Mixing
Proof of Associativity in Boolean Algebra
$begingroup$
I must prove the most basic associativity in boolean algebra and there is two equation to be proved:
(1) a+(b+c) = (a+b)+c (where + indicates OR).
(2) a.(b.c) = (a.b).c (where . indicates AND).
I have a hint to solve this: You can prove that both sides in (1) are equal to [a+(b+c)].[(a+b)+c] (I'm pretty sure that it's coming from idempotency.).
We can use all axioms of boolean algebra: distributivity, commutativity, complements, identity elements, null elements, absorption, idempotency, a = (a')' theorem, a+a'b = a + b theorem (' indicates NOT) except De Morgan's Law. Also duality of boolean algebra for sure.
Please help. Thanks in advance.
boolean-algebra
asked Mar 18 '13 at 9:00
Hazım TürkkanHazım Türkkan
26113
$endgroup$
add a comment |
$begingroup$
I must prove the most basic associativity in boolean algebra and there is two equation to be proved:
(1) a+(b+c) = (a+b)+c (where + indicates OR).
(2) a.(b.c) = (a.b).c (where . indicates AND).
I have a hint to solve this: You can prove that both sides in (1) are equal to [a+(b+c)].[(a+b)+c] (I'm pretty sure that it's coming from idempotency.).
We can use all axioms of boolean algebra: distributivity, commutativity, complements, identity elements, null elements, absorption, idempotency, a = (a')' theorem, a+a'b = a + b theorem (' indicates NOT) except De Morgan's Law. Also duality of boolean algebra for sure.
Please help. Thanks in advance.
boolean-algebra
asked Mar 18 '13 at 9:00
Hazım TürkkanHazım Türkkan
26113
$endgroup$
add a comment |
$begingroup$
I must prove the most basic associativity in boolean algebra and there is two equation to be proved:
(1) a+(b+c) = (a+b)+c (where + indicates OR).
(2) a.(b.c) = (a.b).c (where . indicates AND).
I have a hint to solve this: You can prove that both sides in (1) are equal to [a+(b+c)].[(a+b)+c] (I'm pretty sure that it's coming from idempotency.).
We can use all axioms of boolean algebra: distributivity, commutativity, complements, identity elements, null elements, absorption, idempotency, a = (a')' theorem, a+a'b = a + b theorem (' indicates NOT) except De Morgan's Law. Also duality of boolean algebra for sure.
Please help. Thanks in advance.
boolean-algebra
asked Mar 18 '13 at 9:00
Hazım TürkkanHazım Türkkan
26113
$endgroup$
I must prove the most basic associativity in boolean algebra and there is two equation to be proved:
(1) a+(b+c) = (a+b)+c (where + indicates OR).
(2) a.(b.c) = (a.b).c (where . indicates AND).
I have a hint to solve this: You can prove that both sides in (1) are equal to [a+(b+c)].[(a+b)+c] (I'm pretty sure that it's coming from idempotency.).
We can use all axioms of boolean algebra: distributivity, commutativity, complements, identity elements, null elements, absorption, idempotency, a = (a')' theorem, a+a'b = a + b theorem (' indicates NOT) except De Morgan's Law. Also duality of boolean algebra for sure.
Please help. Thanks in advance.
boolean-algebra
boolean-algebra
asked Mar 18 '13 at 9:00
Hazım TürkkanHazım Türkkan
26113
asked Mar 18 '13 at 9:00
Hazım TürkkanHazım Türkkan
26113
asked Mar 18 '13 at 9:00
Hazım TürkkanHazım Türkkan
26113
asked Mar 18 '13 at 9:00
Hazım TürkkanHazım Türkkan
26113
asked Mar 18 '13 at 9:00
Hazım TürkkanHazım Türkkan
26113
26113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I assume it should be true (and known) that $a + ab = a$.
Assuming this holds, let $x = a+(b+c)$ and $y = (a+b)+c$. We want to show that $x = y$, and following the hint we reduce to showing $x = xy = y$.
I claim that $ax = a, bx = b, cx = c$, and likewise for $y$. We check for $ax$:
$$ ax = aa + a(b+c) = a + a(b+c) = a$$
Likewise, for $bx$:
$$ bx = ba + b(b+c) = ba + (bb+bc) = ba + (b+bc) = ba + b = b$$
The remaining checks are analogous.
