Kyle Sutton

2022-07-13

Proof of commutative property in Boolean algebra
$a\vee b=b\vee a$
$a\wedge b=b\wedge a$

conveneau71

First we prove idempotency $a=a\vee a$, though we might not need it later on.
$a=a\vee 0=a\vee \left(a\wedge {a}^{\prime }\right)=\left(a\vee a\right)\wedge \left(a\vee {a}^{\prime }\right)=\left(a\vee a\right)\wedge 1=a\vee a$
Second, we prove uniqueness of the complement, in the sense that

${a}^{\prime }={a}^{\prime }\wedge 1={a}^{\prime }\wedge \left(b\vee a\right)=\left({a}^{\prime }\wedge b\right)\vee \left({a}^{\prime }\wedge a\right)={a}^{\prime }\wedge b\phantom{\rule{0ex}{0ex}}b=1\wedge b=\left(a\vee {a}^{\prime }\right)\wedge b=\left(a\wedge b\right)\vee \left({a}^{\prime }\wedge b\right)={a}^{\prime }\wedge b$
In particular, it implies ${a}^{″}=a$.
Then certain forms of absorbance follows: $a=a\vee \left(b\wedge a\right)$
${a}^{\prime }\vee \left(a\vee \left(b\wedge a\right)\right)=1\phantom{\rule{0ex}{0ex}}\left(a\vee \left(b\wedge a\right)\right)\wedge {a}^{\prime }=\left(a\wedge {a}^{\prime }\right)\vee \left(b\wedge a\wedge {a}^{\prime }\right)=0$
We similarly get $a=\left(a\wedge b\right)\vee a$, and two other equations by duality.
Then, we get a key lemma: $a\vee b=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{a}^{\prime }={a}^{\prime }\wedge b\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}b\vee a=1$:
Supposed $a\vee b=1$, we get ${a}^{\prime }={a}^{\prime }\wedge 1={a}^{\prime }\wedge \left(a\vee b\right)=\left({a}^{\prime }\wedge a\right)\vee \left({a}^{\prime }\wedge b\right)={a}^{\prime }\wedge b$.
Supposed ${a}^{\prime }={a}^{\prime }\wedge b$, we get $b\vee {a}^{\prime }=b\vee \left({a}^{\prime }\wedge b\right)=b$ by absorbance, so

Note that this implies $a\vee x\vee {a}^{\prime }=1$ for any $x$, as we have .
Using their dual statements as well ($a\wedge b=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}b\wedge a=0$ and ${a}^{\prime }\wedge x\wedge a=0$), we can finally arrive to commutativity by observing that both $a\vee b$ and $b\vee a$ are complements of ${a}^{\prime }\wedge {b}^{\prime }$:
$\left(a\vee b\right)\vee \left({a}^{\prime }\wedge {b}^{\prime }\right)=\left(a\vee b\vee {a}^{\prime }\right)\wedge \left(a\vee b\vee {b}^{\prime }\right)=1\phantom{\rule{0ex}{0ex}}\left(b\vee a\right)\vee \left({a}^{\prime }\wedge {b}^{\prime }\right)=\left(b\vee a\vee {a}^{\prime }\right)\wedge \left(b\vee a\vee {b}^{\prime }\right)=1\phantom{\rule{0ex}{0ex}}-\cdot -\cdot -\phantom{\rule{0ex}{0ex}}\left({a}^{\prime }\wedge {b}^{\prime }\right)\wedge \left(a\vee b\right)=\left({a}^{\prime }\wedge {b}^{\prime }\wedge a\right)\vee \left({a}^{\prime }\wedge {b}^{\prime }\wedge b\right)=0\phantom{\rule{0ex}{0ex}}\left({a}^{\prime }\wedge {b}^{\prime }\right)\wedge \left(b\vee a\right)=\left({a}^{\prime }\wedge {b}^{\prime }\wedge b\right)\vee \left({a}^{\prime }\wedge {b}^{\prime }\wedge a\right)=0$

gaiaecologicaq2