In Discrete Mathematics, is there a difference between $(neg P wedge neg Q)$ and $neg (P wedge Q)$?
$begingroup$
I am wondering, in discrete mathematics, whether there is a difference between
$(neg P wedge neg Q)$ and $neg (P wedge Q)$.
My query comes from a practice problem in a book:
Either John and Bill are both telling the truth, or neither of them is.
Any my solution above corresponds to $P ::=$ John is telling the truth, $Q ::=$ Bill is telling the truth.
Thank you.
discrete-mathematics logic propositional-calculus logic-translation
edited Dec 27 '18 at 21:22
Bram28
64.5k44793
asked Dec 27 '18 at 18:42
Jamie CorkhillJamie Corkhill
1178
$endgroup$
|
show 2 more comments
$begingroup$
I am wondering, in discrete mathematics, whether there is a difference between
$(neg P wedge neg Q)$ and $neg (P wedge Q)$.
My query comes from a practice problem in a book:
Either John and Bill are both telling the truth, or neither of them is.
Any my solution above corresponds to $P ::=$ John is telling the truth, $Q ::=$ Bill is telling the truth.
Thank you.
discrete-mathematics logic propositional-calculus logic-translation
edited Dec 27 '18 at 21:22
Bram28
64.5k44793
asked Dec 27 '18 at 18:42
Jamie CorkhillJamie Corkhill
1178
$endgroup$
3
$begingroup$
Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
$endgroup$
– Bram28
Dec 27 '18 at 18:46
$begingroup$
@Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:48
$begingroup$
Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
$endgroup$
– Bram28
Dec 27 '18 at 18:48
$begingroup$
@Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:49
$begingroup$
OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
$endgroup$
– Bram28
Dec 27 '18 at 18:51
|
show 2 more comments
$begingroup$
I am wondering, in discrete mathematics, whether there is a difference between
$(neg P wedge neg Q)$ and $neg (P wedge Q)$.
My query comes from a practice problem in a book:
Either John and Bill are both telling the truth, or neither of them is.
Any my solution above corresponds to $P ::=$ John is telling the truth, $Q ::=$ Bill is telling the truth.
Thank you.
discrete-mathematics logic propositional-calculus logic-translation
edited Dec 27 '18 at 21:22
Bram28
64.5k44793
asked Dec 27 '18 at 18:42
Jamie CorkhillJamie Corkhill
1178
$endgroup$
I am wondering, in discrete mathematics, whether there is a difference between
$(neg P wedge neg Q)$ and $neg (P wedge Q)$.
My query comes from a practice problem in a book:
Either John and Bill are both telling the truth, or neither of them is.
Any my solution above corresponds to $P ::=$ John is telling the truth, $Q ::=$ Bill is telling the truth.
Thank you.
discrete-mathematics logic propositional-calculus logic-translation
discrete-mathematics logic propositional-calculus logic-translation
edited Dec 27 '18 at 21:22
Bram28
64.5k44793
asked Dec 27 '18 at 18:42
Jamie CorkhillJamie Corkhill
1178
edited Dec 27 '18 at 21:22
Bram28
64.5k44793
asked Dec 27 '18 at 18:42
Jamie CorkhillJamie Corkhill
1178
edited Dec 27 '18 at 21:22
Bram28
64.5k44793
edited Dec 27 '18 at 21:22
Bram28
64.5k44793
edited Dec 27 '18 at 21:22
Bram28
64.5k44793
64.5k44793
asked Dec 27 '18 at 18:42
Jamie CorkhillJamie Corkhill
1178
asked Dec 27 '18 at 18:42
Jamie CorkhillJamie Corkhill
1178
asked Dec 27 '18 at 18:42
Jamie CorkhillJamie Corkhill
1178
1178
3
$begingroup$
Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
$endgroup$
– Bram28
Dec 27 '18 at 18:46
$begingroup$
@Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:48
$begingroup$
Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
$endgroup$
– Bram28
Dec 27 '18 at 18:48
$begingroup$
@Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:49
$begingroup$
OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
$endgroup$
– Bram28
Dec 27 '18 at 18:51
|
show 2 more comments
3
$begingroup$
Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
$endgroup$
– Bram28
Dec 27 '18 at 18:46
$begingroup$
@Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:48
$begingroup$
Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
$endgroup$
– Bram28
Dec 27 '18 at 18:48
$begingroup$
@Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:49
$begingroup$
OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
$endgroup$
– Bram28
Dec 27 '18 at 18:51
3
3
$begingroup$
Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
$endgroup$
– Bram28
Dec 27 '18 at 18:46
$begingroup$
Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
$endgroup$
– Bram28
Dec 27 '18 at 18:46
$begingroup$
@Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:48
$begingroup$
@Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:48
$begingroup$
Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
$endgroup$
– Bram28
Dec 27 '18 at 18:48
$begingroup$
Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
$endgroup$
– Bram28
Dec 27 '18 at 18:48
$begingroup$
@Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:49
$begingroup$
@Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:49
$begingroup$
OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
$endgroup$
– Bram28
Dec 27 '18 at 18:51
$begingroup$
OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
$endgroup$
– Bram28
Dec 27 '18 at 18:51
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Neither John nor Bill telling the truth is:
$$neg P land neg Q$$
or, equivalently:
$$neg (P lor Q)$$
But that is not the same as:
$$neg (P land Q)$$
because that is merely saying that it is not true that they are both telling the truth (i.e. that they are not both telling the truth)... which is compatible with one of the lying, but the other one still telling the truth.