Using these identities, you can derive that anything made up of $a,b,c,+,.$ does not change when multiplied by $x$, in particular $yx = x$:
$$ yx = ((a+b)+c)x = (a+b)x+cx = (ax+bx)+cx = (a+b)+c = y$$
You can use a symmetric argument to conclude that $yx = xy = x$, and hence the claim follows.
For products, you can use a similar trick. Let $x = a.(b.c)$ and $y = (a.b).c$. I claim that $ x = x + y = y$. To see this, first note that $x + a = a$ (because $x+a = a+ a.(...) = a$. Secondly, $x+b = b$, because
$$ x+b = a.(b.c) + b = a.(b.c) + a.b + a'.b = a.(b.c+b) + a'.b = a.b + a'.b = b$$
(I hope this is legit). Likewise, $x+c = c$. Finally, $x + y = y$, because:
$$ y = (a.b).c = ((a+x).(b+x)).(c+x) = (a.b + x).(c+x) = (a.b).c + x = y + x$$
(I used the identity $(u+t).(v+t) = u.v + t.u + t.v + t.t = u.v +t$). The proof that $y+x = x$ is symmetric.
answered Mar 18 '13 at 9:37
Jakub KoniecznyJakub Konieczny
9,32311861
$endgroup$
$begingroup$
Thanks for that beatiful answer and it's totally useful. Thanks a lot again. I must ask that, did we also prove the associativity of multiplication? I mean with this process, did we prove the equation a.(b.c)=(a.b).c ?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 12:17
$begingroup$
I overlooked the part of the problem about the products, sorry about that. I just added that, see if I am making sense there.
$endgroup$
– Jakub Konieczny
Mar 18 '13 at 12:43
$begingroup$
Sorry but I didn't understand a point; here is this: I think (a.b+x).(c+x) = (a.b).(c+x) => by x's property and then (a.b).c + (a.b).x Because (a.b).c + x is destroying parentheses of (c+x). Can you explain it?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:42
$begingroup$
Sorry for my rush, I understood it now it came from (a.b+x).(c+x)=(a.b).c + (a.b).x + x.c + x.x and then idempotent law and two absorptions. Thank you so much that's the final and perfect answer. You are the best.
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:54
add a comment |
$begingroup$
The answer from Jakub Konieczny proves $(1):a+(b+c)=(a+b)+c$ just right, but for $(2):acdot(bcdot c)=(acdot b)cdot c,$ I would argue that you should not use it is more elegant not to use $(1)$. After all, they state the same thing but for different operators, so it makes sense to prove them separately.
In particular when proving that $x+b=b$, the statement $$[acdot (bcdot c)]+[(acdot b)+(a'cdot b)]=[acdot(bcdot c+b)]+(a'cdot b) $$
assumes that $$[acdot (bcdot c)]+[(acdot b)+(a'cdot b)]=[acdot(bcdot c)+acdot b)]+(a'cdot b) $$
So, I think it is better to state that $(2)$ holds because of the duality principle or just be carefull about how to prove each step. For example prove that $b+x=b$ as, $$b+x=b+[acdot(bcdot c)]=[b+a]cdot[b+(bcdot c)]=(b+a)
cdot (b)=b+acdot b=b$$
To conclude, if you do not want to just state that $(2)$ holds because of the duality principle, use the dual of each step of $(1)$ to derive $(2)$.
answered Jan 26 at 2:55
giannisl9giannisl9
579
$endgroup$
add a comment |
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%2f333625%2fproof-of-associativity-in-boolean-algebra%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I assume it should be true (and known) that $a + ab = a$.
Assuming this holds, let $x = a+(b+c)$ and $y = (a+b)+c$. We want to show that $x = y$, and following the hint we reduce to showing $x = xy = y$.
I claim that $ax = a, bx = b, cx = c$, and likewise for $y$. We check for $ax$:
$$ ax = aa + a(b+c) = a + a(b+c) = a$$
Likewise, for $bx$:
$$ bx = ba + b(b+c) = ba + (bb+bc) = ba + (b+bc) = ba + b = b$$
The remaining checks are analogous.
Using these identities, you can derive that anything made up of $a,b,c,+,.$ does not change when multiplied by $x$, in particular $yx = x$:
$$ yx = ((a+b)+c)x = (a+b)x+cx = (ax+bx)+cx = (a+b)+c = y$$
You can use a symmetric argument to conclude that $yx = xy = x$, and hence the claim follows.
