Prove $p ↔ q$ and $(p ∧ q) ∨ (¬p ∧ ¬q)$ are equivalent using logic laws
$begingroup$
I know that we can show this using a truth table but I can't prove it using logic laws.
p ↔ q ≡ (p→q)∧(q→p)
p ↔ q ≡ (¬p∨q)∧(¬q∨p)
p ↔ q ≡ ¬p∨(p∧q) ∧ ¬q∨(q∧p)
I go this far and then I'm stuck.
logic propositional-calculus
edited Jan 10 at 12:55
Key Flex
8,58771233
asked Jan 10 at 11:26
Free SoulFree Soul
102
$endgroup$
add a comment |
$begingroup$
I know that we can show this using a truth table but I can't prove it using logic laws.
p ↔ q ≡ (p→q)∧(q→p)
p ↔ q ≡ (¬p∨q)∧(¬q∨p)
p ↔ q ≡ ¬p∨(p∧q) ∧ ¬q∨(q∧p)
I go this far and then I'm stuck.
logic propositional-calculus
edited Jan 10 at 12:55
Key Flex
8,58771233
asked Jan 10 at 11:26
Free SoulFree Soul
102
$endgroup$
add a comment |
$begingroup$
I know that we can show this using a truth table but I can't prove it using logic laws.
p ↔ q ≡ (p→q)∧(q→p)
p ↔ q ≡ (¬p∨q)∧(¬q∨p)
p ↔ q ≡ ¬p∨(p∧q) ∧ ¬q∨(q∧p)
I go this far and then I'm stuck.
logic propositional-calculus
edited Jan 10 at 12:55
Key Flex
8,58771233
asked Jan 10 at 11:26
Free SoulFree Soul
102
$endgroup$
I know that we can show this using a truth table but I can't prove it using logic laws.
p ↔ q ≡ (p→q)∧(q→p)
p ↔ q ≡ (¬p∨q)∧(¬q∨p)
p ↔ q ≡ ¬p∨(p∧q) ∧ ¬q∨(q∧p)
I go this far and then I'm stuck.
logic propositional-calculus
logic propositional-calculus
edited Jan 10 at 12:55
Key Flex
8,58771233
asked Jan 10 at 11:26
Free SoulFree Soul
102
edited Jan 10 at 12:55
Key Flex
8,58771233
asked Jan 10 at 11:26
Free SoulFree Soul
102
edited Jan 10 at 12:55
Key Flex
8,58771233
edited Jan 10 at 12:55
Key Flex
8,58771233
edited Jan 10 at 12:55
Key Flex
8,58771233
8,58771233
asked Jan 10 at 11:26
Free SoulFree Soul
102
asked Jan 10 at 11:26
Free SoulFree Soul
102
asked Jan 10 at 11:26
Free SoulFree Soul
102
102
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
From the second line on use distributivity and
- $a wedge lnot a = F$
- $a vee F = a$
begin{eqnarray*} (lnot p vee q) wedge (lnot q vee p)
& = & (lnot p wedge lnot q) vee (lnot p wedge p) vee (q wedge lnot q) vee (q wedge p) \
& = & (lnot p wedge lnot q) vee F vee F vee (q wedge p) \
& = & (lnot p wedge lnot q) vee (q wedge p) \
end{eqnarray*}
answered Jan 10 at 11:37
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
$begingroup$
Oh my goodness, you were almost there! You had to take just one more step! You had:
$p ↔ q ≡ (p→q)∧(q→p)$ ... By rewriting $equiv$ ('Equivalence')
$≡ (¬p∨q)∧(¬q∨p)$ ... By rewriting $to$ ('Implication')
$≡ ¬p∨(p∧q) ∧ ¬q∨(q∧p)$ ... By 'Reduction', which states that $¬p∨(p∧q) equiv ¬p∨q$
... OK, and now just do:
$≡ (¬p∧ ¬q)∨(p∧q)$ ... By 'Distribution'!!
... ok , ok, it isn't just Distribution,; you also need to realize that $p∧q$ and $q∧p$ are the same term .. so you need Commutation as well.
