Kyle Sutton

2022-07-13

Proof of commutative property in Boolean algebra

$a\vee b=b\vee a$

$a\wedge b=b\wedge a$

$a\vee b=b\vee a$

$a\wedge b=b\wedge a$

conveneau71

Beginner2022-07-14Added 17 answers

First we prove idempotency $a=a\vee a$, though we might not need it later on.

$a=a\vee 0=a\vee (a\wedge {a}^{\prime})=(a\vee a)\wedge (a\vee {a}^{\prime})=(a\vee a)\wedge 1=a\vee a$

Second, we prove uniqueness of the complement, in the sense that

$a\wedge b=0,\text{}b\vee a=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}b={a}^{\prime}$

${a}^{\prime}={a}^{\prime}\wedge 1={a}^{\prime}\wedge (b\vee a)=({a}^{\prime}\wedge b)\vee ({a}^{\prime}\wedge a)={a}^{\prime}\wedge b\phantom{\rule{0ex}{0ex}}b=1\wedge b=(a\vee {a}^{\prime})\wedge b=(a\wedge b)\vee ({a}^{\prime}\wedge b)={a}^{\prime}\wedge b$

In particular, it implies ${a}^{\u2033}=a$.

Then certain forms of absorbance follows: $a=a\vee (b\wedge a)$

${a}^{\prime}\vee (a\vee (b\wedge a))=1\phantom{\rule{0ex}{0ex}}(a\vee (b\wedge a))\wedge {a}^{\prime}=(a\wedge {a}^{\prime})\vee (b\wedge a\wedge {a}^{\prime})=0$

We similarly get $a=(a\wedge b)\vee a$, and two other equations by duality.

Then, we get a key lemma: $a\vee b=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{a}^{\prime}={a}^{\prime}\wedge b\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}b\vee a=1$:

Supposed $a\vee b=1$, we get ${a}^{\prime}={a}^{\prime}\wedge 1={a}^{\prime}\wedge (a\vee b)=({a}^{\prime}\wedge a)\vee ({a}^{\prime}\wedge b)={a}^{\prime}\wedge b$.

Supposed ${a}^{\prime}={a}^{\prime}\wedge b$, we get $b\vee {a}^{\prime}=b\vee ({a}^{\prime}\wedge b)=b$ by absorbance, so

$\text{}b\vee a=(b\vee {a}^{\prime})\vee a=1$

Note that this implies $a\vee x\vee {a}^{\prime}=1$ for any $x$, as we have $\text{}(x\vee {a}^{\prime})\vee a=1$.

Using their dual statements as well ($a\wedge b=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}b\wedge a=0$ 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 ${a}^{\prime}\wedge {b}^{\prime}$:

$(a\vee b)\vee ({a}^{\prime}\wedge {b}^{\prime})=(a\vee b\vee {a}^{\prime})\wedge (a\vee b\vee {b}^{\prime})=1\phantom{\rule{0ex}{0ex}}(b\vee a)\vee ({a}^{\prime}\wedge {b}^{\prime})=(b\vee a\vee {a}^{\prime})\wedge (b\vee a\vee {b}^{\prime})=1\phantom{\rule{0ex}{0ex}}-\cdot -\cdot -\phantom{\rule{0ex}{0ex}}({a}^{\prime}\wedge {b}^{\prime})\wedge (a\vee b)=({a}^{\prime}\wedge {b}^{\prime}\wedge a)\vee ({a}^{\prime}\wedge {b}^{\prime}\wedge b)=0\phantom{\rule{0ex}{0ex}}({a}^{\prime}\wedge {b}^{\prime})\wedge (b\vee a)=({a}^{\prime}\wedge {b}^{\prime}\wedge b)\vee ({a}^{\prime}\wedge {b}^{\prime}\wedge a)=0$

$a=a\vee 0=a\vee (a\wedge {a}^{\prime})=(a\vee a)\wedge (a\vee {a}^{\prime})=(a\vee a)\wedge 1=a\vee a$

Second, we prove uniqueness of the complement, in the sense that

$a\wedge b=0,\text{}b\vee a=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}b={a}^{\prime}$

${a}^{\prime}={a}^{\prime}\wedge 1={a}^{\prime}\wedge (b\vee a)=({a}^{\prime}\wedge b)\vee ({a}^{\prime}\wedge a)={a}^{\prime}\wedge b\phantom{\rule{0ex}{0ex}}b=1\wedge b=(a\vee {a}^{\prime})\wedge b=(a\wedge b)\vee ({a}^{\prime}\wedge b)={a}^{\prime}\wedge b$

In particular, it implies ${a}^{\u2033}=a$.

