Using the replacement laws to prove that ($a to $b) $vee$ ($a to $c) = $a to ($b $vee$ c)
I have been asked to prove that
($a to $b) $vee$ ($a to $c) = $a to ($b $vee$ c).
I believe it is just the simple case of using the distributive law:
$a wedge ($b $vee$ c)= (a $wedge c) vee ($a $wedge$ b).
But I am not sure.
boolean-algebra
asked Nov 25 at 20:35
David MB
144
add a comment |
I have been asked to prove that
($a to $b) $vee$ ($a to $c) = $a to ($b $vee$ c).
I believe it is just the simple case of using the distributive law:
$a wedge ($b $vee$ c)= (a $wedge c) vee ($a $wedge$ b).
But I am not sure.
boolean-algebra
asked Nov 25 at 20:35
David MB
144
add a comment |
I have been asked to prove that
($a to $b) $vee$ ($a to $c) = $a to ($b $vee$ c).
I believe it is just the simple case of using the distributive law:
$a wedge ($b $vee$ c)= (a $wedge c) vee ($a $wedge$ b).
But I am not sure.
boolean-algebra
asked Nov 25 at 20:35
David MB
144
I have been asked to prove that
($a to $b) $vee$ ($a to $c) = $a to ($b $vee$ c).
I believe it is just the simple case of using the distributive law:
$a wedge ($b $vee$ c)= (a $wedge c) vee ($a $wedge$ b).
But I am not sure.
boolean-algebra
boolean-algebra
asked Nov 25 at 20:35
David MB
144
asked Nov 25 at 20:35
David MB
144
asked Nov 25 at 20:35
David MB
144
asked Nov 25 at 20:35
David MB
144
asked Nov 25 at 20:35
David MB
144
144
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
I believe it is just the simple case of using the distributive law:
$a wedge ($b $vee$ c)= (a $wedge c) vee ($a $wedge$ b).
But I am not sure.
You are right to not be sure.
First of all, there are distribution laws involving the conditional, but you have to be very careful. This is what is the case:
Distribution Laws for Conditionals
$a rightarrow (b land c) = (a rightarrow b) land (a rightarrow c)$
$a rightarrow (b lor c) = (a rightarrow b) lor (a rightarrow c)$
$(b land c) rightarrow a = (b rightarrow a) lor (c rightarrow a)$
$(b lor c) rightarrow a = (b rightarrow a) land (c rightarrow a)$
Note that for the last two, the $land$ and the $lor$ switch! ... So, it is easy to make mistakes with these. Therefore, it is typically safer to rewrite the conditionals as disjunctions.
That is, go from:
$(a rightarrow b) lor (a rightarrow c)$
to:
$(neg a lor b) lor (neg a lor c)$
Now, can we do Distribution on this? No, because all the operators are $lor$'s.
OK, but what you can do, given that all operators are the same, is Association. Using Association, you can either move parentheses around or, if you are allowed to do so, drop parentheses altogether. Assuming we can indeed just drop parentheses, we thus get:
$neg a lor b lor neg a lor c$
Now, at this point notice that you have two of the same terms $neg a$. By Idempotnece (which says that $p lor p = p$), you can drop one of them:
$neg a lor b lor c$
OK, and now you can rewrite the disjunction back as a conditional:
$a rightarrow (b lor c)$
answered Nov 26 at 14:21
Bram28
59.7k44186
Brilliantly explained. Much appreciated.
– David MB
Nov 26 at 17:05
@DavidMB You're welcome! :)
– Bram28
Nov 26 at 17:06
add a comment |
Left hand side is: $(a rightarrow b) vee (a rightarrow c)=(neg a vee b)vee(neg a vee c)=neg a vee b vee c.$
also
Right hand side is :$a rightarrow (b vee c)= neg a vee (b vee c)= neg a vee bvee c$.
So they are equal.
answered Nov 25 at 20:41
mathnoob
1,761422
add a comment |
Hint: you want to prove $(neg alor b)lor(neg alor c)$ is equivalent by replacement to $neg alor (blor c)$.
answered Nov 25 at 20:42
J.G.