In short: 'both not' is not the same as 'not both'
answered Dec 27 '18 at 18:55
Bram28Bram28
64.5k44793
$endgroup$
add a comment |
$begingroup$
Try this for yourself:
Suppose $P$ is true and $Q$ is false. Then:
- Is $(neg P wedge neg Q)$ true or false?
- Is $neg(Pwedge Q)$ true or false?
answered Dec 27 '18 at 18:48
TonyKTonyK
44.1k358137
$endgroup$
add a comment |
$begingroup$
These two statements does not have the same truth value as
By Demorgan's law, $ neg(p wedge q) = neg p vee neg q$.
Here,
John and Bill are both telling the truth corresponds to $p wedge q$.
The negation of the statement is either John is not telling the truth or Bill is not telling the truth($neg p vee neg q$).
It is not the same as neither of them is telling the truth($neg p vee neg q$).
answered Dec 27 '18 at 18:48
mathpadawanmathpadawan
2,021422
$endgroup$
add a comment |
Your Answer
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%2f3054246%2fin-discrete-mathematics-is-there-a-difference-between-neg-p-wedge-neg-q%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Neither John nor Bill telling the truth is:
$$neg P land neg Q$$
or, equivalently:
$$neg (P lor Q)$$
But that is not the same as:
$$neg (P land Q)$$
because that is merely saying that it is not true that they are both telling the truth (i.e. that they are not both telling the truth)... which is compatible with one of the lying, but the other one still telling the truth.
In short: 'both not' is not the same as 'not both'
answered Dec 27 '18 at 18:55
Bram28Bram28
64.5k44793
$endgroup$
add a comment |
$begingroup$
Neither John nor Bill telling the truth is:
$$neg P land neg Q$$
or, equivalently:
$$neg (P lor Q)$$
But that is not the same as:
$$neg (P land Q)$$
because that is merely saying that it is not true that they are both telling the truth (i.e. that they are not both telling the truth)... which is compatible with one of the lying, but the other one still telling the truth.
In short: 'both not' is not the same as 'not both'
answered Dec 27 '18 at 18:55
Bram28Bram28
64.5k44793
$endgroup$
add a comment |
$begingroup$
Neither John nor Bill telling the truth is:
$$neg P land neg Q$$
or, equivalently:
$$neg (P lor Q)$$
But that is not the same as:
$$neg (P land Q)$$
because that is merely saying that it is not true that they are both telling the truth (i.e. that they are not both telling the truth)... which is compatible with one of the lying, but the other one still telling the truth.
In short: 'both not' is not the same as 'not both'
answered Dec 27 '18 at 18:55
Bram28Bram28
64.5k44793
$endgroup$
Neither John nor Bill telling the truth is:
$$neg P land neg Q$$
or, equivalently:
$$neg (P lor Q)$$
But that is not the same as:
$$neg (P land Q)$$
because that is merely saying that it is not true that they are both telling the truth (i.e. that they are not both telling the truth)... which is compatible with one of the lying, but the other one still telling the truth.
In short: 'both not' is not the same as 'not both'
answered Dec 27 '18 at 18:55
Bram28Bram28
64.5k44793
answered Dec 27 '18 at 18:55
Bram28Bram28
64.5k44793
answered Dec 27 '18 at 18:55
Bram28Bram28
64.5k44793
answered Dec 27 '18 at 18:55
Bram28Bram28
64.5k44793
64.5k44793
add a comment |
add a comment |
$begingroup$
Try this for yourself:
Suppose $P$ is true and $Q$ is false. Then:
- Is $(neg P wedge neg Q)$ true or false?
- Is $neg(Pwedge Q)$ true or false?
answered Dec 27 '18 at 18:48
TonyKTonyK
44.1k358137
$endgroup$
add a comment |
$begingroup$
Try this for yourself:
Suppose $P$ is true and $Q$ is false. Then:
- Is $(neg P wedge neg Q)$ true or false?
