Combinatorial Circuit












1












$begingroup$


I'm trying to learn more about logical statement. I have this question:
(q --> -p) --> r. Im supposed to create a combinatorial circuit out of this statement. What i did was to convert the logical statement to something easier. So i got:
$(q rightarrow -p ) rightarrow r$ ... $(-q cup -p ) rightarrow r$...$-(-q cup -p ) cup r$ and then i got $(q cup p ) cup r
$



Im really not sure if this is correct, could anyone please check and correct if I did anything wrong? Thanks!






discrete-mathematics logic







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are the unions supposed to be $lor$? This doesn’t look quite right, as $q=T, p=F, r=F$ satisfies the second expression but not the first.
    $endgroup$
    – platty
    Dec 15 '18 at 16:10










  • $begingroup$
    Does the minus sign mean $lnot?$ You can use lnot for $lnot$ vee for $vee$
    $endgroup$
    – saulspatz
    Dec 15 '18 at 16:12










  • $begingroup$
    Yes it does mean that! Thanks for the tip
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:13
















1












$begingroup$


I'm trying to learn more about logical statement. I have this question:
(q --> -p) --> r. Im supposed to create a combinatorial circuit out of this statement. What i did was to convert the logical statement to something easier. So i got:
$(q rightarrow -p ) rightarrow r$ ... $(-q cup -p ) rightarrow r$...$-(-q cup -p ) cup r$ and then i got $(q cup p ) cup r
$



Im really not sure if this is correct, could anyone please check and correct if I did anything wrong? Thanks!






discrete-mathematics logic







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are the unions supposed to be $lor$? This doesn’t look quite right, as $q=T, p=F, r=F$ satisfies the second expression but not the first.
    $endgroup$
    – platty
    Dec 15 '18 at 16:10










  • $begingroup$
    Does the minus sign mean $lnot?$ You can use lnot for $lnot$ vee for $vee$
    $endgroup$
    – saulspatz
    Dec 15 '18 at 16:12










  • $begingroup$
    Yes it does mean that! Thanks for the tip
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:13














1












1








1





$begingroup$


I'm trying to learn more about logical statement. I have this question:
(q --> -p) --> r. Im supposed to create a combinatorial circuit out of this statement. What i did was to convert the logical statement to something easier. So i got:
$(q rightarrow -p ) rightarrow r$ ... $(-q cup -p ) rightarrow r$...$-(-q cup -p ) cup r$ and then i got $(q cup p ) cup r
$



Im really not sure if this is correct, could anyone please check and correct if I did anything wrong? Thanks!






discrete-mathematics logic







share|cite|improve this question









$endgroup$




I'm trying to learn more about logical statement. I have this question:
(q --> -p) --> r. Im supposed to create a combinatorial circuit out of this statement. What i did was to convert the logical statement to something easier. So i got:
$(q rightarrow -p ) rightarrow r$ ... $(-q cup -p ) rightarrow r$...$-(-q cup -p ) cup r$ and then i got $(q cup p ) cup r
$



Im really not sure if this is correct, could anyone please check and correct if I did anything wrong? Thanks!






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










asked Dec 15 '18 at 16:06









SheryaSherya

152




152












  • $begingroup$
    Are the unions supposed to be $lor$? This doesn’t look quite right, as $q=T, p=F, r=F$ satisfies the second expression but not the first.
    $endgroup$
    – platty
    Dec 15 '18 at 16:10










  • $begingroup$
    Does the minus sign mean $lnot?$ You can use lnot for $lnot$ vee for $vee$
    $endgroup$
    – saulspatz
    Dec 15 '18 at 16:12










  • $begingroup$
    Yes it does mean that! Thanks for the tip
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:13


















  • $begingroup$
    Are the unions supposed to be $lor$? This doesn’t look quite right, as $q=T, p=F, r=F$ satisfies the second expression but not the first.
    $endgroup$
    – platty
    Dec 15 '18 at 16:10










  • $begingroup$
    Does the minus sign mean $lnot?$ You can use lnot for $lnot$ vee for $vee$
    $endgroup$
    – saulspatz
    Dec 15 '18 at 16:12










  • $begingroup$
    Yes it does mean that! Thanks for the tip
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:13
















$begingroup$
Are the unions supposed to be $lor$? This doesn’t look quite right, as $q=T, p=F, r=F$ satisfies the second expression but not the first.
$endgroup$
– platty
Dec 15 '18 at 16:10




$begingroup$
Are the unions supposed to be $lor$? This doesn’t look quite right, as $q=T, p=F, r=F$ satisfies the second expression but not the first.
$endgroup$
– platty
Dec 15 '18 at 16:10