22.1k22034
add a comment |
Yes, I believe you're correct. By expanding, using the distributive laws and simplifying you can prove this.
$$
begin{align}
&quad (a Rightarrow b) lor (a Rightarrow c) \
&Leftrightarrow (neg a lor b) lor (neg a lor c) \
&Leftrightarrow (neg a lor b) lor (neg a lor b lor c) \
&Leftrightarrow (neg a lor b) lor c \
&Leftrightarrow neg a lor (blor c) \
&Leftrightarrow a Rightarrow (blor c)
end{align}
$$
answered Nov 25 at 20:42
Thomas Lang
1624
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%2f3013355%2fusing-the-replacement-laws-to-prove-that-a-to-b-vee-a-to-c-a-to%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
I believe it is just the simple case of using the distributive law:
$a wedge ($b $vee$ c)= (a $wedge c) vee ($a $wedge$ b).
But I am not sure.
You are right to not be sure.
First of all, there are distribution laws involving the conditional, but you have to be very careful. This is what is the case:
Distribution Laws for Conditionals
$a rightarrow (b land c) = (a rightarrow b) land (a rightarrow c)$
$a rightarrow (b lor c) = (a rightarrow b) lor (a rightarrow c)$
$(b land c) rightarrow a = (b rightarrow a) lor (c rightarrow a)$
$(b lor c) rightarrow a = (b rightarrow a) land (c rightarrow a)$
Note that for the last two, the $land$ and the $lor$ switch! ... So, it is easy to make mistakes with these. Therefore, it is typically safer to rewrite the conditionals as disjunctions.
That is, go from:
$(a rightarrow b) lor (a rightarrow c)$
to:
$(neg a lor b) lor (neg a lor c)$
Now, can we do Distribution on this? No, because all the operators are $lor$'s.
OK, but what you can do, given that all operators are the same, is Association. Using Association, you can either move parentheses around or, if you are allowed to do so, drop parentheses altogether. Assuming we can indeed just drop parentheses, we thus get:
$neg a lor b lor neg a lor c$
Now, at this point notice that you have two of the same terms $neg a$. By Idempotnece (which says that $p lor p = p$), you can drop one of them:
$neg a lor b lor c$
OK, and now you can rewrite the disjunction back as a conditional:
$a rightarrow (b lor c)$
answered Nov 26 at 14:21
Bram28
59.7k44186
Brilliantly explained. Much appreciated.
– David MB
Nov 26 at 17:05
@DavidMB You're welcome! :)
– Bram28
Nov 26 at 17:06
add a comment |
I believe it is just the simple case of using the distributive law:
$a wedge ($b $vee$ c)= (a $wedge c) vee ($a $wedge$ b).
But I am not sure.
You are right to not be sure.
First of all, there are distribution laws involving the conditional, but you have to be very careful. This is what is the case:
Distribution Laws for Conditionals
$a rightarrow (b land c) = (a rightarrow b) land (a rightarrow c)$
$a rightarrow (b lor c) = (a rightarrow b) lor (a rightarrow c)$
$(b land c) rightarrow a = (b rightarrow a) lor (c rightarrow a)$
$(b lor c) rightarrow a = (b rightarrow a) land (c rightarrow a)$
Note that for the last two, the $land$ and the $lor$ switch! ... So, it is easy to make mistakes with these. Therefore, it is typically safer to rewrite the conditionals as disjunctions.
That is, go from:
$(a rightarrow b) lor (a rightarrow c)$
to:
$(neg a lor b) lor (neg a lor c)$
Now, can we do Distribution on this? No, because all the operators are $lor$'s.
OK, but what you can do, given that all operators are the same, is Association. Using Association, you can either move parentheses around or, if you are allowed to do so, drop parentheses altogether. Assuming we can indeed just drop parentheses, we thus get:
$neg a lor b lor neg a lor c$
Now, at this point notice that you have two of the same terms $neg a$. By Idempotnece (which says that $p lor p = p$), you can drop one of them:
$neg a lor b lor c$
OK, and now you can rewrite the disjunction back as a conditional:
$a rightarrow (b lor c)$
answered Nov 26 at 14:21
Bram28
59.7k44186
Brilliantly explained. Much appreciated.
– David MB
Nov 26 at 17:05
@DavidMB You're welcome! :)
– Bram28
Nov 26 at 17:06
add a comment |
I believe it is just the simple case of using the distributive law:
$a wedge ($b $vee$ c)= (a $wedge c) vee ($a $wedge$ b).
But I am not sure.
You are right to not be sure.
First of all, there are distribution laws involving the conditional, but you have to be very careful. This is what is the case:
Distribution Laws for Conditionals
$a rightarrow (b land c) = (a rightarrow b) land (a rightarrow c)$
$a rightarrow (b lor c) = (a rightarrow b) lor (a rightarrow c)$
$(b land c) rightarrow a = (b rightarrow a) lor (c rightarrow a)$
$(b lor c) rightarrow a = (b rightarrow a) land (c rightarrow a)$
Note that for the last two, the $land$ and the $lor$ switch! ... So, it is easy to make mistakes with these. Therefore, it is typically safer to rewrite the conditionals as disjunctions.
