Logical equivalence of ¬p→q
$begingroup$
Just wondering what other ways $neg p to q$ can be expressed.
I know that $pto q$ is logically equivalent to $neg plor q$, hence I think that $neg pto q$ has the same logical equivalence as $plor q$.
propositional-calculus
edited Nov 11 '18 at 18:41
Mutantoe
604513
asked Nov 11 '18 at 8:23
Jordan SolomonsJordan Solomons
197
$endgroup$
add a comment |
$begingroup$
Just wondering what other ways $neg p to q$ can be expressed.
I know that $pto q$ is logically equivalent to $neg plor q$, hence I think that $neg pto q$ has the same logical equivalence as $plor q$.
propositional-calculus
edited Nov 11 '18 at 18:41
Mutantoe
604513
asked Nov 11 '18 at 8:23
Jordan SolomonsJordan Solomons
197
$endgroup$
add a comment |
$begingroup$
Just wondering what other ways $neg p to q$ can be expressed.
I know that $pto q$ is logically equivalent to $neg plor q$, hence I think that $neg pto q$ has the same logical equivalence as $plor q$.
propositional-calculus
edited Nov 11 '18 at 18:41
Mutantoe
604513
asked Nov 11 '18 at 8:23
Jordan SolomonsJordan Solomons
197
$endgroup$
Just wondering what other ways $neg p to q$ can be expressed.
I know that $pto q$ is logically equivalent to $neg plor q$, hence I think that $neg pto q$ has the same logical equivalence as $plor q$.
propositional-calculus
propositional-calculus
edited Nov 11 '18 at 18:41
Mutantoe
604513
asked Nov 11 '18 at 8:23
Jordan SolomonsJordan Solomons
197
edited Nov 11 '18 at 18:41
Mutantoe
604513
asked Nov 11 '18 at 8:23
Jordan SolomonsJordan Solomons
197
edited Nov 11 '18 at 18:41
Mutantoe
604513
edited Nov 11 '18 at 18:41
Mutantoe
604513
edited Nov 11 '18 at 18:41
Mutantoe
604513
604513
asked Nov 11 '18 at 8:23
Jordan SolomonsJordan Solomons
197
asked Nov 11 '18 at 8:23
Jordan SolomonsJordan Solomons
197
asked Nov 11 '18 at 8:23
Jordan SolomonsJordan Solomons
197
197
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
As others have pointed out, truth tables will confirm the equivalence of $neg p to q$ and $p lor q$.
But also you'll get additional insight by thinking informally. Suppose you are given that either $p$ or $q$. Then, if you rule out $p$, then you are left with $q$. So, in symbols, from $p lor q$ you can infer $neg p to q$.
Conversely, suppose you are given that if not-$p$ then $q$. Then assuming you accept excluded middle, the principle that either $p$ or not $p$, it follows that either $p$ or (using that conditional) $q$. So, in symbols, from $neg p to q$ you can infer $p lor q$.
answered Nov 11 '18 at 8:49
Peter SmithPeter Smith
40.6k340120
$endgroup$
$begingroup$
Thanks for confirming!
$endgroup$
– Jordan Solomons
Nov 11 '18 at 10:55
add a comment |
$begingroup$
You have pretty much given the answer yourself already. Logical equivalence of $p$ and $q$ is given if $p$ is true if and only if $q$ is true (and hence $p$ is false iff $q$ is false). Since $p$ and $q$ can only be true or false, you can use truth tables and check whether logical equivalence is given. From there you can derive permissible manipulations of propositions.
In this case, $(lnot p) rightarrow q equiv lnot(lnot p) lor q equiv p lor q$, as you stated correctly.
answered Nov 11 '18 at 8:34
MacRanceMacRance
1226
$endgroup$
add a comment |
$begingroup$
As you said, $pRightarrow q$ is logically equivalent to $neg pvee q$.
Then by double negation, $neg p Rightarrow q$ is logically equivalent to $neg(neg p)vee q$, which amounts to $pvee q$.
answered Nov 11 '18 at 8:32
WuestenfuxWuestenfux
4,2691413
$endgroup$
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%2f2993594%2flogical-equivalence-of-%25c2%25acp%25e2%2586%2592q%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
$begingroup$
As others have pointed out, truth tables will confirm the equivalence of $neg p to q$ and $p lor q$.