For products, you can use a similar trick. Let $x = a.(b.c)$ and $y = (a.b).c$. I claim that $ x = x + y = y$. To see this, first note that $x + a = a$ (because $x+a = a+ a.(...) = a$. Secondly, $x+b = b$, because
$$ x+b = a.(b.c) + b = a.(b.c) + a.b + a'.b = a.(b.c+b) + a'.b = a.b + a'.b = b$$
(I hope this is legit). Likewise, $x+c = c$. Finally, $x + y = y$, because:
$$ y = (a.b).c = ((a+x).(b+x)).(c+x) = (a.b + x).(c+x) = (a.b).c + x = y + x$$
(I used the identity $(u+t).(v+t) = u.v + t.u + t.v + t.t = u.v +t$). The proof that $y+x = x$ is symmetric.
answered Mar 18 '13 at 9:37
Jakub KoniecznyJakub Konieczny
9,32311861
$endgroup$
$begingroup$
Thanks for that beatiful answer and it's totally useful. Thanks a lot again. I must ask that, did we also prove the associativity of multiplication? I mean with this process, did we prove the equation a.(b.c)=(a.b).c ?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 12:17
$begingroup$
I overlooked the part of the problem about the products, sorry about that. I just added that, see if I am making sense there.
$endgroup$
– Jakub Konieczny
Mar 18 '13 at 12:43
$begingroup$
Sorry but I didn't understand a point; here is this: I think (a.b+x).(c+x) = (a.b).(c+x) => by x's property and then (a.b).c + (a.b).x Because (a.b).c + x is destroying parentheses of (c+x). Can you explain it?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:42
$begingroup$
Sorry for my rush, I understood it now it came from (a.b+x).(c+x)=(a.b).c + (a.b).x + x.c + x.x and then idempotent law and two absorptions. Thank you so much that's the final and perfect answer. You are the best.
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:54
add a comment |
$begingroup$
I assume it should be true (and known) that $a + ab = a$.
Assuming this holds, let $x = a+(b+c)$ and $y = (a+b)+c$. We want to show that $x = y$, and following the hint we reduce to showing $x = xy = y$.
I claim that $ax = a, bx = b, cx = c$, and likewise for $y$. We check for $ax$:
$$ ax = aa + a(b+c) = a + a(b+c) = a$$
Likewise, for $bx$:
$$ bx = ba + b(b+c) = ba + (bb+bc) = ba + (b+bc) = ba + b = b$$
The remaining checks are analogous.
Using these identities, you can derive that anything made up of $a,b,c,+,.$ does not change when multiplied by $x$, in particular $yx = x$:
$$ yx = ((a+b)+c)x = (a+b)x+cx = (ax+bx)+cx = (a+b)+c = y$$
You can use a symmetric argument to conclude that $yx = xy = x$, and hence the claim follows.
For products, you can use a similar trick. Let $x = a.(b.c)$ and $y = (a.b).c$. I claim that $ x = x + y = y$. To see this, first note that $x + a = a$ (because $x+a = a+ a.(...) = a$. Secondly, $x+b = b$, because
$$ x+b = a.(b.c) + b = a.(b.c) + a.b + a'.b = a.(b.c+b) + a'.b = a.b + a'.b = b$$
(I hope this is legit). Likewise, $x+c = c$. Finally, $x + y = y$, because:
$$ y = (a.b).c = ((a+x).(b+x)).(c+x) = (a.b + x).(c+x) = (a.b).c + x = y + x$$
(I used the identity $(u+t).(v+t) = u.v + t.u + t.v + t.t = u.v +t$). The proof that $y+x = x$ is symmetric.
answered Mar 18 '13 at 9:37
Jakub KoniecznyJakub Konieczny
9,32311861
$endgroup$
$begingroup$
Thanks for that beatiful answer and it's totally useful. Thanks a lot again. I must ask that, did we also prove the associativity of multiplication? I mean with this process, did we prove the equation a.(b.c)=(a.b).c ?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 12:17
$begingroup$
I overlooked the part of the problem about the products, sorry about that. I just added that, see if I am making sense there.
$endgroup$
– Jakub Konieczny
Mar 18 '13 at 12:43
$begingroup$
Sorry but I didn't understand a point; here is this: I think (a.b+x).(c+x) = (a.b).(c+x) => by x's property and then (a.b).c + (a.b).x Because (a.b).c + x is destroying parentheses of (c+x). Can you explain it?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:42
$begingroup$
Sorry for my rush, I understood it now it came from (a.b+x).(c+x)=(a.b).c + (a.b).x + x.c + x.x and then idempotent law and two absorptions. Thank you so much that's the final and perfect answer. You are the best.