That is, go from:
$(a rightarrow b) lor (a rightarrow c)$
to:
$(neg a lor b) lor (neg a lor c)$
Now, can we do Distribution on this? No, because all the operators are $lor$'s.
OK, but what you can do, given that all operators are the same, is Association. Using Association, you can either move parentheses around or, if you are allowed to do so, drop parentheses altogether. Assuming we can indeed just drop parentheses, we thus get:
$neg a lor b lor neg a lor c$
Now, at this point notice that you have two of the same terms $neg a$. By Idempotnece (which says that $p lor p = p$), you can drop one of them:
$neg a lor b lor c$
OK, and now you can rewrite the disjunction back as a conditional:
$a rightarrow (b lor c)$
answered Nov 26 at 14:21
Bram28
59.7k44186
I believe it is just the simple case of using the distributive law:
$a wedge ($b $vee$ c)= (a $wedge c) vee ($a $wedge$ b).
But I am not sure.
You are right to not be sure.
First of all, there are distribution laws involving the conditional, but you have to be very careful. This is what is the case:
Distribution Laws for Conditionals
$a rightarrow (b land c) = (a rightarrow b) land (a rightarrow c)$
$a rightarrow (b lor c) = (a rightarrow b) lor (a rightarrow c)$
$(b land c) rightarrow a = (b rightarrow a) lor (c rightarrow a)$
$(b lor c) rightarrow a = (b rightarrow a) land (c rightarrow a)$
Note that for the last two, the $land$ and the $lor$ switch! ... So, it is easy to make mistakes with these. Therefore, it is typically safer to rewrite the conditionals as disjunctions.
That is, go from:
$(a rightarrow b) lor (a rightarrow c)$
to:
$(neg a lor b) lor (neg a lor c)$
Now, can we do Distribution on this? No, because all the operators are $lor$'s.
OK, but what you can do, given that all operators are the same, is Association. Using Association, you can either move parentheses around or, if you are allowed to do so, drop parentheses altogether. Assuming we can indeed just drop parentheses, we thus get:
$neg a lor b lor neg a lor c$
Now, at this point notice that you have two of the same terms $neg a$. By Idempotnece (which says that $p lor p = p$), you can drop one of them:
$neg a lor b lor c$
OK, and now you can rewrite the disjunction back as a conditional:
$a rightarrow (b lor c)$
answered Nov 26 at 14:21
Bram28
59.7k44186
answered Nov 26 at 14:21
Bram28
59.7k44186
answered Nov 26 at 14:21
Bram28
59.7k44186
answered Nov 26 at 14:21
Bram28
59.7k44186
59.7k44186
Brilliantly explained. Much appreciated.
– David MB
Nov 26 at 17:05
@DavidMB You're welcome! :)
– Bram28
Nov 26 at 17:06
add a comment |
Brilliantly explained. Much appreciated.
– David MB
Nov 26 at 17:05
@DavidMB You're welcome! :)
– Bram28
Nov 26 at 17:06
Brilliantly explained. Much appreciated.
– David MB
Nov 26 at 17:05
Brilliantly explained. Much appreciated.
– David MB
Nov 26 at 17:05
@DavidMB You're welcome! :)
– Bram28
Nov 26 at 17:06
@DavidMB You're welcome! :)
– Bram28
Nov 26 at 17:06
add a comment |
Left hand side is: $(a rightarrow b) vee (a rightarrow c)=(neg a vee b)vee(neg a vee c)=neg a vee b vee c.$
also
Right hand side is :$a rightarrow (b vee c)= neg a vee (b vee c)= neg a vee bvee c$.
So they are equal.
answered Nov 25 at 20:41
mathnoob
1,761422
add a comment |
Left hand side is: $(a rightarrow b) vee (a rightarrow c)=(neg a vee b)vee(neg a vee c)=neg a vee b vee c.$
also
Right hand side is :$a rightarrow (b vee c)= neg a vee (b vee c)= neg a vee bvee c$.
So they are equal.
answered Nov 25 at 20:41
mathnoob
1,761422
add a comment |
Left hand side is: $(a rightarrow b) vee (a rightarrow c)=(neg a vee b)vee(neg a vee c)=neg a vee b vee c.$
also
Right hand side is :$a rightarrow (b vee c)= neg a vee (b vee c)= neg a vee bvee c$.
