Need help formalising simple propositional logic sentences
$begingroup$
I'm a beginner learning about propositional logic and how to formalise sentences.
I'm currently working through the following sentences and translating them into logical statements.
$p$ means “Emily is happy”
$q$ means “Emily paints a picture”
$r$ means “David is happy”
1. If Emily is happy then Emily paints a picture.
This is $p implies q$.
2. If Emily is happy and paints a picture then David is not happy.
This is $[p ∧ q] implies ¬r$.
3. David and Emily cannot both be happy.
I'm stuck on this last one.
I'm finding it difficult to understand how to go about formalising the last sentence. I was thinking it may include negating both $p$ and $r$ but the word "cannot" kinda throws me a bit. Any suggestions? Sorry if it's really obvious but I really am just starting out!
logic propositional-calculus
New contributor
$endgroup$
add a comment |
$begingroup$
I'm a beginner learning about propositional logic and how to formalise sentences.
I'm currently working through the following sentences and translating them into logical statements.
$p$ means “Emily is happy”
$q$ means “Emily paints a picture”
$r$ means “David is happy”
1. If Emily is happy then Emily paints a picture.
This is $p implies q$.
2. If Emily is happy and paints a picture then David is not happy.
This is $[p ∧ q] implies ¬r$.
3. David and Emily cannot both be happy.
I'm stuck on this last one.
I'm finding it difficult to understand how to go about formalising the last sentence. I was thinking it may include negating both $p$ and $r$ but the word "cannot" kinda throws me a bit. Any suggestions? Sorry if it's really obvious but I really am just starting out!
logic propositional-calculus
New contributor
$endgroup$
add a comment |
$begingroup$
I'm a beginner learning about propositional logic and how to formalise sentences.
I'm currently working through the following sentences and translating them into logical statements.
$p$ means “Emily is happy”
$q$ means “Emily paints a picture”
$r$ means “David is happy”
1. If Emily is happy then Emily paints a picture.
This is $p implies q$.
2. If Emily is happy and paints a picture then David is not happy.
This is $[p ∧ q] implies ¬r$.
3. David and Emily cannot both be happy.
I'm stuck on this last one.
I'm finding it difficult to understand how to go about formalising the last sentence. I was thinking it may include negating both $p$ and $r$ but the word "cannot" kinda throws me a bit. Any suggestions? Sorry if it's really obvious but I really am just starting out!
logic propositional-calculus
New contributor
$endgroup$
I'm a beginner learning about propositional logic and how to formalise sentences.
I'm currently working through the following sentences and translating them into logical statements.
$p$ means “Emily is happy”
$q$ means “Emily paints a picture”
$r$ means “David is happy”
1. If Emily is happy then Emily paints a picture.
This is $p implies q$.
2. If Emily is happy and paints a picture then David is not happy.
This is $[p ∧ q] implies ¬r$.
3. David and Emily cannot both be happy.
I'm stuck on this last one.
I'm finding it difficult to understand how to go about formalising the last sentence. I was thinking it may include negating both $p$ and $r$ but the word "cannot" kinda throws me a bit. Any suggestions? Sorry if it's really obvious but I really am just starting out!
logic propositional-calculus
logic propositional-calculus
New contributor
New contributor
edited 1 hour ago
Eevee Trainer
6,0631936
6,0631936
New contributor
asked 1 hour ago
new2Logicnew2Logic
161
161
New contributor
New contributor
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The last sentence means that the statement "David is happy and Emily is happy" is false. Thus, it is equivalent to $overline{p wedge r}$, or $bar p vee bar r$.
$endgroup$
add a comment |
$begingroup$
If David and Emily cannot both be happy, then this would be one of the following means of writing it.
$$neg [p land r]$$
i.e. the negation of both David and Emily being happy. Since they cannot be happy, then the statement they're both happy is false, and thus we use the negation.