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:54
add a comment |
$begingroup$
I assume it should be true (and known) that $a + ab = a$.
Assuming this holds, let $x = a+(b+c)$ and $y = (a+b)+c$. We want to show that $x = y$, and following the hint we reduce to showing $x = xy = y$.
I claim that $ax = a, bx = b, cx = c$, and likewise for $y$. We check for $ax$:
$$ ax = aa + a(b+c) = a + a(b+c) = a$$
Likewise, for $bx$:
$$ bx = ba + b(b+c) = ba + (bb+bc) = ba + (b+bc) = ba + b = b$$
The remaining checks are analogous.
Using these identities, you can derive that anything made up of $a,b,c,+,.$ does not change when multiplied by $x$, in particular $yx = x$:
$$ yx = ((a+b)+c)x = (a+b)x+cx = (ax+bx)+cx = (a+b)+c = y$$
You can use a symmetric argument to conclude that $yx = xy = x$, and hence the claim follows.
For products, you can use a similar trick. Let $x = a.(b.c)$ and $y = (a.b).c$. I claim that $ x = x + y = y$. To see this, first note that $x + a = a$ (because $x+a = a+ a.(...) = a$. Secondly, $x+b = b$, because
$$ x+b = a.(b.c) + b = a.(b.c) + a.b + a'.b = a.(b.c+b) + a'.b = a.b + a'.b = b$$
(I hope this is legit). Likewise, $x+c = c$. Finally, $x + y = y$, because:
$$ y = (a.b).c = ((a+x).(b+x)).(c+x) = (a.b + x).(c+x) = (a.b).c + x = y + x$$
(I used the identity $(u+t).(v+t) = u.v + t.u + t.v + t.t = u.v +t$). The proof that $y+x = x$ is symmetric.
answered Mar 18 '13 at 9:37
Jakub KoniecznyJakub Konieczny
9,32311861
$endgroup$
I assume it should be true (and known) that $a + ab = a$.
Assuming this holds, let $x = a+(b+c)$ and $y = (a+b)+c$. We want to show that $x = y$, and following the hint we reduce to showing $x = xy = y$.
I claim that $ax = a, bx = b, cx = c$, and likewise for $y$. We check for $ax$:
$$ ax = aa + a(b+c) = a + a(b+c) = a$$
Likewise, for $bx$:
$$ bx = ba + b(b+c) = ba + (bb+bc) = ba + (b+bc) = ba + b = b$$
The remaining checks are analogous.
Using these identities, you can derive that anything made up of $a,b,c,+,.$ does not change when multiplied by $x$, in particular $yx = x$:
$$ yx = ((a+b)+c)x = (a+b)x+cx = (ax+bx)+cx = (a+b)+c = y$$
You can use a symmetric argument to conclude that $yx = xy = x$, and hence the claim follows.
For products, you can use a similar trick. Let $x = a.(b.c)$ and $y = (a.b).c$. I claim that $ x = x + y = y$. To see this, first note that $x + a = a$ (because $x+a = a+ a.(...) = a$. Secondly, $x+b = b$, because
$$ x+b = a.(b.c) + b = a.(b.c) + a.b + a'.b = a.(b.c+b) + a'.b = a.b + a'.b = b$$
(I hope this is legit). Likewise, $x+c = c$. Finally, $x + y = y$, because:
$$ y = (a.b).c = ((a+x).(b+x)).(c+x) = (a.b + x).(c+x) = (a.b).c + x = y + x$$
(I used the identity $(u+t).(v+t) = u.v + t.u + t.v + t.t = u.v +t$). The proof that $y+x = x$ is symmetric.
answered Mar 18 '13 at 9:37
Jakub KoniecznyJakub Konieczny
9,32311861
edited Mar 18 '13 at 12:42
answered Mar 18 '13 at 9:37
Jakub KoniecznyJakub Konieczny
9,32311861
answered Mar 18 '13 at 9:37
Jakub KoniecznyJakub Konieczny
9,32311861
answered Mar 18 '13 at 9:37
Jakub KoniecznyJakub Konieczny
9,32311861
9,32311861
$begingroup$
Thanks for that beatiful answer and it's totally useful. Thanks a lot again. I must ask that, did we also prove the associativity of multiplication? I mean with this process, did we prove the equation a.(b.c)=(a.b).c ?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 12:17
$begingroup$
I overlooked the part of the problem about the products, sorry about that. I just added that, see if I am making sense there.
