Proving Logical Equivalence - $lnot(( p leftrightarrow q )) equiv (p leftrightarrow ¬( q )))$












1












$begingroup$


I'm trying to prove that the LHS and RHS of the following are logically equivalent:



$lnot(( p leftrightarrow q ) equiv (p leftrightarrow ¬( q )))$



This is what I've managed to get so far, but I'm not sure where I can go from there...



$≡ ¬(( pto q ) ∧ ( qto p ))]$



$≡ ¬(((¬p) ∨ q ) ∧ ((¬q) ∨ p )))$



$≡ ¬((¬p) ∨ q ) ∨ ¬((¬q) ∨ p ))$



$≡ (p ∧ (¬q)) ∨ (q ∧ (¬p))$






discrete-mathematics logic







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
    $endgroup$
    – dantopa
    Jan 18 at 16:24
















1












$begingroup$


I'm trying to prove that the LHS and RHS of the following are logically equivalent:



$lnot(( p leftrightarrow q ) equiv (p leftrightarrow ¬( q )))$



This is what I've managed to get so far, but I'm not sure where I can go from there...



$≡ ¬(( pto q ) ∧ ( qto p ))]$



$≡ ¬(((¬p) ∨ q ) ∧ ((¬q) ∨ p )))$



$≡ ¬((¬p) ∨ q ) ∨ ¬((¬q) ∨ p ))$



$≡ (p ∧ (¬q)) ∨ (q ∧ (¬p))$






discrete-mathematics logic







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
    $endgroup$
    – dantopa
    Jan 18 at 16:24














1












1








1





$begingroup$


I'm trying to prove that the LHS and RHS of the following are logically equivalent:



$lnot(( p leftrightarrow q ) equiv (p leftrightarrow ¬( q )))$



This is what I've managed to get so far, but I'm not sure where I can go from there...



$≡ ¬(( pto q ) ∧ ( qto p ))]$



$≡ ¬(((¬p) ∨ q ) ∧ ((¬q) ∨ p )))$



$≡ ¬((¬p) ∨ q ) ∨ ¬((¬q) ∨ p ))$



$≡ (p ∧ (¬q)) ∨ (q ∧ (¬p))$






discrete-mathematics logic







share|cite|improve this question











$endgroup$




I'm trying to prove that the LHS and RHS of the following are logically equivalent:



$lnot(( p leftrightarrow q ) equiv (p leftrightarrow ¬( q )))$



This is what I've managed to get so far, but I'm not sure where I can go from there...



$≡ ¬(( pto q ) ∧ ( qto p ))]$



$≡ ¬(((¬p) ∨ q ) ∧ ((¬q) ∨ p )))$



$≡ ¬((¬p) ∨ q ) ∨ ¬((¬q) ∨ p ))$



$≡ (p ∧ (¬q)) ∨ (q ∧ (¬p))$






discrete-mathematics logic




discrete-mathematics logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 18:04









jordan_glen

1




1










asked Jan 18 at 16:19









BandoleroBandolero

243




243












  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
    $endgroup$
    – dantopa
    Jan 18 at 16:24


















  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
    $endgroup$
    – dantopa
    Jan 18 at 16:24
















$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 18 at 16:24




$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 18 at 16:24










1 Answer
1






active

oldest

votes


















1












$begingroup$

All your work, thus far, is correct.



You might want to work on the RHS of the proposed equivalence:



$$(piff (-q)) equiv ((pto (lnot q)) land ((lnot q) to p)$$



$$equiv ((lnot p) lor (lnot q)) land (q lor p)$$



$$equiv ((lnot p) land (qlor p))lor ((lnot q)land (qlor p))tag{distributive law}$$



$$equiv (((lnot p) land q) lor ((lnot p)land p))) lor (((lnot q) land q) lor((lnot q) land p)))tag{dist. law $times 2$}$$



$$equiv ((lnot p) land q) lor F lor F lor ((lnot q) land p$$



$$equiv ((lnot p) land q)lor ((lnot q) land p)$$



$$equiv (q land (lnot p)) lor (p land (lnot q)) tag{commutativity}$$



$$equiv (p land (lnot q)) lor (q land (lnot p))tag{commutativity}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Good. Yes, sometimes you want to work on both the LHS and the RHS, and try and 'meet in the middle'.
    $endgroup$
    – Bram28
    Jan 18 at 17:37










