Logic Equivalences
I am having trouble finding the logical equivalence to $neg P Rightarrow neg Q$. I have tried factoring out the negation and going from their but it is not working out. Any ideas on how to g about doing this?
logic
edited Nov 29 at 22:30
Eevee Trainer
3,588326
asked Nov 29 at 21:35
laxattack
64
add a comment |
I am having trouble finding the logical equivalence to $neg P Rightarrow neg Q$. I have tried factoring out the negation and going from their but it is not working out. Any ideas on how to g about doing this?
logic
edited Nov 29 at 22:30
Eevee Trainer
3,588326
asked Nov 29 at 21:35
laxattack
64
1
Are you looking for all things equivalent to it? [there are several such.]
– coffeemath
Nov 29 at 21:37
While I'm not completely familiar with doing operations in logic, $neg P Rightarrow neg Q$ is the contrapositive (and thus logically equivalent) to $Q Rightarrow P$. That's one possible logical equivalence, but there are others. So the key point would be to give use the appropriate context - are you seeking a specific kind of equivalence? What have you tried, and what exactly are you seeking?
– Eevee Trainer
Nov 29 at 21:39
I am trying to prove that P → Q is not the same as ¬P → ¬Q
– laxattack
Nov 29 at 21:40
Can you use truth tables? $P to Q$ will be true if $P$ is false but $Q$ is true. And $Pto Q$ will definitely be false if $P$ is true and $Q$ is false. $lnot P to lnot Q$ will be the exact opposite. However $Qto P$ and $lnot P to lnot Q$ will have the same truth values in all cases.
– fleablood
Nov 29 at 22:31
add a comment |
I am having trouble finding the logical equivalence to $neg P Rightarrow neg Q$. I have tried factoring out the negation and going from their but it is not working out. Any ideas on how to g about doing this?
logic
edited Nov 29 at 22:30
Eevee Trainer
3,588326
asked Nov 29 at 21:35
laxattack
64
I am having trouble finding the logical equivalence to $neg P Rightarrow neg Q$. I have tried factoring out the negation and going from their but it is not working out. Any ideas on how to g about doing this?
logic
logic
edited Nov 29 at 22:30
Eevee Trainer
3,588326
asked Nov 29 at 21:35
laxattack
64
edited Nov 29 at 22:30
Eevee Trainer
3,588326
asked Nov 29 at 21:35
laxattack
64
edited Nov 29 at 22:30
Eevee Trainer
3,588326
edited Nov 29 at 22:30
Eevee Trainer
3,588326
edited Nov 29 at 22:30
Eevee Trainer
3,588326
3,588326
asked Nov 29 at 21:35
laxattack
64
asked Nov 29 at 21:35
laxattack
64
asked Nov 29 at 21:35
laxattack
64
64
1
Are you looking for all things equivalent to it? [there are several such.]
– coffeemath
Nov 29 at 21:37
While I'm not completely familiar with doing operations in logic, $neg P Rightarrow neg Q$ is the contrapositive (and thus logically equivalent) to $Q Rightarrow P$. That's one possible logical equivalence, but there are others. So the key point would be to give use the appropriate context - are you seeking a specific kind of equivalence? What have you tried, and what exactly are you seeking?
– Eevee Trainer
Nov 29 at 21:39
I am trying to prove that P → Q is not the same as ¬P → ¬Q
– laxattack
Nov 29 at 21:40
Can you use truth tables? $P to Q$ will be true if $P$ is false but $Q$ is true. And $Pto Q$ will definitely be false if $P$ is true and $Q$ is false. $lnot P to lnot Q$ will be the exact opposite. However $Qto P$ and $lnot P to lnot Q$ will have the same truth values in all cases.
– fleablood
Nov 29 at 22:31
add a comment |
1
Are you looking for all things equivalent to it? [there are several such.]
– coffeemath
Nov 29 at 21:37
While I'm not completely familiar with doing operations in logic, $neg P Rightarrow neg Q$ is the contrapositive (and thus logically equivalent) to $Q Rightarrow P$. That's one possible logical equivalence, but there are others. So the key point would be to give use the appropriate context - are you seeking a specific kind of equivalence? What have you tried, and what exactly are you seeking?
– Eevee Trainer
Nov 29 at 21:39
I am trying to prove that P → Q is not the same as ¬P → ¬Q
– laxattack
Nov 29 at 21:40
Can you use truth tables? $P to Q$ will be true if $P$ is false but $Q$ is true. And $Pto Q$ will definitely be false if $P$ is true and $Q$ is false. $lnot P to lnot Q$ will be the exact opposite. However $Qto P$ and $lnot P to lnot Q$ will have the same truth values in all cases.
– fleablood
Nov 29 at 22:31
1
1
Are you looking for all things equivalent to it? [there are several such.]
– coffeemath
Nov 29 at 21:37
Are you looking for all things equivalent to it? [there are several such.]
– coffeemath
Nov 29 at 21:37
While I'm not completely familiar with doing operations in logic, $neg P Rightarrow neg Q$ is the contrapositive (and thus logically equivalent) to $Q Rightarrow P$. That's one possible logical equivalence, but there are others. So the key point would be to give use the appropriate context - are you seeking a specific kind of equivalence? What have you tried, and what exactly are you seeking?
– Eevee Trainer
Nov 29 at 21:39
While I'm not completely familiar with doing operations in logic, $neg P Rightarrow neg Q$ is the contrapositive (and thus logically equivalent) to $Q Rightarrow P$. That's one possible logical equivalence, but there are others. So the key point would be to give use the appropriate context - are you seeking a specific kind of equivalence? What have you tried, and what exactly are you seeking?
