Why $p lor (lnot p land q)$ is not equal to $(p lor lnot p) land q$ ? I had trying making Truth Table but,...
$begingroup$
Why $p lor (lnot p land q)$ is not equal to $(p lor lnot p) land q$ ? I had trying making Truth Table but, still can't figure it out ?
logic propositional-calculus
$endgroup$
add a comment |
$begingroup$
Why $p lor (lnot p land q)$ is not equal to $(p lor lnot p) land q$ ? I had trying making Truth Table but, still can't figure it out ?
logic propositional-calculus
$endgroup$
$begingroup$
By $sim$ do you mean $lnot$?
$endgroup$
– Yanko
Jan 5 at 12:43
1
$begingroup$
Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
$endgroup$
– Mauro ALLEGRANZA
Jan 5 at 12:44
$begingroup$
But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
$endgroup$
– Alvin Carter
Jan 5 at 12:53
2
$begingroup$
Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
$endgroup$
– Yanko
Jan 5 at 13:02
$begingroup$
Got it. It helps a lot.
$endgroup$
– Alvin Carter
Jan 5 at 19:19
add a comment |
$begingroup$
Why $p lor (lnot p land q)$ is not equal to $(p lor lnot p) land q$ ? I had trying making Truth Table but, still can't figure it out ?
logic propositional-calculus
$endgroup$
Why $p lor (lnot p land q)$ is not equal to $(p lor lnot p) land q$ ? I had trying making Truth Table but, still can't figure it out ?
logic propositional-calculus
logic propositional-calculus
edited Jan 5 at 20:43
Ruggiero Rilievi
179112
179112
asked Jan 5 at 12:41
Alvin CarterAlvin Carter
114
114
$begingroup$
By $sim$ do you mean $lnot$?
$endgroup$
– Yanko
Jan 5 at 12:43
1
$begingroup$
Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
$endgroup$
– Mauro ALLEGRANZA
Jan 5 at 12:44
$begingroup$
But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
$endgroup$
– Alvin Carter
Jan 5 at 12:53
2
$begingroup$
Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
$endgroup$
– Yanko
Jan 5 at 13:02
$begingroup$
Got it. It helps a lot.
$endgroup$
– Alvin Carter
Jan 5 at 19:19
add a comment |
$begingroup$
By $sim$ do you mean $lnot$?
$endgroup$
– Yanko
Jan 5 at 12:43
1
$begingroup$
Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
$endgroup$
– Mauro ALLEGRANZA
Jan 5 at 12:44
$begingroup$
But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
$endgroup$
– Alvin Carter
Jan 5 at 12:53
2
$begingroup$
Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
$endgroup$
– Yanko
Jan 5 at 13:02
$begingroup$
Got it. It helps a lot.
$endgroup$
– Alvin Carter
Jan 5 at 19:19
$begingroup$
By $sim$ do you mean $lnot$?
$endgroup$
– Yanko
Jan 5 at 12:43
$begingroup$
By $sim$ do you mean $lnot$?
$endgroup$
– Yanko
Jan 5 at 12:43
1
1
$begingroup$
Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
$endgroup$
– Mauro ALLEGRANZA
Jan 5 at 12:44
$begingroup$
Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
$endgroup$
– Mauro ALLEGRANZA
Jan 5 at 12:44
$begingroup$
But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
$endgroup$
– Alvin Carter
Jan 5 at 12:53
$begingroup$
But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
$endgroup$
– Alvin Carter
Jan 5 at 12:53
2
2
$begingroup$
Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
$endgroup$
– Yanko
Jan 5 at 13:02
$begingroup$
Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
$endgroup$
– Yanko
Jan 5 at 13:02
$begingroup$
Got it. It helps a lot.
$endgroup$
– Alvin Carter
Jan 5 at 19:19
$begingroup$
Got it. It helps a lot.
$endgroup$
– Alvin Carter
Jan 5 at 19:19
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
If $p$ is true then the first is automatically true. If $q$ is false then the second is automatically false.
I have no idea what you mean by $sim$ but it doesn't even matter.
$endgroup$
add a comment |
$begingroup$
The statement $pveeneg p$ is a true statement in Boolean algebra.
Then:
$$pvee(neg pwedge q)=(pveeneg p)wedge(pvee q)=pvee q$$by distribution and:$$(pveeneg p)wedge q=q$$
If you are approaching with truth tables then you encounter the difference for $q=0$ and $p=1$.
$endgroup$
add a comment |
$begingroup$
$(pveesim p)wedge q=1wedge q=qbecause pveesim p=1$
$p∨(sim p∧q)=pvee q$ by the Absorption law.
Try $p=1,q=0$.
$endgroup$
add a comment |
$begingroup$
The term $(pveeneg p)vee q$ is equivalent to $q$ since term in parenthesis is logically true.
The term $pvee (neg pwedge q)$ has the truth value $1$ if $p=1$ independent of the value of $q$.
$endgroup$
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%2f3062677%2fwhy-p-lor-lnot-p-land-q-is-not-equal-to-p-lor-lnot-p-land-q-i-ha%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $p$ is true then the first is automatically true. If $q$ is false then the second is automatically false.
I have no idea what you mean by $sim$ but it doesn't even matter.