  • $begingroup$
    Thanks for the answer, did it with RHS like you did, but got a final answer of [ ¬ ( q <-> p ) ] instead of [ ¬ ( p <-> q ) ]. Is it reversible (i.e. basically the same) or maybe I might've done something wrong?
    $endgroup$
    – Bandolero
    Jan 18 at 17:38












  • $begingroup$
    Hi, Bandolero. Sorry for taking so long to get back to you. Your result is correct, because $(qiff p)equiv (piff q)$ by commutativity. So the negation of one is equivalent to the negation of the other.
    $endgroup$
    – jordan_glen
    Jan 19 at 17:21












  • $begingroup$
    What I tried to do wrt this question, was accept your correct work for which you found the LHS $equiv big((p land (lnot q)) lor (q land (lnot p))big)$, and I verified that the RHS $equivbig( (p land (lnot q)) lor (q land (lnot p))big)$. Therefore, the LHS $equiv$ RHS.
    $endgroup$
    – jordan_glen
    Jan 19 at 18:54













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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3078443%2fproving-logical-equivalence-lnot-p-leftrightarrow-q-equiv-p-leftrig%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









1












$begingroup$

All your work, thus far, is correct.



You might want to work on the RHS of the proposed equivalence:



$$(piff (-q)) equiv ((pto (lnot q)) land ((lnot q) to p)$$



$$equiv ((lnot p) lor (lnot q)) land (q lor p)$$



$$equiv ((lnot p) land (qlor p))lor ((lnot q)land (qlor p))tag{distributive law}$$



$$equiv (((lnot p) land q) lor ((lnot p)land p))) lor (((lnot q) land q) lor((lnot q) land p)))tag{dist. law $times 2$}$$



$$equiv ((lnot p) land q) lor F lor F lor ((lnot q) land p$$



$$equiv ((lnot p) land q)lor ((lnot q) land p)$$



$$equiv (q land (lnot p)) lor (p land (lnot q)) tag{commutativity}$$



$$equiv (p land (lnot q)) lor (q land (lnot p))tag{commutativity}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Good. Yes, sometimes you want to work on both the LHS and the RHS, and try and 'meet in the middle'.
    $endgroup$
    – Bram28
    Jan 18 at 17:37










  • $begingroup$
    Thanks for the answer, did it with RHS like you did, but got a final answer of [ ¬ ( q <-> p ) ] instead of [ ¬ ( p <-> q ) ]. Is it reversible (i.e. basically the same) or maybe I might've done something wrong?
    $endgroup$
    – Bandolero
    Jan 18 at 17:38












  • $begingroup$
    Hi, Bandolero. Sorry for taking so long to get back to you. Your result is correct, because $(qiff p)equiv (piff q)$ by commutativity. So the negation of one is equivalent to the negation of the other.
    $endgroup$
    – jordan_glen
    Jan 19 at 17:21












  • $begingroup$
    What I tried to do wrt this question, was accept your correct work for which you found the LHS $equiv big((p land (lnot q)) lor (q land (lnot p))big)$, and I verified that the RHS $equivbig( (p land (lnot q)) lor (q land (lnot p))big)$. Therefore, the LHS $equiv$ RHS.
    $endgroup$
    – jordan_glen
    Jan 19 at 18:54


















1












$begingroup$

All your work, thus far, is correct.



You might want to work on the RHS of the proposed equivalence:



$$(piff (-q)) equiv ((pto (lnot q)) land ((lnot q) to p)$$



$$equiv ((lnot p) lor (lnot q)) land (q lor p)$$



$$equiv ((lnot p) land (qlor p))lor ((lnot q)land (qlor p))tag{distributive law}$$



$$equiv (((lnot p) land q) lor ((lnot p)land p))) lor (((lnot q) land q) lor((lnot q) land p)))tag{dist. law $times 2$}$$



$$equiv ((lnot p) land q) lor F lor F lor ((lnot q) land p$$



$$equiv ((lnot p) land q)lor ((lnot q) land p)$$



$$equiv (q land (lnot p)) lor (p land (lnot q)) tag{commutativity}$$



$$equiv (p land (lnot q)) lor (q land (lnot p))tag{commutativity}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Good. Yes, sometimes you want to work on both the LHS and the RHS, and try and 'meet in the middle'.
    $endgroup$
    – Bram28
    Jan 18 at 17:37










  • $begingroup$
    Thanks for the answer, did it with RHS like you did, but got a final answer of [ ¬ ( q <-> p ) ] instead of [ ¬ ( p <-> q ) ]. Is it reversible (i.e. basically the same) or maybe I might've done something wrong?
    $endgroup$
    – Bandolero
    Jan 18 at 17:38












