬E(M&N) v E (M&N) v E(A)*¬E(M&N) ≡ 1
   ¬E(R) v E(M&N) v E(A)≡ 1


Links
1. http://kpolyakov.spb.ru/download/mea18bit.pdf