- Is $neg(Pwedge Q)$ true or false?
answered Dec 27 '18 at 18:48
TonyKTonyK
44.1k358137
$endgroup$
add a comment |
$begingroup$
Try this for yourself:
Suppose $P$ is true and $Q$ is false. Then:
- Is $(neg P wedge neg Q)$ true or false?
- Is $neg(Pwedge Q)$ true or false?
answered Dec 27 '18 at 18:48
TonyKTonyK
44.1k358137
$endgroup$
Try this for yourself:
Suppose $P$ is true and $Q$ is false. Then:
- Is $(neg P wedge neg Q)$ true or false?
- Is $neg(Pwedge Q)$ true or false?
answered Dec 27 '18 at 18:48
TonyKTonyK
44.1k358137
answered Dec 27 '18 at 18:48
TonyKTonyK
44.1k358137
answered Dec 27 '18 at 18:48
TonyKTonyK
44.1k358137
answered Dec 27 '18 at 18:48
TonyKTonyK
44.1k358137
44.1k358137
add a comment |
add a comment |
$begingroup$
These two statements does not have the same truth value as
By Demorgan's law, $ neg(p wedge q) = neg p vee neg q$.
Here,
John and Bill are both telling the truth corresponds to $p wedge q$.
The negation of the statement is either John is not telling the truth or Bill is not telling the truth($neg p vee neg q$).
It is not the same as neither of them is telling the truth($neg p vee neg q$).
answered Dec 27 '18 at 18:48
mathpadawanmathpadawan
2,021422
$endgroup$
add a comment |
$begingroup$
These two statements does not have the same truth value as
By Demorgan's law, $ neg(p wedge q) = neg p vee neg q$.
Here,
John and Bill are both telling the truth corresponds to $p wedge q$.
The negation of the statement is either John is not telling the truth or Bill is not telling the truth($neg p vee neg q$).
It is not the same as neither of them is telling the truth($neg p vee neg q$).
answered Dec 27 '18 at 18:48
mathpadawanmathpadawan
2,021422
$endgroup$
add a comment |
$begingroup$
These two statements does not have the same truth value as
By Demorgan's law, $ neg(p wedge q) = neg p vee neg q$.
Here,
John and Bill are both telling the truth corresponds to $p wedge q$.
The negation of the statement is either John is not telling the truth or Bill is not telling the truth($neg p vee neg q$).
It is not the same as neither of them is telling the truth($neg p vee neg q$).
answered Dec 27 '18 at 18:48
mathpadawanmathpadawan
2,021422
$endgroup$
These two statements does not have the same truth value as
By Demorgan's law, $ neg(p wedge q) = neg p vee neg q$.
Here,
John and Bill are both telling the truth corresponds to $p wedge q$.
The negation of the statement is either John is not telling the truth or Bill is not telling the truth($neg p vee neg q$).
It is not the same as neither of them is telling the truth($neg p vee neg q$).
answered Dec 27 '18 at 18:48
mathpadawanmathpadawan
2,021422
edited Dec 27 '18 at 18:54
answered Dec 27 '18 at 18:48
mathpadawanmathpadawan
2,021422
answered Dec 27 '18 at 18:48
mathpadawanmathpadawan
2,021422
answered Dec 27 '18 at 18:48
mathpadawanmathpadawan
2,021422
2,021422
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%2f3054246%2fin-discrete-mathematics-is-there-a-difference-between-neg-p-wedge-neg-q%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
3
$begingroup$
Yes, these two expressions are different (work out the truth-table and you will see) ... but I don't quite understand the connection of these expressions to your exercise .. can you please explain why you are looking at these two expressions in relation to that practice problem?
$endgroup$
– Bram28
Dec 27 '18 at 18:46
$begingroup$
@Bram28 Well, that is just converting the English sentence into logic and connectives. I took the English sentence, and re-wrote it in terms of $P$ and $Q$, using the appropriate connectives. Thanks. But the fact they are different is all I need. I didn't think to work the Truth Table.
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:48
$begingroup$
Well .... you didn't do the symbolization correctly ... OK, just take the part that they are both telling the truth. How would you symbolize that?
$endgroup$
– Bram28
Dec 27 '18 at 18:48
$begingroup$
@Bram28 That is not my complete answer. My complete answer is: $(P ∧ Q) ∨ (neg P ∧ neg Q)$
$endgroup$
– Jamie Corkhill
Dec 27 '18 at 18:49
$begingroup$
OK, that looks good ... but there is no $neg (P land Q)$ in there ... so why are you asking about $neg (P land Q)$ in your original question? Were you maybe thinking of writing it as $(P land Q) lor neg (P land Q)$?
$endgroup$
– Bram28
Dec 27 '18 at 18:51