You could hypothetically use each of the possible cases in a sort of list separated by "or" operators. If David and Emily cannot both be happy, then either Emily is happy but David isn't, David is happy but Emily isn't, or neither are. In which case this could be listed as
$$[neg p land r] lor [p land neg r] lor [neg p land neg r]$$
A third way to write it: since Emily or David cannot be both happy, it means at least one is unhappy. That is, either David is unhappy, or Emily is unhappy, or possibly both (but we need not account for this in this way of writing it). Thus another take on this is
$$neg p lor neg r$$
I imagine the first and third would be the "intended" answers for an exercise of this sort since they're the most compact. They also show a nice equality worth keeping in mind:
$$neg [p land r] = neg p lor neg r$$
(Footnote: throughout this post, $neg$ is used as the "negation" operator, since I've seen a few different notations for it.)
$endgroup$
1
$begingroup$
You should change $q$ to $r$.
$endgroup$
– Haris Gusic
1 hour ago
$begingroup$
Yeah I noticed that a minute ago, already fixed it. xD
$endgroup$
– Eevee Trainer
1 hour ago
add a comment |
$begingroup$
I can understand that the use of 'cannot' is a bit confusing ... it seems to be stronger than just saying that David and Emily are both not happy.
In fact, in modal logic you can express these kinds of stronger claims, where:
$square P$ means "It is necessary that P is true"
$Diamond P$ means "It is possible that P is true"
Using those symbols, translating "David and Emily cannot both be happy" can be done as:
$neg Diamond (r land p)$
or, equivalently:
$square neg (r land p)$
But, I assume you are currently not doing any model logic at all, since you are just starting with propositional logic. As such, you should really just treat the sentence as "David and Emily are not both happy"
Good for you for noticing that those two sentences are not quite the same thing though!!
$endgroup$
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
});
}
});
new2Logic is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3115736%2fneed-help-formalising-simple-propositional-logic-sentences%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
$begingroup$
The last sentence means that the statement "David is happy and Emily is happy" is false. Thus, it is equivalent to $overline{p wedge r}$, or $bar p vee bar r$.
$endgroup$
add a comment |
$begingroup$
The last sentence means that the statement "David is happy and Emily is happy" is false. Thus, it is equivalent to $overline{p wedge r}$, or $bar p vee bar r$.
$endgroup$
add a comment |
$begingroup$
The last sentence means that the statement "David is happy and Emily is happy" is false. Thus, it is equivalent to $overline{p wedge r}$, or $bar p vee bar r$.
$endgroup$
The last sentence means that the statement "David is happy and Emily is happy" is false. Thus, it is equivalent to $overline{p wedge r}$, or $bar p vee bar r$.
answered 1 hour ago
Haris GusicHaris Gusic
946
946
add a comment |
add a comment |
$begingroup$
If David and Emily cannot both be happy, then this would be one of the following means of writing it.
$$neg [p land r]$$
i.e. the negation of both David and Emily being happy. Since they cannot be happy, then the statement they're both happy is false, and thus we use the negation.
You could hypothetically use each of the possible cases in a sort of list separated by "or" operators. If David and Emily cannot both be happy, then either Emily is happy but David isn't, David is happy but Emily isn't, or neither are. In which case this could be listed as
$$[neg p land r] lor [p land neg r] lor [neg p land neg r]$$
A third way to write it: since Emily or David cannot be both happy, it means at least one is unhappy. That is, either David is unhappy, or Emily is unhappy, or possibly both (but we need not account for this in this way of writing it). Thus another take on this is
$$neg p lor neg r$$
I imagine the first and third would be the "intended" answers for an exercise of this sort since they're the most compact. They also show a nice equality worth keeping in mind:
$$neg [p land r] = neg p lor neg r$$
(Footnote: throughout this post, $neg$ is used as the "negation" operator, since I've seen a few different notations for it.)
$endgroup$
1
$begingroup$
You should change $q$ to $r$.