So they are equal.
answered Nov 25 at 20:41
mathnoob
1,761422
Left hand side is: $(a rightarrow b) vee (a rightarrow c)=(neg a vee b)vee(neg a vee c)=neg a vee b vee c.$
also
Right hand side is :$a rightarrow (b vee c)= neg a vee (b vee c)= neg a vee bvee c$.
So they are equal.
answered Nov 25 at 20:41
mathnoob
1,761422
answered Nov 25 at 20:41
mathnoob
1,761422
answered Nov 25 at 20:41
mathnoob
1,761422
answered Nov 25 at 20:41
mathnoob
1,761422
1,761422
add a comment |
add a comment |
Hint: you want to prove $(neg alor b)lor(neg alor c)$ is equivalent by replacement to $neg alor (blor c)$.
answered Nov 25 at 20:42
J.G.
22.1k22034
add a comment |
Hint: you want to prove $(neg alor b)lor(neg alor c)$ is equivalent by replacement to $neg alor (blor c)$.
answered Nov 25 at 20:42
J.G.
22.1k22034
add a comment |
Hint: you want to prove $(neg alor b)lor(neg alor c)$ is equivalent by replacement to $neg alor (blor c)$.
answered Nov 25 at 20:42
J.G.
22.1k22034
Hint: you want to prove $(neg alor b)lor(neg alor c)$ is equivalent by replacement to $neg alor (blor c)$.
answered Nov 25 at 20:42
J.G.
22.1k22034
answered Nov 25 at 20:42
J.G.
22.1k22034
answered Nov 25 at 20:42
J.G.
22.1k22034
answered Nov 25 at 20:42
J.G.
22.1k22034
22.1k22034
add a comment |
add a comment |
Yes, I believe you're correct. By expanding, using the distributive laws and simplifying you can prove this.
$$
begin{align}
&quad (a Rightarrow b) lor (a Rightarrow c) \
&Leftrightarrow (neg a lor b) lor (neg a lor c) \
&Leftrightarrow (neg a lor b) lor (neg a lor b lor c) \
&Leftrightarrow (neg a lor b) lor c \
&Leftrightarrow neg a lor (blor c) \
&Leftrightarrow a Rightarrow (blor c)
end{align}
$$
answered Nov 25 at 20:42
Thomas Lang
1624
add a comment |
Yes, I believe you're correct. By expanding, using the distributive laws and simplifying you can prove this.
$$
begin{align}
&quad (a Rightarrow b) lor (a Rightarrow c) \
&Leftrightarrow (neg a lor b) lor (neg a lor c) \
&Leftrightarrow (neg a lor b) lor (neg a lor b lor c) \
&Leftrightarrow (neg a lor b) lor c \
&Leftrightarrow neg a lor (blor c) \
&Leftrightarrow a Rightarrow (blor c)
end{align}
$$
answered Nov 25 at 20:42
Thomas Lang
1624
add a comment |
Yes, I believe you're correct. By expanding, using the distributive laws and simplifying you can prove this.
$$
begin{align}
&quad (a Rightarrow b) lor (a Rightarrow c) \
&Leftrightarrow (neg a lor b) lor (neg a lor c) \
&Leftrightarrow (neg a lor b) lor (neg a lor b lor c) \
&Leftrightarrow (neg a lor b) lor c \
&Leftrightarrow neg a lor (blor c) \
&Leftrightarrow a Rightarrow (blor c)
end{align}
$$
answered Nov 25 at 20:42
Thomas Lang
1624
Yes, I believe you're correct. By expanding, using the distributive laws and simplifying you can prove this.
$$
begin{align}
&quad (a Rightarrow b) lor (a Rightarrow c) \
&Leftrightarrow (neg a lor b) lor (neg a lor c) \
&Leftrightarrow (neg a lor b) lor (neg a lor b lor c) \
&Leftrightarrow (neg a lor b) lor c \
&Leftrightarrow neg a lor (blor c) \
&Leftrightarrow a Rightarrow (blor c)
end{align}
$$
answered Nov 25 at 20:42
Thomas Lang
1624
answered Nov 25 at 20:42
Thomas Lang
1624
answered Nov 25 at 20:42
Thomas Lang
1624
answered Nov 25 at 20:42
Thomas Lang
1624
1624
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3013355%2fusing-the-replacement-laws-to-prove-that-a-to-b-vee-a-to-c-a-to%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