$begingroup$
Does the minus sign mean $lnot?$ You can use lnot for $lnot$ vee for $vee$
$endgroup$
– saulspatz
Dec 15 '18 at 16:12




$begingroup$
Does the minus sign mean $lnot?$ You can use lnot for $lnot$ vee for $vee$
$endgroup$
– saulspatz
Dec 15 '18 at 16:12












$begingroup$
Yes it does mean that! Thanks for the tip
$endgroup$
– Sherya
Dec 15 '18 at 16:13




$begingroup$
Yes it does mean that! Thanks for the tip
$endgroup$
– Sherya
Dec 15 '18 at 16:13










1 Answer
1






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oldest

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1












$begingroup$

Up until the last step, your reasoning works. However, we have to use De Morgan’s law to distribute negation over conjunction/disjunction. Instead, we would get:
$$
neg (neg q lor neg p) lor r equiv (q land p) lor r
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay thanks for the help!!
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:14










  • $begingroup$
    Just another question, was it right for me to convert the original statement if I wanted to create a combinatorial circuit?
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:15












  • $begingroup$
    Depending on what gates you can use, it is probably the most straightforward way to handle this one.
    $endgroup$
    – platty
    Dec 15 '18 at 16:16











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1 Answer
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active

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1 Answer
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active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

Up until the last step, your reasoning works. However, we have to use De Morgan’s law to distribute negation over conjunction/disjunction. Instead, we would get:
$$
neg (neg q lor neg p) lor r equiv (q land p) lor r
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay thanks for the help!!
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:14










  • $begingroup$
    Just another question, was it right for me to convert the original statement if I wanted to create a combinatorial circuit?
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:15












  • $begingroup$
    Depending on what gates you can use, it is probably the most straightforward way to handle this one.
    $endgroup$
    – platty
    Dec 15 '18 at 16:16
















1












$begingroup$

Up until the last step, your reasoning works. However, we have to use De Morgan’s law to distribute negation over conjunction/disjunction. Instead, we would get:
$$
neg (neg q lor neg p) lor r equiv (q land p) lor r
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay thanks for the help!!
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:14










  • $begingroup$
    Just another question, was it right for me to convert the original statement if I wanted to create a combinatorial circuit?
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:15












  • $begingroup$
    Depending on what gates you can use, it is probably the most straightforward way to handle this one.
    $endgroup$
    – platty
    Dec 15 '18 at 16:16














1












1








1





$begingroup$

Up until the last step, your reasoning works. However, we have to use De Morgan’s law to distribute negation over conjunction/disjunction. Instead, we would get:
$$
neg (neg q lor neg p) lor r equiv (q land p) lor r
$$






share|cite|improve this answer









$endgroup$



Up until the last step, your reasoning works. However, we have to use De Morgan’s law to distribute negation over conjunction/disjunction. Instead, we would get:
$$
neg (neg q lor neg p) lor r equiv (q land p) lor r
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 15 '18 at 16:13









plattyplatty

3,370320




3,370320












  • $begingroup$
    Okay thanks for the help!!
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:14










  • $begingroup$
    Just another question, was it right for me to convert the original statement if I wanted to create a combinatorial circuit?
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:15












  • $begingroup$
    Depending on what gates you can use, it is probably the most straightforward way to handle this one.
    $endgroup$
    – platty
    Dec 15 '18 at 16:16


















  • $begingroup$
    Okay thanks for the help!!
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:14










  • $begingroup$
    Just another question, was it right for me to convert the original statement if I wanted to create a combinatorial circuit?
    $endgroup$
    – Sherya
    Dec 15 '18 at 16:15












  • $begingroup$
    Depending on what gates you can use, it is probably the most straightforward way to handle this one.
    $endgroup$
    – platty
    Dec 15 '18 at 16:16
















$begingroup$
Okay thanks for the help!!
$endgroup$
– Sherya
Dec 15 '18 at 16:14




$begingroup$
Okay thanks for the help!!
$endgroup$
– Sherya
Dec 15 '18 at 16:14












$begingroup$
Just another question, was it right for me to convert the original statement if I wanted to create a combinatorial circuit?
$endgroup$
– Sherya
Dec 15 '18 at 16:15






$begingroup$
Just another question, was it right for me to convert the original statement if I wanted to create a combinatorial circuit?
$endgroup$
– Sherya
Dec 15 '18 at 16:15














$begingroup$
Depending on what gates you can use, it is probably the most straightforward way to handle this one.
$endgroup$
– platty
Dec 15 '18 at 16:16




$begingroup$
Depending on what gates you can use, it is probably the most straightforward way to handle this one.
$endgroup$
– platty
Dec 15 '18 at 16:16


















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