$endgroup$
– Haris Gusic
1 hour ago
$begingroup$
Yeah I noticed that a minute ago, already fixed it. xD
$endgroup$
– Eevee Trainer
1 hour ago
add a comment |
$begingroup$
If David and Emily cannot both be happy, then this would be one of the following means of writing it.
$$neg [p land r]$$
i.e. the negation of both David and Emily being happy. Since they cannot be happy, then the statement they're both happy is false, and thus we use the negation.
You could hypothetically use each of the possible cases in a sort of list separated by "or" operators. If David and Emily cannot both be happy, then either Emily is happy but David isn't, David is happy but Emily isn't, or neither are. In which case this could be listed as
$$[neg p land r] lor [p land neg r] lor [neg p land neg r]$$
A third way to write it: since Emily or David cannot be both happy, it means at least one is unhappy. That is, either David is unhappy, or Emily is unhappy, or possibly both (but we need not account for this in this way of writing it). Thus another take on this is
$$neg p lor neg r$$
I imagine the first and third would be the "intended" answers for an exercise of this sort since they're the most compact. They also show a nice equality worth keeping in mind:
$$neg [p land r] = neg p lor neg r$$
(Footnote: throughout this post, $neg$ is used as the "negation" operator, since I've seen a few different notations for it.)
$endgroup$
1
$begingroup$
You should change $q$ to $r$.
$endgroup$
– Haris Gusic
1 hour ago
$begingroup$
Yeah I noticed that a minute ago, already fixed it. xD
$endgroup$
– Eevee Trainer
1 hour ago
add a comment |
$begingroup$
If David and Emily cannot both be happy, then this would be one of the following means of writing it.
$$neg [p land r]$$
i.e. the negation of both David and Emily being happy. Since they cannot be happy, then the statement they're both happy is false, and thus we use the negation.
You could hypothetically use each of the possible cases in a sort of list separated by "or" operators. If David and Emily cannot both be happy, then either Emily is happy but David isn't, David is happy but Emily isn't, or neither are. In which case this could be listed as
$$[neg p land r] lor [p land neg r] lor [neg p land neg r]$$
A third way to write it: since Emily or David cannot be both happy, it means at least one is unhappy. That is, either David is unhappy, or Emily is unhappy, or possibly both (but we need not account for this in this way of writing it). Thus another take on this is
$$neg p lor neg r$$
I imagine the first and third would be the "intended" answers for an exercise of this sort since they're the most compact. They also show a nice equality worth keeping in mind:
$$neg [p land r] = neg p lor neg r$$
(Footnote: throughout this post, $neg$ is used as the "negation" operator, since I've seen a few different notations for it.)
$endgroup$
If David and Emily cannot both be happy, then this would be one of the following means of writing it.
$$neg [p land r]$$
i.e. the negation of both David and Emily being happy. Since they cannot be happy, then the statement they're both happy is false, and thus we use the negation.
You could hypothetically use each of the possible cases in a sort of list separated by "or" operators. If David and Emily cannot both be happy, then either Emily is happy but David isn't, David is happy but Emily isn't, or neither are. In which case this could be listed as
$$[neg p land r] lor [p land neg r] lor [neg p land neg r]$$
A third way to write it: since Emily or David cannot be both happy, it means at least one is unhappy. That is, either David is unhappy, or Emily is unhappy, or possibly both (but we need not account for this in this way of writing it). Thus another take on this is
$$neg p lor neg r$$
I imagine the first and third would be the "intended" answers for an exercise of this sort since they're the most compact. They also show a nice equality worth keeping in mind:
$$neg [p land r] = neg p lor neg r$$
(Footnote: throughout this post, $neg$ is used as the "negation" operator, since I've seen a few different notations for it.)
edited 1 hour ago
answered 1 hour ago
Eevee TrainerEevee Trainer
6,0631936
6,0631936
1
$begingroup$
You should change $q$ to $r$.
$endgroup$
– Haris Gusic
1 hour ago
$begingroup$
Yeah I noticed that a minute ago, already fixed it. xD
$endgroup$
– Eevee Trainer
1 hour ago
add a comment |
1
$begingroup$
You should change $q$ to $r$.
