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

Links

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