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Let $B=\{0,1\}$ and the binary operations $\oplus,\cdot$

We define a bijection $\varphi$ s.t.:

$$ \varphi:B \longrightarrow L=\{\mathbf{False},\mathbf{True}\}, $$ $$ \varphi(x):= \begin{cases} \mathbf{True}, \quad x=1 \\ \mathbf{False}, \quad x=0 \\ \end{cases} $$ In addition, it satisfies $$ \varphi(a \oplus b)=\varphi(a) \lor \varphi(b) $$ $$ \varphi(a \cdot b)=\varphi(a) \land \varphi(b) $$ Thus, $\varphi$ is an isomorphism between $(B,\oplus,\cdot)$ and $(L,\lor,\land)$.

Does this prove that $B$ equipped with $\oplus$ and $\cdot$ is a Boolean algebra?

Update: As for negation, we might need to introduce a unary operation $(\bullet)^*$ on $B$, s.t. $1^*=0$ and $0^*=1$, which is isomoprhic to $\neg$ on $L$ for the proof to be complete.

Also: $a \oplus b:=a+b-ab$

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  • $\begingroup$ Usually boolean algebras also have $\lnot$ operation, but other than that your isomorphism fixes both operations, so yes, their properties are satisfied. $\endgroup$
    – Artur Riazanov
    Commented Dec 18, 2018 at 15:27
  • $\begingroup$ What does $\oplus$ mean here? If it is the usual addition modulo $2$, then clearly $\varphi(a\oplus b)=\varphi(a)\vee \varphi(b)$ is false. $\endgroup$
    – user593746
    Commented Dec 18, 2018 at 15:45
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    $\begingroup$ @Zvi You're right. Let's then define $\oplus$ as: $$ a \oplus b := a+b-ab $$ $\endgroup$
    – Andrew
    Commented Dec 18, 2018 at 15:55

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