Combinatorial Circuit
$begingroup$
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
asked Dec 15 '18 at 16:06
SheryaSherya
152
$endgroup$
add a comment |
$begingroup$
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
asked Dec 15 '18 at 16:06
SheryaSherya
152
$endgroup$
$begingroup$
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 uselnotfor $lnot$veefor $vee$
$endgroup$
– 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
add a comment |
$begingroup$
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
asked Dec 15 '18 at 16:06
SheryaSherya
152
$endgroup$
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
asked Dec 15 '18 at 16:06
SheryaSherya
152
asked Dec 15 '18 at 16:06
SheryaSherya
152
asked Dec 15 '18 at 16:06
SheryaSherya
152
asked Dec 15 '18 at 16:06
SheryaSherya
152
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 uselnotfor $lnot$veefor $vee$
$endgroup$
– 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
add a comment |
$begingroup$
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 uselnotfor $lnot$veefor $vee$
$endgroup$
– 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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:
$$
neg (neg q lor neg p) lor r equiv (q land p) lor r
$$
answered Dec 15 '18 at 16:13
plattyplatty
3,370320
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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:
$$
neg (neg q lor neg p) lor r equiv (q land p) lor r
$$
answered Dec 15 '18 at 16:13
plattyplatty
3,370320
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$
neg (neg q lor neg p) lor r equiv (q land p) lor r
$$
answered Dec 15 '18 at 16:13
plattyplatty
3,370320
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$
neg (neg q lor neg p) lor r equiv (q land p) lor r
$$
answered Dec 15 '18 at 16:13
plattyplatty
3,370320
$endgroup$
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:
$$
neg (neg q lor neg p) lor r equiv (q land p) lor r
$$
answered Dec 15 '18 at 16:13
plattyplatty
3,370320
answered Dec 15 '18 at 16:13
plattyplatty
3,370320
answered Dec 15 '18 at 16:13
plattyplatty
3,370320
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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
lnotfor $lnot$veefor $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