More importantly though, you really needed to add some parentheses to line 3:
$≡ (¬p∨(p∧q)) ∧ (¬q∨(q∧p))$
And, had you done, that, I think it would have been more likely that you would have noticed the possibility or recognizing that the two terms here both share the term $p land q$ (again, after commutation), and hence that Distribution would have gotten you there!
answered Jan 12 at 20:23
Bram28Bram28
64.2k44793
$endgroup$
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%2f3068522%2fprove-p-%25e2%2586%2594-q-and-p-%25e2%2588%25a7-q-%25e2%2588%25a8-%25c2%25acp-%25e2%2588%25a7-%25c2%25acq-are-equivalent-using-logic-laws%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From the second line on use distributivity and
- $a wedge lnot a = F$
- $a vee F = a$
begin{eqnarray*} (lnot p vee q) wedge (lnot q vee p)
& = & (lnot p wedge lnot q) vee (lnot p wedge p) vee (q wedge lnot q) vee (q wedge p) \
& = & (lnot p wedge lnot q) vee F vee F vee (q wedge p) \
& = & (lnot p wedge lnot q) vee (q wedge p) \
end{eqnarray*}
answered Jan 10 at 11:37
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
$begingroup$
From the second line on use distributivity and
- $a wedge lnot a = F$
- $a vee F = a$
begin{eqnarray*} (lnot p vee q) wedge (lnot q vee p)
& = & (lnot p wedge lnot q) vee (lnot p wedge p) vee (q wedge lnot q) vee (q wedge p) \
& = & (lnot p wedge lnot q) vee F vee F vee (q wedge p) \
& = & (lnot p wedge lnot q) vee (q wedge p) \
end{eqnarray*}
answered Jan 10 at 11:37
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
$begingroup$
From the second line on use distributivity and
- $a wedge lnot a = F$
- $a vee F = a$
begin{eqnarray*} (lnot p vee q) wedge (lnot q vee p)
& = & (lnot p wedge lnot q) vee (lnot p wedge p) vee (q wedge lnot q) vee (q wedge p) \
& = & (lnot p wedge lnot q) vee F vee F vee (q wedge p) \
& = & (lnot p wedge lnot q) vee (q wedge p) \
end{eqnarray*}
answered Jan 10 at 11:37
trancelocationtrancelocation
13.6k1829
$endgroup$
From the second line on use distributivity and
- $a wedge lnot a = F$
- $a vee F = a$
begin{eqnarray*} (lnot p vee q) wedge (lnot q vee p)
& = & (lnot p wedge lnot q) vee (lnot p wedge p) vee (q wedge lnot q) vee (q wedge p) \
& = & (lnot p wedge lnot q) vee F vee F vee (q wedge p) \
& = & (lnot p wedge lnot q) vee (q wedge p) \
end{eqnarray*}
answered Jan 10 at 11:37
trancelocationtrancelocation
13.6k1829
answered Jan 10 at 11:37
trancelocationtrancelocation
13.6k1829
answered Jan 10 at 11:37
trancelocationtrancelocation
13.6k1829
answered Jan 10 at 11:37
trancelocationtrancelocation
13.6k1829
13.6k1829
add a comment |
add a comment |
$begingroup$
Oh my goodness, you were almost there! You had to take just one more step! You had:
$p ↔ q ≡ (p→q)∧(q→p)$ ... By rewriting $equiv$ ('Equivalence')
$≡ (¬p∨q)∧(¬q∨p)$ ... By rewriting $to$ ('Implication')
$≡ ¬p∨(p∧q) ∧ ¬q∨(q∧p)$ ... By 'Reduction', which states that $¬p∨(p∧q) equiv ¬p∨q$
... OK, and now just do:
$≡ (¬p∧ ¬q)∨(p∧q)$ ... By 'Distribution'!!
... ok , ok, it isn't just Distribution,; you also need to realize that $p∧q$ and $q∧p$ are the same term .. so you need Commutation as well.
