Is this a Boolean algebra? (proof)
$begingroup$
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: $a oplus b:=a+b-ab$
discrete-mathematics proof-verification propositional-calculus boolean-algebra
asked Dec 18 '18 at 15:01
JevautJevaut
1,168312
$endgroup$
add a comment |
$begingroup$
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: $a oplus b:=a+b-ab$
discrete-mathematics proof-verification propositional-calculus boolean-algebra
asked Dec 18 '18 at 15:01
JevautJevaut
1,168312
$endgroup$
$begingroup$
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
Dec 18 '18 at 15:27
$begingroup$
What does $oplus$ mean here? If it is the usual addition modulo $2$, then clearly $varphi(aoplus b)=varphi(a)vee varphi(b)$ is false.
$endgroup$
– user593746
Dec 18 '18 at 15:45
1
$begingroup$
@Zvi You're right. Let's then define $oplus$ as: $$ a oplus b := a+b-ab $$
$endgroup$
– Jevaut
Dec 18 '18 at 15:55
add a comment |
$begingroup$
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: $a oplus b:=a+b-ab$
discrete-mathematics proof-verification propositional-calculus boolean-algebra
asked Dec 18 '18 at 15:01
JevautJevaut
1,168312
$endgroup$
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: $a oplus b:=a+b-ab$
discrete-mathematics proof-verification propositional-calculus boolean-algebra
discrete-mathematics proof-verification propositional-calculus boolean-algebra
asked Dec 18 '18 at 15:01
JevautJevaut
1,168312
asked Dec 18 '18 at 15:01
JevautJevaut
1,168312
edited Dec 18 '18 at 15:56
Jevaut
asked Dec 18 '18 at 15:01
JevautJevaut
1,168312
asked Dec 18 '18 at 15:01
JevautJevaut
1,168312
asked Dec 18 '18 at 15:01
JevautJevaut
1,168312
1,168312
$begingroup$
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
Dec 18 '18 at 15:27
$begingroup$
What does $oplus$ mean here? If it is the usual addition modulo $2$, then clearly $varphi(aoplus b)=varphi(a)vee varphi(b)$ is false.
$endgroup$
– user593746
Dec 18 '18 at 15:45
1
$begingroup$
@Zvi You're right. Let's then define $oplus$ as: $$ a oplus b := a+b-ab $$
$endgroup$
– Jevaut
Dec 18 '18 at 15:55
add a comment |
$begingroup$
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
Dec 18 '18 at 15:27
$begingroup$
What does $oplus$ mean here? If it is the usual addition modulo $2$, then clearly $varphi(aoplus b)=varphi(a)vee varphi(b)$ is false.
$endgroup$
– user593746
Dec 18 '18 at 15:45
1
$begingroup$
@Zvi You're right. Let's then define $oplus$ as: $$ a oplus b := a+b-ab $$
$endgroup$
– Jevaut
Dec 18 '18 at 15:55
$begingroup$
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
Dec 18 '18 at 15:27
$begingroup$
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
Dec 18 '18 at 15:27
$begingroup$
What does $oplus$ mean here? If it is the usual addition modulo $2$, then clearly $varphi(aoplus b)=varphi(a)vee varphi(b)$ is false.
$endgroup$
– user593746
Dec 18 '18 at 15:45
$begingroup$
What does $oplus$ mean here? If it is the usual addition modulo $2$, then clearly $varphi(aoplus b)=varphi(a)vee varphi(b)$ is false.
$endgroup$
– user593746
Dec 18 '18 at 15:45
1
1
$begingroup$
@Zvi You're right. Let's then define $oplus$ as: $$ a oplus b := a+b-ab $$
$endgroup$
– Jevaut
Dec 18 '18 at 15:55
$begingroup$
@Zvi You're right. Let's then define $oplus$ as: $$ a oplus b := a+b-ab $$
$endgroup$
– Jevaut
Dec 18 '18 at 15:55
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045258%2fis-this-a-boolean-algebra-proof%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045258%2fis-this-a-boolean-algebra-proof%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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
Dec 18 '18 at 15:27
$begingroup$
What does $oplus$ mean here? If it is the usual addition modulo $2$, then clearly $varphi(aoplus b)=varphi(a)vee varphi(b)$ is false.
$endgroup$
– user593746
Dec 18 '18 at 15:45
1
$begingroup$
@Zvi You're right. Let's then define $oplus$ as: $$ a oplus b := a+b-ab $$
$endgroup$
– Jevaut
Dec 18 '18 at 15:55