# Proof of commutative property in Boolean algebra a

Proof of commutative property in Boolean algebra
$a\vee b=b\vee a$
$a\wedge b=b\wedge a$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

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 ${a}^{\prime }\wedge {b}^{\prime }$:
$\left(a\vee b\right)\vee \left({a}^{\prime }\wedge {b}^{\prime }\right)=\left(a\vee b\vee {a}^{\prime }\right)\wedge \left(a\vee b\vee {b}^{\prime }\right)=1\phantom{\rule{0ex}{0ex}}\left(b\vee a\right)\vee \left({a}^{\prime }\wedge {b}^{\prime }\right)=\left(b\vee a\vee {a}^{\prime }\right)\wedge \left(b\vee a\vee {b}^{\prime }\right)=1\phantom{\rule{0ex}{0ex}}-\cdot -\cdot -\phantom{\rule{0ex}{0ex}}\left({a}^{\prime }\wedge {b}^{\prime }\right)\wedge \left(a\vee b\right)=\left({a}^{\prime }\wedge {b}^{\prime }\wedge a\right)\vee \left({a}^{\prime }\wedge {b}^{\prime }\wedge b\right)=0\phantom{\rule{0ex}{0ex}}\left({a}^{\prime }\wedge {b}^{\prime }\right)\wedge \left(b\vee a\right)=\left({a}^{\prime }\wedge {b}^{\prime }\wedge b\right)\vee \left({a}^{\prime }\wedge {b}^{\prime }\wedge a\right)=0$
###### Not exactly what you’re looking for?
gaiaecologicaq2
Your proof of join-idempotency uses a form of distributivity that is not part of the axioms. Of course that one will follow from the ones we have and the absorption laws, which could be proven first. On the other hand, if I didn't miss anything, you didn't really use any idempotency law.