Mixing
if $a'b+cd'=0$, then prove that $ab+c'(a'+d')=ab+bd+b'd'+a'c'd$
$begingroup$
Let $a'b+cd'=0$, then prove that $$ab+c'(a'+d')=ab+bd+b'd'+a'c'd$$
I would like to know how to solve this expression, not able to make any headway. I have tried canonical form expansion and reduction but the terms in the if condition does not match and also not able to generate the terms in the given expression.
boolean-algebra
edited Jan 16 at 12:01
drhab
102k545136
asked Jan 16 at 10:27
harindra bhatiharindra bhati
1
$endgroup$
add a comment |
$begingroup$
Let $a'b+cd'=0$, then prove that $$ab+c'(a'+d')=ab+bd+b'd'+a'c'd$$
I would like to know how to solve this expression, not able to make any headway. I have tried canonical form expansion and reduction but the terms in the if condition does not match and also not able to generate the terms in the given expression.
boolean-algebra
edited Jan 16 at 12:01
drhab
102k545136
asked Jan 16 at 10:27
harindra bhatiharindra bhati
1
$endgroup$
1
$begingroup$
What is the symbol $'$?
$endgroup$
– Alex Silva
Jan 16 at 10:31
$begingroup$
Hi! And welcome to MSE. I think that you are missing some relations. If you start considering $ab+c'(a'+d')$, you can not make appear $d$, since your unique relation $a'b+cd'=0$ does not contain any $d$, for example. Hence, I guess you have some missing relations. It could be between those $a,b,c,d$ and their respective $a',b',c',d'$. Let us know. Edit the post.
$endgroup$
– idriskameni
Jan 16 at 10:42
add a comment |
$begingroup$
Let $a'b+cd'=0$, then prove that $$ab+c'(a'+d')=ab+bd+b'd'+a'c'd$$
I would like to know how to solve this expression, not able to make any headway. I have tried canonical form expansion and reduction but the terms in the if condition does not match and also not able to generate the terms in the given expression.
boolean-algebra
edited Jan 16 at 12:01
drhab
102k545136
asked Jan 16 at 10:27
harindra bhatiharindra bhati
1
$endgroup$
Let $a'b+cd'=0$, then prove that $$ab+c'(a'+d')=ab+bd+b'd'+a'c'd$$
I would like to know how to solve this expression, not able to make any headway. I have tried canonical form expansion and reduction but the terms in the if condition does not match and also not able to generate the terms in the given expression.
boolean-algebra
boolean-algebra
edited Jan 16 at 12:01
drhab
102k545136
asked Jan 16 at 10:27
harindra bhatiharindra bhati
1
edited Jan 16 at 12:01
drhab
102k545136
asked Jan 16 at 10:27
harindra bhatiharindra bhati
1
edited Jan 16 at 12:01
drhab
102k545136
edited Jan 16 at 12:01
drhab
102k545136
edited Jan 16 at 12:01
drhab
102k545136
102k545136
asked Jan 16 at 10:27
harindra bhatiharindra bhati
1
asked Jan 16 at 10:27
harindra bhatiharindra bhati
1
asked Jan 16 at 10:27
harindra bhatiharindra bhati
1
1
1
$begingroup$
What is the symbol $'$?
$endgroup$
– Alex Silva
Jan 16 at 10:31
$begingroup$
Hi! And welcome to MSE. I think that you are missing some relations. If you start considering $ab+c'(a'+d')$, you can not make appear $d$, since your unique relation $a'b+cd'=0$ does not contain any $d$, for example. Hence, I guess you have some missing relations. It could be between those $a,b,c,d$ and their respective $a',b',c',d'$. Let us know. Edit the post.
$endgroup$
– idriskameni
Jan 16 at 10:42
add a comment |
1
$begingroup$
What is the symbol $'$?
$endgroup$
– Alex Silva
Jan 16 at 10:31
$begingroup$
Hi! And welcome to MSE. I think that you are missing some relations. If you start considering $ab+c'(a'+d')$, you can not make appear $d$, since your unique relation $a'b+cd'=0$ does not contain any $d$, for example. Hence, I guess you have some missing relations. It could be between those $a,b,c,d$ and their respective $a',b',c',d'$. Let us know. Edit the post.
$endgroup$
– idriskameni
Jan 16 at 10:42
1
1
$begingroup$
What is the symbol $'$?
$endgroup$
– Alex Silva
Jan 16 at 10:31
$begingroup$
What is the symbol $'$?
$endgroup$
– Alex Silva
Jan 16 at 10:31
$begingroup$
Hi! And welcome to MSE. I think that you are missing some relations. If you start considering $ab+c'(a'+d')$, you can not make appear $d$, since your unique relation $a'b+cd'=0$ does not contain any $d$, for example. Hence, I guess you have some missing relations. It could be between those $a,b,c,d$ and their respective $a',b',c',d'$. Let us know. Edit the post.
$endgroup$
– idriskameni
Jan 16 at 10:42
$begingroup$
Hi! And welcome to MSE. I think that you are missing some relations. If you start considering $ab+c'(a'+d')$, you can not make appear $d$, since your unique relation $a'b+cd'=0$ does not contain any $d$, for example. Hence, I guess you have some missing relations. It could be between those $a,b,c,d$ and their respective $a',b',c',d'$. Let us know. Edit the post.
$endgroup$
– idriskameni
Jan 16 at 10:42
add a comment |
1 Answer
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$begingroup$
A truth table shows that left-hand-side and righ-hand-side are in fact equivalent if the constraint is fulfilled (indicated by background color):
answered Jan 16 at 14:18
Axel KemperAxel Kemper
3,33611418
$endgroup$
add a comment |
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$begingroup$
A truth table shows that left-hand-side and righ-hand-side are in fact equivalent if the constraint is fulfilled (indicated by background color):
answered Jan 16 at 14:18
Axel KemperAxel Kemper
3,33611418
$endgroup$
add a comment |
$begingroup$
A truth table shows that left-hand-side and righ-hand-side are in fact equivalent if the constraint is fulfilled (indicated by background color):
answered Jan 16 at 14:18
Axel KemperAxel Kemper
3,33611418
$endgroup$
add a comment |
$begingroup$
A truth table shows that left-hand-side and righ-hand-side are in fact equivalent if the constraint is fulfilled (indicated by background color):
answered Jan 16 at 14:18
Axel KemperAxel Kemper
3,33611418
$endgroup$
A truth table shows that left-hand-side and righ-hand-side are in fact equivalent if the constraint is fulfilled (indicated by background color):
answered Jan 16 at 14:18
Axel KemperAxel Kemper
3,33611418
answered Jan 16 at 14:18
Axel KemperAxel Kemper
3,33611418
answered Jan 16 at 14:18
Axel KemperAxel Kemper
3,33611418
answered Jan 16 at 14:18
Axel KemperAxel Kemper
3,33611418
3,33611418
add a comment |
add a comment |
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$begingroup$
What is the symbol $'$?
$endgroup$
– Alex Silva
Jan 16 at 10:31
$begingroup$
Hi! And welcome to MSE. I think that you are missing some relations. If you start considering $ab+c'(a'+d')$, you can not make appear $d$, since your unique relation $a'b+cd'=0$ does not contain any $d$, for example. Hence, I guess you have some missing relations. It could be between those $a,b,c,d$ and their respective $a',b',c',d'$. Let us know. Edit the post.
$endgroup$
– idriskameni
Jan 16 at 10:42