  • $begingroup$
    Hi, Bandolero. Sorry for taking so long to get back to you. Your result is correct, because $(qiff p)equiv (piff q)$ by commutativity. So the negation of one is equivalent to the negation of the other.
    $endgroup$
    – jordan_glen
    Jan 19 at 17:21












  • $begingroup$
    What I tried to do wrt this question, was accept your correct work for which you found the LHS $equiv big((p land (lnot q)) lor (q land (lnot p))big)$, and I verified that the RHS $equivbig( (p land (lnot q)) lor (q land (lnot p))big)$. Therefore, the LHS $equiv$ RHS.
    $endgroup$
    – jordan_glen
    Jan 19 at 18:54
















1












1








1





$begingroup$

All your work, thus far, is correct.



You might want to work on the RHS of the proposed equivalence:



$$(piff (-q)) equiv ((pto (lnot q)) land ((lnot q) to p)$$



$$equiv ((lnot p) lor (lnot q)) land (q lor p)$$



$$equiv ((lnot p) land (qlor p))lor ((lnot q)land (qlor p))tag{distributive law}$$



$$equiv (((lnot p) land q) lor ((lnot p)land p))) lor (((lnot q) land q) lor((lnot q) land p)))tag{dist. law $times 2$}$$



$$equiv ((lnot p) land q) lor F lor F lor ((lnot q) land p$$



$$equiv ((lnot p) land q)lor ((lnot q) land p)$$



$$equiv (q land (lnot p)) lor (p land (lnot q)) tag{commutativity}$$



$$equiv (p land (lnot q)) lor (q land (lnot p))tag{commutativity}$$






share|cite|improve this answer











$endgroup$



All your work, thus far, is correct.



You might want to work on the RHS of the proposed equivalence:



$$(piff (-q)) equiv ((pto (lnot q)) land ((lnot q) to p)$$



$$equiv ((lnot p) lor (lnot q)) land (q lor p)$$



$$equiv ((lnot p) land (qlor p))lor ((lnot q)land (qlor p))tag{distributive law}$$



$$equiv (((lnot p) land q) lor ((lnot p)land p))) lor (((lnot q) land q) lor((lnot q) land p)))tag{dist. law $times 2$}$$



$$equiv ((lnot p) land q) lor F lor F lor ((lnot q) land p$$



$$equiv ((lnot p) land q)lor ((lnot q) land p)$$



$$equiv (q land (lnot p)) lor (p land (lnot q)) tag{commutativity}$$



$$equiv (p land (lnot q)) lor (q land (lnot p))tag{commutativity}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 18 at 16:44

























answered Jan 18 at 16:29









jordan_glenjordan_glen

1




1












  • $begingroup$
    Good. Yes, sometimes you want to work on both the LHS and the RHS, and try and 'meet in the middle'.
    $endgroup$
    – Bram28
    Jan 18 at 17:37










  • $begingroup$
    Thanks for the answer, did it with RHS like you did, but got a final answer of [ ¬ ( q <-> p ) ] instead of [ ¬ ( p <-> q ) ]. Is it reversible (i.e. basically the same) or maybe I might've done something wrong?
    $endgroup$
    – Bandolero
    Jan 18 at 17:38












  • $begingroup$
    Hi, Bandolero. Sorry for taking so long to get back to you. Your result is correct, because $(qiff p)equiv (piff q)$ by commutativity. So the negation of one is equivalent to the negation of the other.
    $endgroup$
    – jordan_glen
    Jan 19 at 17:21












  • $begingroup$
    What I tried to do wrt this question, was accept your correct work for which you found the LHS $equiv big((p land (lnot q)) lor (q land (lnot p))big)$, and I verified that the RHS $equivbig( (p land (lnot q)) lor (q land (lnot p))big)$. Therefore, the LHS $equiv$ RHS.
    $endgroup$
    – jordan_glen
    Jan 19 at 18:54




















  • $begingroup$
    Good. Yes, sometimes you want to work on both the LHS and the RHS, and try and 'meet in the middle'.
    $endgroup$
    – Bram28
    Jan 18 at 17:37










  • $begingroup$
    Thanks for the answer, did it with RHS like you did, but got a final answer of [ ¬ ( q <-> p ) ] instead of [ ¬ ( p <-> q ) ]. Is it reversible (i.e. basically the same) or maybe I might've done something wrong?
    $endgroup$
    – Bandolero
    Jan 18 at 17:38