– Eevee Trainer
Nov 29 at 21:39
I am trying to prove that P → Q is not the same as ¬P → ¬Q
– laxattack
Nov 29 at 21:40
I am trying to prove that P → Q is not the same as ¬P → ¬Q
– laxattack
Nov 29 at 21:40
Can you use truth tables? $P to Q$ will be true if $P$ is false but $Q$ is true. And $Pto Q$ will definitely be false if $P$ is true and $Q$ is false. $lnot P to lnot Q$ will be the exact opposite. However $Qto P$ and $lnot P to lnot Q$ will have the same truth values in all cases.
– fleablood
Nov 29 at 22:31
Can you use truth tables? $P to Q$ will be true if $P$ is false but $Q$ is true. And $Pto Q$ will definitely be false if $P$ is true and $Q$ is false. $lnot P to lnot Q$ will be the exact opposite. However $Qto P$ and $lnot P to lnot Q$ will have the same truth values in all cases.
– fleablood
Nov 29 at 22:31
add a comment |
1 Answer
1
active
oldest
votes
eIt depends on the rules, but $A Rightarrow B$ is normally defined as $neg A lor B$. Then
$$
neg P Rightarrow neg Q equiv (neg neg P) lor neg Q equiv
P lor neg Q equiv Q Rightarrow P
$$
All those espressions are logically equivalent.
If you want to prove that $P Rightarrow Q$ and $neg P Rightarrow neg Q$ are not equivalent you can just look at a counterexample, e.g. if $P$ is false and $Q$ is true then $P Rightarrow Q$ is true, but $neg P Rightarrow neg Q$ is false.
answered Nov 29 at 21:53
mlerma54
1,087138
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%2f3019250%2flogic-equivalences%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
eIt depends on the rules, but $A Rightarrow B$ is normally defined as $neg A lor B$. Then
$$
neg P Rightarrow neg Q equiv (neg neg P) lor neg Q equiv
P lor neg Q equiv Q Rightarrow P
$$
All those espressions are logically equivalent.
If you want to prove that $P Rightarrow Q$ and $neg P Rightarrow neg Q$ are not equivalent you can just look at a counterexample, e.g. if $P$ is false and $Q$ is true then $P Rightarrow Q$ is true, but $neg P Rightarrow neg Q$ is false.
answered Nov 29 at 21:53
mlerma54
1,087138
add a comment |
eIt depends on the rules, but $A Rightarrow B$ is normally defined as $neg A lor B$. Then
$$
neg P Rightarrow neg Q equiv (neg neg P) lor neg Q equiv
P lor neg Q equiv Q Rightarrow P
$$
All those espressions are logically equivalent.
If you want to prove that $P Rightarrow Q$ and $neg P Rightarrow neg Q$ are not equivalent you can just look at a counterexample, e.g. if $P$ is false and $Q$ is true then $P Rightarrow Q$ is true, but $neg P Rightarrow neg Q$ is false.
answered Nov 29 at 21:53
mlerma54
1,087138
add a comment |
eIt depends on the rules, but $A Rightarrow B$ is normally defined as $neg A lor B$. Then
$$
neg P Rightarrow neg Q equiv (neg neg P) lor neg Q equiv
P lor neg Q equiv Q Rightarrow P
$$
All those espressions are logically equivalent.
If you want to prove that $P Rightarrow Q$ and $neg P Rightarrow neg Q$ are not equivalent you can just look at a counterexample, e.g. if $P$ is false and $Q$ is true then $P Rightarrow Q$ is true, but $neg P Rightarrow neg Q$ is false.
answered Nov 29 at 21:53
mlerma54
1,087138
eIt depends on the rules, but $A Rightarrow B$ is normally defined as $neg A lor B$. Then
$$
neg P Rightarrow neg Q equiv (neg neg P) lor neg Q equiv
P lor neg Q equiv Q Rightarrow P
$$
All those espressions are logically equivalent.
If you want to prove that $P Rightarrow Q$ and $neg P Rightarrow neg Q$ are not equivalent you can just look at a counterexample, e.g. if $P$ is false and $Q$ is true then $P Rightarrow Q$ is true, but $neg P Rightarrow neg Q$ is false.
answered Nov 29 at 21:53
mlerma54
1,087138
edited Nov 29 at 22:58
answered Nov 29 at 21:53
mlerma54
1,087138
answered Nov 29 at 21:53
mlerma54
1,087138
answered Nov 29 at 21:53
mlerma54
1,087138
1,087138
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%2f3019250%2flogic-equivalences%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
1
Are you looking for all things equivalent to it? [there are several such.]
– coffeemath
Nov 29 at 21:37
While I'm not completely familiar with doing operations in logic, $neg P Rightarrow neg Q$ is the contrapositive (and thus logically equivalent) to $Q Rightarrow P$. That's one possible logical equivalence, but there are others. So the key point would be to give use the appropriate context - are you seeking a specific kind of equivalence? What have you tried, and what exactly are you seeking?
– Eevee Trainer
Nov 29 at 21:39
I am trying to prove that P → Q is not the same as ¬P → ¬Q
– laxattack
Nov 29 at 21:40
Can you use truth tables? $P to Q$ will be true if $P$ is false but $Q$ is true. And $Pto Q$ will definitely be false if $P$ is true and $Q$ is false. $lnot P to lnot Q$ will be the exact opposite. However $Qto P$ and $lnot P to lnot Q$ will have the same truth values in all cases.
– fleablood
Nov 29 at 22:31