But also you'll get additional insight by thinking informally. Suppose you are given that either $p$ or $q$. Then, if you rule out $p$, then you are left with $q$. So, in symbols, from $p lor q$ you can infer $neg p to q$.
Conversely, suppose you are given that if not-$p$ then $q$. Then assuming you accept excluded middle, the principle that either $p$ or not $p$, it follows that either $p$ or (using that conditional) $q$. So, in symbols, from $neg p to q$ you can infer $p lor q$.
answered Nov 11 '18 at 8:49
Peter SmithPeter Smith
40.6k340120
$endgroup$
$begingroup$
Thanks for confirming!
$endgroup$
– Jordan Solomons
Nov 11 '18 at 10:55
add a comment |
$begingroup$
As others have pointed out, truth tables will confirm the equivalence of $neg p to q$ and $p lor q$.
But also you'll get additional insight by thinking informally. Suppose you are given that either $p$ or $q$. Then, if you rule out $p$, then you are left with $q$. So, in symbols, from $p lor q$ you can infer $neg p to q$.
Conversely, suppose you are given that if not-$p$ then $q$. Then assuming you accept excluded middle, the principle that either $p$ or not $p$, it follows that either $p$ or (using that conditional) $q$. So, in symbols, from $neg p to q$ you can infer $p lor q$.
answered Nov 11 '18 at 8:49
Peter SmithPeter Smith
40.6k340120
$endgroup$
$begingroup$
Thanks for confirming!
$endgroup$
– Jordan Solomons
Nov 11 '18 at 10:55
add a comment |
$begingroup$
As others have pointed out, truth tables will confirm the equivalence of $neg p to q$ and $p lor q$.
But also you'll get additional insight by thinking informally. Suppose you are given that either $p$ or $q$. Then, if you rule out $p$, then you are left with $q$. So, in symbols, from $p lor q$ you can infer $neg p to q$.
Conversely, suppose you are given that if not-$p$ then $q$. Then assuming you accept excluded middle, the principle that either $p$ or not $p$, it follows that either $p$ or (using that conditional) $q$. So, in symbols, from $neg p to q$ you can infer $p lor q$.
answered Nov 11 '18 at 8:49
Peter SmithPeter Smith
40.6k340120
$endgroup$
As others have pointed out, truth tables will confirm the equivalence of $neg p to q$ and $p lor q$.
But also you'll get additional insight by thinking informally. Suppose you are given that either $p$ or $q$. Then, if you rule out $p$, then you are left with $q$. So, in symbols, from $p lor q$ you can infer $neg p to q$.
Conversely, suppose you are given that if not-$p$ then $q$. Then assuming you accept excluded middle, the principle that either $p$ or not $p$, it follows that either $p$ or (using that conditional) $q$. So, in symbols, from $neg p to q$ you can infer $p lor q$.
answered Nov 11 '18 at 8:49
Peter SmithPeter Smith
40.6k340120
answered Nov 11 '18 at 8:49
Peter SmithPeter Smith
40.6k340120
answered Nov 11 '18 at 8:49
Peter SmithPeter Smith
40.6k340120
answered Nov 11 '18 at 8:49
Peter SmithPeter Smith
40.6k340120
40.6k340120
$begingroup$
Thanks for confirming!
$endgroup$
– Jordan Solomons
Nov 11 '18 at 10:55
add a comment |
$begingroup$
Thanks for confirming!
$endgroup$
– Jordan Solomons
Nov 11 '18 at 10:55
$begingroup$
Thanks for confirming!
$endgroup$
– Jordan Solomons
Nov 11 '18 at 10:55
$begingroup$
Thanks for confirming!
$endgroup$
– Jordan Solomons
Nov 11 '18 at 10:55
add a comment |
$begingroup$
You have pretty much given the answer yourself already. Logical equivalence of $p$ and $q$ is given if $p$ is true if and only if $q$ is true (and hence $p$ is false iff $q$ is false). Since $p$ and $q$ can only be true or false, you can use truth tables and check whether logical equivalence is given. From there you can derive permissible manipulations of propositions.