$endgroup$
add a comment |
$begingroup$
If $p$ is true then the first is automatically true. If $q$ is false then the second is automatically false.
I have no idea what you mean by $sim$ but it doesn't even matter.
$endgroup$
add a comment |
$begingroup$
If $p$ is true then the first is automatically true. If $q$ is false then the second is automatically false.
I have no idea what you mean by $sim$ but it doesn't even matter.
$endgroup$
If $p$ is true then the first is automatically true. If $q$ is false then the second is automatically false.
I have no idea what you mean by $sim$ but it doesn't even matter.
answered Jan 5 at 12:44
YankoYanko
7,4801729
7,4801729
add a comment |
add a comment |
$begingroup$
The statement $pveeneg p$ is a true statement in Boolean algebra.
Then:
$$pvee(neg pwedge q)=(pveeneg p)wedge(pvee q)=pvee q$$by distribution and:$$(pveeneg p)wedge q=q$$
If you are approaching with truth tables then you encounter the difference for $q=0$ and $p=1$.
$endgroup$
add a comment |
$begingroup$
The statement $pveeneg p$ is a true statement in Boolean algebra.
Then:
$$pvee(neg pwedge q)=(pveeneg p)wedge(pvee q)=pvee q$$by distribution and:$$(pveeneg p)wedge q=q$$
If you are approaching with truth tables then you encounter the difference for $q=0$ and $p=1$.
$endgroup$
add a comment |
$begingroup$
The statement $pveeneg p$ is a true statement in Boolean algebra.
Then:
$$pvee(neg pwedge q)=(pveeneg p)wedge(pvee q)=pvee q$$by distribution and:$$(pveeneg p)wedge q=q$$
If you are approaching with truth tables then you encounter the difference for $q=0$ and $p=1$.
$endgroup$
The statement $pveeneg p$ is a true statement in Boolean algebra.
Then:
$$pvee(neg pwedge q)=(pveeneg p)wedge(pvee q)=pvee q$$by distribution and:$$(pveeneg p)wedge q=q$$
If you are approaching with truth tables then you encounter the difference for $q=0$ and $p=1$.
answered Jan 5 at 12:49
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
$begingroup$
$(pveesim p)wedge q=1wedge q=qbecause pveesim p=1$
$p∨(sim p∧q)=pvee q$ by the Absorption law.
Try $p=1,q=0$.
$endgroup$
add a comment |
$begingroup$
$(pveesim p)wedge q=1wedge q=qbecause pveesim p=1$
$p∨(sim p∧q)=pvee q$ by the Absorption law.
Try $p=1,q=0$.
$endgroup$
add a comment |
$begingroup$
$(pveesim p)wedge q=1wedge q=qbecause pveesim p=1$
$p∨(sim p∧q)=pvee q$ by the Absorption law.
Try $p=1,q=0$.
$endgroup$
$(pveesim p)wedge q=1wedge q=qbecause pveesim p=1$
$p∨(sim p∧q)=pvee q$ by the Absorption law.
Try $p=1,q=0$.
answered Jan 5 at 12:45
Shubham JohriShubham Johri
5,204718
5,204718
add a comment |
add a comment |
$begingroup$
The term $(pveeneg p)vee q$ is equivalent to $q$ since term in parenthesis is logically true.
The term $pvee (neg pwedge q)$ has the truth value $1$ if $p=1$ independent of the value of $q$.
$endgroup$
add a comment |
$begingroup$
The term $(pveeneg p)vee q$ is equivalent to $q$ since term in parenthesis is logically true.
The term $pvee (neg pwedge q)$ has the truth value $1$ if $p=1$ independent of the value of $q$.
$endgroup$
add a comment |
$begingroup$
The term $(pveeneg p)vee q$ is equivalent to $q$ since term in parenthesis is logically true.
The term $pvee (neg pwedge q)$ has the truth value $1$ if $p=1$ independent of the value of $q$.
$endgroup$
The term $(pveeneg p)vee q$ is equivalent to $q$ since term in parenthesis is logically true.
The term $pvee (neg pwedge q)$ has the truth value $1$ if $p=1$ independent of the value of $q$.
answered Jan 5 at 12:45
WuestenfuxWuestenfux
4,7921513
4,7921513
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%2f3062677%2fwhy-p-lor-lnot-p-land-q-is-not-equal-to-p-lor-lnot-p-land-q-i-ha%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
$begingroup$
By $sim$ do you mean $lnot$?
$endgroup$
– Yanko
Jan 5 at 12:43
1
$begingroup$
Because the first one is equiv to $(p lor lnot p) land (p lor q)$. See Distributivity.
$endgroup$
– Mauro ALLEGRANZA
Jan 5 at 12:44
$begingroup$
But by, Assosiative Law we can change that Bracket thing ??? so why they are not true???
$endgroup$
– Alvin Carter
Jan 5 at 12:53
2
$begingroup$
Note that you have two different operations $land$ and $lor$. We don't have associativity with these two different operations. It might help you to compare it with $+,cdot$. For example $(4+2)cdot 3 not = 4+(2cdot 3)$.
$endgroup$
– Yanko
Jan 5 at 13:02
$begingroup$
Got it. It helps a lot.
$endgroup$
– Alvin Carter
Jan 5 at 19:19