$endgroup$
– Jakub Konieczny
Mar 18 '13 at 12:43
$begingroup$
Sorry but I didn't understand a point; here is this: I think (a.b+x).(c+x) = (a.b).(c+x) => by x's property and then (a.b).c + (a.b).x Because (a.b).c + x is destroying parentheses of (c+x). Can you explain it?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:42
$begingroup$
Sorry for my rush, I understood it now it came from (a.b+x).(c+x)=(a.b).c + (a.b).x + x.c + x.x and then idempotent law and two absorptions. Thank you so much that's the final and perfect answer. You are the best.
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:54
add a comment |
$begingroup$
Thanks for that beatiful answer and it's totally useful. Thanks a lot again. I must ask that, did we also prove the associativity of multiplication? I mean with this process, did we prove the equation a.(b.c)=(a.b).c ?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 12:17
$begingroup$
I overlooked the part of the problem about the products, sorry about that. I just added that, see if I am making sense there.
$endgroup$
– Jakub Konieczny
Mar 18 '13 at 12:43
$begingroup$
Sorry but I didn't understand a point; here is this: I think (a.b+x).(c+x) = (a.b).(c+x) => by x's property and then (a.b).c + (a.b).x Because (a.b).c + x is destroying parentheses of (c+x). Can you explain it?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:42
$begingroup$
Sorry for my rush, I understood it now it came from (a.b+x).(c+x)=(a.b).c + (a.b).x + x.c + x.x and then idempotent law and two absorptions. Thank you so much that's the final and perfect answer. You are the best.
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:54
$begingroup$
Thanks for that beatiful answer and it's totally useful. Thanks a lot again. I must ask that, did we also prove the associativity of multiplication? I mean with this process, did we prove the equation a.(b.c)=(a.b).c ?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 12:17
$begingroup$
Thanks for that beatiful answer and it's totally useful. Thanks a lot again. I must ask that, did we also prove the associativity of multiplication? I mean with this process, did we prove the equation a.(b.c)=(a.b).c ?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 12:17
$begingroup$
I overlooked the part of the problem about the products, sorry about that. I just added that, see if I am making sense there.
$endgroup$
– Jakub Konieczny
Mar 18 '13 at 12:43
$begingroup$
I overlooked the part of the problem about the products, sorry about that. I just added that, see if I am making sense there.
$endgroup$
– Jakub Konieczny
Mar 18 '13 at 12:43
$begingroup$
Sorry but I didn't understand a point; here is this: I think (a.b+x).(c+x) = (a.b).(c+x) => by x's property and then (a.b).c + (a.b).x Because (a.b).c + x is destroying parentheses of (c+x). Can you explain it?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:42
$begingroup$
Sorry but I didn't understand a point; here is this: I think (a.b+x).(c+x) = (a.b).(c+x) => by x's property and then (a.b).c + (a.b).x Because (a.b).c + x is destroying parentheses of (c+x). Can you explain it?
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:42
$begingroup$
Sorry for my rush, I understood it now it came from (a.b+x).(c+x)=(a.b).c + (a.b).x + x.c + x.x and then idempotent law and two absorptions. Thank you so much that's the final and perfect answer. You are the best.
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:54
$begingroup$
Sorry for my rush, I understood it now it came from (a.b+x).(c+x)=(a.b).c + (a.b).x + x.c + x.x and then idempotent law and two absorptions. Thank you so much that's the final and perfect answer. You are the best.
$endgroup$
– Hazım Türkkan
Mar 18 '13 at 14:54
add a comment |
$begingroup$
The answer from Jakub Konieczny proves $(1):a+(b+c)=(a+b)+c$ just right, but for $(2):acdot(bcdot c)=(acdot b)cdot c,$ I would argue that you should not use it is more elegant not to use $(1)$. After all, they state the same thing but for different operators, so it makes sense to prove them separately.