In this case, $(lnot p) rightarrow q equiv lnot(lnot p) lor q equiv p lor q$, as you stated correctly.
answered Nov 11 '18 at 8:34
MacRanceMacRance
1226
$endgroup$
add a comment |
$begingroup$
You have pretty much given the answer yourself already. Logical equivalence of $p$ and $q$ is given if $p$ is true if and only if $q$ is true (and hence $p$ is false iff $q$ is false). Since $p$ and $q$ can only be true or false, you can use truth tables and check whether logical equivalence is given. From there you can derive permissible manipulations of propositions.
In this case, $(lnot p) rightarrow q equiv lnot(lnot p) lor q equiv p lor q$, as you stated correctly.
answered Nov 11 '18 at 8:34
MacRanceMacRance
1226
$endgroup$
add a comment |
$begingroup$
You have pretty much given the answer yourself already. Logical equivalence of $p$ and $q$ is given if $p$ is true if and only if $q$ is true (and hence $p$ is false iff $q$ is false). Since $p$ and $q$ can only be true or false, you can use truth tables and check whether logical equivalence is given. From there you can derive permissible manipulations of propositions.
In this case, $(lnot p) rightarrow q equiv lnot(lnot p) lor q equiv p lor q$, as you stated correctly.
answered Nov 11 '18 at 8:34
MacRanceMacRance
1226
$endgroup$
You have pretty much given the answer yourself already. Logical equivalence of $p$ and $q$ is given if $p$ is true if and only if $q$ is true (and hence $p$ is false iff $q$ is false). Since $p$ and $q$ can only be true or false, you can use truth tables and check whether logical equivalence is given. From there you can derive permissible manipulations of propositions.
In this case, $(lnot p) rightarrow q equiv lnot(lnot p) lor q equiv p lor q$, as you stated correctly.
answered Nov 11 '18 at 8:34
MacRanceMacRance
1226
answered Nov 11 '18 at 8:34
MacRanceMacRance
1226
answered Nov 11 '18 at 8:34
MacRanceMacRance
1226
answered Nov 11 '18 at 8:34
MacRanceMacRance
1226
1226
add a comment |
add a comment |
$begingroup$
As you said, $pRightarrow q$ is logically equivalent to $neg pvee q$.
Then by double negation, $neg p Rightarrow q$ is logically equivalent to $neg(neg p)vee q$, which amounts to $pvee q$.
answered Nov 11 '18 at 8:32
WuestenfuxWuestenfux
4,2691413
$endgroup$
add a comment |
$begingroup$
As you said, $pRightarrow q$ is logically equivalent to $neg pvee q$.
Then by double negation, $neg p Rightarrow q$ is logically equivalent to $neg(neg p)vee q$, which amounts to $pvee q$.
answered Nov 11 '18 at 8:32
WuestenfuxWuestenfux
4,2691413
$endgroup$
add a comment |
$begingroup$
As you said, $pRightarrow q$ is logically equivalent to $neg pvee q$.
Then by double negation, $neg p Rightarrow q$ is logically equivalent to $neg(neg p)vee q$, which amounts to $pvee q$.
answered Nov 11 '18 at 8:32
WuestenfuxWuestenfux
4,2691413
$endgroup$
As you said, $pRightarrow q$ is logically equivalent to $neg pvee q$.
Then by double negation, $neg p Rightarrow q$ is logically equivalent to $neg(neg p)vee q$, which amounts to $pvee q$.
answered Nov 11 '18 at 8:32
WuestenfuxWuestenfux
4,2691413
answered Nov 11 '18 at 8:32
WuestenfuxWuestenfux
4,2691413
answered Nov 11 '18 at 8:32
WuestenfuxWuestenfux
4,2691413
answered Nov 11 '18 at 8:32
WuestenfuxWuestenfux
4,2691413
4,2691413
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%2f2993594%2flogical-equivalence-of-%25c2%25acp%25e2%2586%2592q%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