  • $begingroup$
    Hi, Bandolero. Sorry for taking so long to get back to you. Your result is correct, because $(qiff p)equiv (piff q)$ by commutativity. So the negation of one is equivalent to the negation of the other.
    $endgroup$
    – jordan_glen
    Jan 19 at 17:21












  • $begingroup$
    What I tried to do wrt this question, was accept your correct work for which you found the LHS $equiv big((p land (lnot q)) lor (q land (lnot p))big)$, and I verified that the RHS $equivbig( (p land (lnot q)) lor (q land (lnot p))big)$. Therefore, the LHS $equiv$ RHS.
    $endgroup$
    – jordan_glen
    Jan 19 at 18:54


















$begingroup$
Good. Yes, sometimes you want to work on both the LHS and the RHS, and try and 'meet in the middle'.
$endgroup$
– Bram28
Jan 18 at 17:37




$begingroup$
Good. Yes, sometimes you want to work on both the LHS and the RHS, and try and 'meet in the middle'.
$endgroup$
– Bram28
Jan 18 at 17:37












$begingroup$
Thanks for the answer, did it with RHS like you did, but got a final answer of [ ¬ ( q <-> p ) ] instead of [ ¬ ( p <-> q ) ]. Is it reversible (i.e. basically the same) or maybe I might've done something wrong?
$endgroup$
– Bandolero
Jan 18 at 17:38






$begingroup$
Thanks for the answer, did it with RHS like you did, but got a final answer of [ ¬ ( q <-> p ) ] instead of [ ¬ ( p <-> q ) ]. Is it reversible (i.e. basically the same) or maybe I might've done something wrong?
$endgroup$
– Bandolero
Jan 18 at 17:38














$begingroup$
Hi, Bandolero. Sorry for taking so long to get back to you. Your result is correct, because $(qiff p)equiv (piff q)$ by commutativity. So the negation of one is equivalent to the negation of the other.
$endgroup$
– jordan_glen
Jan 19 at 17:21






$begingroup$
Hi, Bandolero. Sorry for taking so long to get back to you. Your result is correct, because $(qiff p)equiv (piff q)$ by commutativity. So the negation of one is equivalent to the negation of the other.
$endgroup$
– jordan_glen
Jan 19 at 17:21














$begingroup$
What I tried to do wrt this question, was accept your correct work for which you found the LHS $equiv big((p land (lnot q)) lor (q land (lnot p))big)$, and I verified that the RHS $equivbig( (p land (lnot q)) lor (q land (lnot p))big)$. Therefore, the LHS $equiv$ RHS.
$endgroup$
– jordan_glen
Jan 19 at 18:54






$begingroup$
What I tried to do wrt this question, was accept your correct work for which you found the LHS $equiv big((p land (lnot q)) lor (q land (lnot p))big)$, and I verified that the RHS $equivbig( (p land (lnot q)) lor (q land (lnot p))big)$. Therefore, the LHS $equiv$ RHS.
$endgroup$
– jordan_glen
Jan 19 at 18:54




















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3078443%2fproving-logical-equivalence-lnot-p-leftrightarrow-q-equiv-p-leftrig%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
Read more

Partial Derivative Guidance.

Image
0 $begingroup$ I was given this tricky problem in which I'm confuse on how to solve it and I would like some help: Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$ . Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives. My attempt (which is wrong): Using chain rule: $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$ $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$ So, $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$ $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$ Thanks, calculus multivariable-calculus share | cite | improve this question edited Jan 19 at 19:38...
Read more

Understanding the size os this class of aleatory events

Image
0 1 $begingroup$ Hi I am having problems to understand this: I have: $Omega = [0,1]$ , $Lambda =$ all subsets with defined length Let $A_{0}$ = {A $subset$ [0,1]: A is finite union of intervals} 1º: My doubt is that $A_{0}$ is a algebra but its not a sigma algebra. Why? Also, A = $(0,1/2) cup (1/2,3/4) cup (3/4,7/8) cup ... cup(1-1/2^{n},1-1/2^{n+1})...$ 2º: So the class of aleatory events would be bigger than $A_{0}$ . Why? How the set A is greater than $A_{0}$ ? measure-theory share | cite | improve this question asked Jan 8 at 19:17 Diogo Bastos Diogo Bastos 116 6 ...
Read more