More importantly though, you really needed to add some parentheses to line 3:
$≡ (¬p∨(p∧q)) ∧ (¬q∨(q∧p))$
And, had you done, that, I think it would have been more likely that you would have noticed the possibility or recognizing that the two terms here both share the term $p land q$ (again, after commutation), and hence that Distribution would have gotten you there!
answered Jan 12 at 20:23
Bram28Bram28
64.2k44793
$endgroup$
add a comment |
$begingroup$
Oh my goodness, you were almost there! You had to take just one more step! You had:
$p ↔ q ≡ (p→q)∧(q→p)$ ... By rewriting $equiv$ ('Equivalence')
$≡ (¬p∨q)∧(¬q∨p)$ ... By rewriting $to$ ('Implication')
$≡ ¬p∨(p∧q) ∧ ¬q∨(q∧p)$ ... By 'Reduction', which states that $¬p∨(p∧q) equiv ¬p∨q$
... OK, and now just do:
$≡ (¬p∧ ¬q)∨(p∧q)$ ... By 'Distribution'!!
... ok , ok, it isn't just Distribution,; you also need to realize that $p∧q$ and $q∧p$ are the same term .. so you need Commutation as well.
More importantly though, you really needed to add some parentheses to line 3:
$≡ (¬p∨(p∧q)) ∧ (¬q∨(q∧p))$
And, had you done, that, I think it would have been more likely that you would have noticed the possibility or recognizing that the two terms here both share the term $p land q$ (again, after commutation), and hence that Distribution would have gotten you there!
answered Jan 12 at 20:23
Bram28Bram28
64.2k44793
$endgroup$
add a comment |
$begingroup$
Oh my goodness, you were almost there! You had to take just one more step! You had:
$p ↔ q ≡ (p→q)∧(q→p)$ ... By rewriting $equiv$ ('Equivalence')
$≡ (¬p∨q)∧(¬q∨p)$ ... By rewriting $to$ ('Implication')
$≡ ¬p∨(p∧q) ∧ ¬q∨(q∧p)$ ... By 'Reduction', which states that $¬p∨(p∧q) equiv ¬p∨q$
... OK, and now just do:
$≡ (¬p∧ ¬q)∨(p∧q)$ ... By 'Distribution'!!
... ok , ok, it isn't just Distribution,; you also need to realize that $p∧q$ and $q∧p$ are the same term .. so you need Commutation as well.
More importantly though, you really needed to add some parentheses to line 3:
$≡ (¬p∨(p∧q)) ∧ (¬q∨(q∧p))$
And, had you done, that, I think it would have been more likely that you would have noticed the possibility or recognizing that the two terms here both share the term $p land q$ (again, after commutation), and hence that Distribution would have gotten you there!
answered Jan 12 at 20:23
Bram28Bram28
64.2k44793
$endgroup$
Oh my goodness, you were almost there! You had to take just one more step! You had:
$p ↔ q ≡ (p→q)∧(q→p)$ ... By rewriting $equiv$ ('Equivalence')
$≡ (¬p∨q)∧(¬q∨p)$ ... By rewriting $to$ ('Implication')
$≡ ¬p∨(p∧q) ∧ ¬q∨(q∧p)$ ... By 'Reduction', which states that $¬p∨(p∧q) equiv ¬p∨q$
... OK, and now just do:
$≡ (¬p∧ ¬q)∨(p∧q)$ ... By 'Distribution'!!
... ok , ok, it isn't just Distribution,; you also need to realize that $p∧q$ and $q∧p$ are the same term .. so you need Commutation as well.
More importantly though, you really needed to add some parentheses to line 3:
$≡ (¬p∨(p∧q)) ∧ (¬q∨(q∧p))$
And, had you done, that, I think it would have been more likely that you would have noticed the possibility or recognizing that the two terms here both share the term $p land q$ (again, after commutation), and hence that Distribution would have gotten you there!
answered Jan 12 at 20:23
Bram28Bram28
64.2k44793
answered Jan 12 at 20:23
Bram28Bram28
64.2k44793
answered Jan 12 at 20:23
Bram28Bram28
64.2k44793
answered Jan 12 at 20:23
Bram28Bram28
64.2k44793
64.2k44793
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%2f3068522%2fprove-p-%25e2%2586%2594-q-and-p-%25e2%2588%25a7-q-%25e2%2588%25a8-%25c2%25acp-%25e2%2588%25a7-%25c2%25acq-are-equivalent-using-logic-laws%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