Then certain forms of absorbance follows: $a=a\vee (b\wedge a)$

${a}^{\prime}\vee (a\vee (b\wedge a))=1\phantom{\rule{0ex}{0ex}}(a\vee (b\wedge a))\wedge {a}^{\prime}=(a\wedge {a}^{\prime})\vee (b\wedge a\wedge {a}^{\prime})=0$

We similarly get $a=(a\wedge b)\vee a$, and two other equations by duality.

Then, we get a key lemma: $a\vee b=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{a}^{\prime}={a}^{\prime}\wedge b\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}b\vee a=1$:

Supposed $a\vee b=1$, we get ${a}^{\prime}={a}^{\prime}\wedge 1={a}^{\prime}\wedge (a\vee b)=({a}^{\prime}\wedge a)\vee ({a}^{\prime}\wedge b)={a}^{\prime}\wedge b$.

Supposed ${a}^{\prime}={a}^{\prime}\wedge b$, we get $b\vee {a}^{\prime}=b\vee ({a}^{\prime}\wedge b)=b$ by absorbance, so

$\text{}b\vee a=(b\vee {a}^{\prime})\vee a=1$

Note that this implies $a\vee x\vee {a}^{\prime}=1$ for any $x$, as we have $\text{}(x\vee {a}^{\prime})\vee a=1$.

Using their dual statements as well ($a\wedge b=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}b\wedge a=0$ 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 ${a}^{\prime}\wedge {b}^{\prime}$:

$(a\vee b)\vee ({a}^{\prime}\wedge {b}^{\prime})=(a\vee b\vee {a}^{\prime})\wedge (a\vee b\vee {b}^{\prime})=1\phantom{\rule{0ex}{0ex}}(b\vee a)\vee ({a}^{\prime}\wedge {b}^{\prime})=(b\vee a\vee {a}^{\prime})\wedge (b\vee a\vee {b}^{\prime})=1\phantom{\rule{0ex}{0ex}}-\cdot -\cdot -\phantom{\rule{0ex}{0ex}}({a}^{\prime}\wedge {b}^{\prime})\wedge (a\vee b)=({a}^{\prime}\wedge {b}^{\prime}\wedge a)\vee ({a}^{\prime}\wedge {b}^{\prime}\wedge b)=0\phantom{\rule{0ex}{0ex}}({a}^{\prime}\wedge {b}^{\prime})\wedge (b\vee a)=({a}^{\prime}\wedge {b}^{\prime}\wedge b)\vee ({a}^{\prime}\wedge {b}^{\prime}\wedge a)=0$

gaiaecologicaq2

Beginner2022-07-15Added 6 answers

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.

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Show that the image of ${\mathbb{P}}^{n}\times {\mathbb{P}}^{m}$ under the Segre embedding $\psi $ is actually irreducible.

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My attempt: Since $\mathfrak{p}\subseteq \mathfrak{q}$, we have $\mathfrak{m}={S}^{-1}\mathfrak{p}\subseteq {S}^{-1}\mathfrak{q}=\mathfrak{n}\subseteq {B}_{\mathfrak{p}}$. But m is maximal in ${A}_{\mathfrak{p}}$, which is not necessarily maximal in ${B}_{\mathfrak{p}}$. I can't get $\mathfrak{m}=\mathfrak{n}$ by this.

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and

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(ii) Let $\phi $ be a multiplicative functional on a ${C}^{\ast}$-algebra $\mathcal{A}$. Show that $\phi $ is a pure state on $\mathcal{A}$.

(iii) Show that the pure and multiplicative states coincide for commutative $\mathcal{A}$.

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