In particular when proving that $x+b=b$, the statement $$[acdot (bcdot c)]+[(acdot b)+(a'cdot b)]=[acdot(bcdot c+b)]+(a'cdot b) $$
assumes that $$[acdot (bcdot c)]+[(acdot b)+(a'cdot b)]=[acdot(bcdot c)+acdot b)]+(a'cdot b) $$
So, I think it is better to state that $(2)$ holds because of the duality principle or just be carefull about how to prove each step. For example prove that $b+x=b$ as, $$b+x=b+[acdot(bcdot c)]=[b+a]cdot[b+(bcdot c)]=(b+a)
cdot (b)=b+acdot b=b$$
To conclude, if you do not want to just state that $(2)$ holds because of the duality principle, use the dual of each step of $(1)$ to derive $(2)$.
answered Jan 26 at 2:55
giannisl9giannisl9
579
$endgroup$
add a comment |
$begingroup$
The answer from Jakub Konieczny proves $(1):a+(b+c)=(a+b)+c$ just right, but for $(2):acdot(bcdot c)=(acdot b)cdot c,$ I would argue that you should not use it is more elegant not to use $(1)$. After all, they state the same thing but for different operators, so it makes sense to prove them separately.
In particular when proving that $x+b=b$, the statement $$[acdot (bcdot c)]+[(acdot b)+(a'cdot b)]=[acdot(bcdot c+b)]+(a'cdot b) $$
assumes that $$[acdot (bcdot c)]+[(acdot b)+(a'cdot b)]=[acdot(bcdot c)+acdot b)]+(a'cdot b) $$
So, I think it is better to state that $(2)$ holds because of the duality principle or just be carefull about how to prove each step. For example prove that $b+x=b$ as, $$b+x=b+[acdot(bcdot c)]=[b+a]cdot[b+(bcdot c)]=(b+a)
cdot (b)=b+acdot b=b$$
To conclude, if you do not want to just state that $(2)$ holds because of the duality principle, use the dual of each step of $(1)$ to derive $(2)$.
answered Jan 26 at 2:55
giannisl9giannisl9
579
$endgroup$
add a comment |
$begingroup$
The answer from Jakub Konieczny proves $(1):a+(b+c)=(a+b)+c$ just right, but for $(2):acdot(bcdot c)=(acdot b)cdot c,$ I would argue that you should not use it is more elegant not to use $(1)$. After all, they state the same thing but for different operators, so it makes sense to prove them separately.
In particular when proving that $x+b=b$, the statement $$[acdot (bcdot c)]+[(acdot b)+(a'cdot b)]=[acdot(bcdot c+b)]+(a'cdot b) $$
assumes that $$[acdot (bcdot c)]+[(acdot b)+(a'cdot b)]=[acdot(bcdot c)+acdot b)]+(a'cdot b) $$
So, I think it is better to state that $(2)$ holds because of the duality principle or just be carefull about how to prove each step. For example prove that $b+x=b$ as, $$b+x=b+[acdot(bcdot c)]=[b+a]cdot[b+(bcdot c)]=(b+a)
cdot (b)=b+acdot b=b$$
To conclude, if you do not want to just state that $(2)$ holds because of the duality principle, use the dual of each step of $(1)$ to derive $(2)$.
answered Jan 26 at 2:55
giannisl9giannisl9
579
$endgroup$
The answer from Jakub Konieczny proves $(1):a+(b+c)=(a+b)+c$ just right, but for $(2):acdot(bcdot c)=(acdot b)cdot c,$ I would argue that you should not use it is more elegant not to use $(1)$. After all, they state the same thing but for different operators, so it makes sense to prove them separately.
In particular when proving that $x+b=b$, the statement $$[acdot (bcdot c)]+[(acdot b)+(a'cdot b)]=[acdot(bcdot c+b)]+(a'cdot b) $$
assumes that $$[acdot (bcdot c)]+[(acdot b)+(a'cdot b)]=[acdot(bcdot c)+acdot b)]+(a'cdot b) $$
So, I think it is better to state that $(2)$ holds because of the duality principle or just be carefull about how to prove each step. For example prove that $b+x=b$ as, $$b+x=b+[acdot(bcdot c)]=[b+a]cdot[b+(bcdot c)]=(b+a)
cdot (b)=b+acdot b=b$$
To conclude, if you do not want to just state that $(2)$ holds because of the duality principle, use the dual of each step of $(1)$ to derive $(2)$.
answered Jan 26 at 2:55
giannisl9giannisl9
579
edited Jan 26 at 16:02
answered Jan 26 at 2:55
giannisl9giannisl9
579
answered Jan 26 at 2:55
giannisl9giannisl9
579
answered Jan 26 at 2:55
giannisl9giannisl9
579
579
add a comment |
add a comment |
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%2f333625%2fproof-of-associativity-in-boolean-algebra%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