$endgroup$
– Haris Gusic
1 hour ago
$begingroup$
Yeah I noticed that a minute ago, already fixed it. xD
$endgroup$
– Eevee Trainer
1 hour ago
1
1
$begingroup$
You should change $q$ to $r$.
$endgroup$
– Haris Gusic
1 hour ago
$begingroup$
You should change $q$ to $r$.
$endgroup$
– Haris Gusic
1 hour ago
$begingroup$
Yeah I noticed that a minute ago, already fixed it. xD
$endgroup$
– Eevee Trainer
1 hour ago
$begingroup$
Yeah I noticed that a minute ago, already fixed it. xD
$endgroup$
– Eevee Trainer
1 hour ago
add a comment |
$begingroup$
I can understand that the use of 'cannot' is a bit confusing ... it seems to be stronger than just saying that David and Emily are both not happy.
In fact, in modal logic you can express these kinds of stronger claims, where:
$square P$ means "It is necessary that P is true"
$Diamond P$ means "It is possible that P is true"
Using those symbols, translating "David and Emily cannot both be happy" can be done as:
$neg Diamond (r land p)$
or, equivalently:
$square neg (r land p)$
But, I assume you are currently not doing any model logic at all, since you are just starting with propositional logic. As such, you should really just treat the sentence as "David and Emily are not both happy"
Good for you for noticing that those two sentences are not quite the same thing though!!
$endgroup$
add a comment |
$begingroup$
I can understand that the use of 'cannot' is a bit confusing ... it seems to be stronger than just saying that David and Emily are both not happy.
In fact, in modal logic you can express these kinds of stronger claims, where:
$square P$ means "It is necessary that P is true"
$Diamond P$ means "It is possible that P is true"
Using those symbols, translating "David and Emily cannot both be happy" can be done as:
$neg Diamond (r land p)$
or, equivalently:
$square neg (r land p)$
But, I assume you are currently not doing any model logic at all, since you are just starting with propositional logic. As such, you should really just treat the sentence as "David and Emily are not both happy"
Good for you for noticing that those two sentences are not quite the same thing though!!
$endgroup$
add a comment |
$begingroup$
I can understand that the use of 'cannot' is a bit confusing ... it seems to be stronger than just saying that David and Emily are both not happy.
In fact, in modal logic you can express these kinds of stronger claims, where:
$square P$ means "It is necessary that P is true"
$Diamond P$ means "It is possible that P is true"
Using those symbols, translating "David and Emily cannot both be happy" can be done as:
$neg Diamond (r land p)$
or, equivalently:
$square neg (r land p)$
But, I assume you are currently not doing any model logic at all, since you are just starting with propositional logic. As such, you should really just treat the sentence as "David and Emily are not both happy"
Good for you for noticing that those two sentences are not quite the same thing though!!
$endgroup$
I can understand that the use of 'cannot' is a bit confusing ... it seems to be stronger than just saying that David and Emily are both not happy.
In fact, in modal logic you can express these kinds of stronger claims, where:
$square P$ means "It is necessary that P is true"
$Diamond P$ means "It is possible that P is true"
Using those symbols, translating "David and Emily cannot both be happy" can be done as:
$neg Diamond (r land p)$
or, equivalently:
$square neg (r land p)$
But, I assume you are currently not doing any model logic at all, since you are just starting with propositional logic. As such, you should really just treat the sentence as "David and Emily are not both happy"
Good for you for noticing that those two sentences are not quite the same thing though!!
answered 1 hour ago
Bram28Bram28
62.7k44793
62.7k44793
add a comment |
add a comment |
new2Logic is a new contributor. Be nice, and check out our Code of Conduct.
new2Logic is a new contributor. Be nice, and check out our Code of Conduct.
new2Logic is a new contributor. Be nice, and check out our Code of Conduct.
new2Logic is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3115736%2fneed-help-formalising-simple-propositional-logic-sentences%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