How to eliminate redundant logic term using simplification rules and expressions.
0
$begingroup$
This expression is right according to Karnaugh Map
Is it possible to eliminate extra term using Boolean algebra transformation and simplification rules? In other words - what is calculation process of this expression, which eliminates extra term.
Map:
combinatorics logic
asked Dec 12 '18 at 17:38
andreikashinandreikashin
1013
$endgroup$
add a comment |
0
$begingroup$
This expression is right according to Karnaugh Map
Is it possible to eliminate extra term using Boolean algebra transformation and simplification rules? In other words - what is calculation process of this expression, which eliminates extra term.
Map:
combinatorics logic
asked Dec 12 '18 at 17:38
andreikashinandreikashin
1013
$endgroup$
$begingroup$
What do you mean "by combinatorics"? The Karnaugh map itself may be considered as a concept based on combinatorics.
$endgroup$
– Vincenzo
Dec 12 '18 at 21:50
$begingroup$
I mean using Boolean Algebra simplification rules
$endgroup$
– andreikashin
Dec 13 '18 at 17:39
$begingroup$
Look up the consensus theorem.
$endgroup$
– Fabio Somenzi
Dec 13 '18 at 17:46
$begingroup$
@Fabio, yes, that is the answer - thank you!
$endgroup$
– andreikashin
Dec 13 '18 at 18:04
$begingroup$
math.stackexchange.com/a/60724/599981
$endgroup$
– andreikashin
Dec 14 '18 at 9:44
add a comment |
0
0
0
$begingroup$
This expression is right according to Karnaugh Map
Is it possible to eliminate extra term using Boolean algebra transformation and simplification rules? In other words - what is calculation process of this expression, which eliminates extra term.
Map:
combinatorics logic
asked Dec 12 '18 at 17:38
andreikashinandreikashin
1013
$endgroup$
This expression is right according to Karnaugh Map
Is it possible to eliminate extra term using Boolean algebra transformation and simplification rules? In other words - what is calculation process of this expression, which eliminates extra term.
Map:
combinatorics logic
combinatorics logic
asked Dec 12 '18 at 17:38
andreikashinandreikashin
1013
asked Dec 12 '18 at 17:38
andreikashinandreikashin
1013
edited Dec 13 '18 at 17:41
andreikashin
asked Dec 12 '18 at 17:38
andreikashinandreikashin
1013
asked Dec 12 '18 at 17:38
andreikashinandreikashin
1013
asked Dec 12 '18 at 17:38
andreikashinandreikashin
1013
1013
$begingroup$
What do you mean "by combinatorics"? The Karnaugh map itself may be considered as a concept based on combinatorics.
$endgroup$
– Vincenzo
Dec 12 '18 at 21:50
$begingroup$
I mean using Boolean Algebra simplification rules
$endgroup$
– andreikashin
Dec 13 '18 at 17:39
$begingroup$
Look up the consensus theorem.
$endgroup$
– Fabio Somenzi
Dec 13 '18 at 17:46
$begingroup$
@Fabio, yes, that is the answer - thank you!
$endgroup$
– andreikashin
Dec 13 '18 at 18:04
$begingroup$
math.stackexchange.com/a/60724/599981
$endgroup$
– andreikashin
Dec 14 '18 at 9:44
add a comment |
$begingroup$
What do you mean "by combinatorics"? The Karnaugh map itself may be considered as a concept based on combinatorics.
$endgroup$
– Vincenzo
Dec 12 '18 at 21:50
$begingroup$
I mean using Boolean Algebra simplification rules
$endgroup$
– andreikashin
Dec 13 '18 at 17:39
$begingroup$
Look up the consensus theorem.
$endgroup$
– Fabio Somenzi
Dec 13 '18 at 17:46
$begingroup$
@Fabio, yes, that is the answer - thank you!
$endgroup$
– andreikashin
Dec 13 '18 at 18:04
$begingroup$
math.stackexchange.com/a/60724/599981
$endgroup$
– andreikashin
Dec 14 '18 at 9:44
$begingroup$
What do you mean "by combinatorics"? The Karnaugh map itself may be considered as a concept based on combinatorics.
$endgroup$
– Vincenzo
Dec 12 '18 at 21:50
$begingroup$
What do you mean "by combinatorics"? The Karnaugh map itself may be considered as a concept based on combinatorics.
$endgroup$
– Vincenzo
Dec 12 '18 at 21:50
$begingroup$
I mean using Boolean Algebra simplification rules
$endgroup$
– andreikashin
Dec 13 '18 at 17:39
$begingroup$
I mean using Boolean Algebra simplification rules
$endgroup$
– andreikashin
Dec 13 '18 at 17:39
$begingroup$
Look up the consensus theorem.
$endgroup$
– Fabio Somenzi
Dec 13 '18 at 17:46
$begingroup$
Look up the consensus theorem.
$endgroup$
– Fabio Somenzi
Dec 13 '18 at 17:46
$begingroup$
@Fabio, yes, that is the answer - thank you!
$endgroup$
– andreikashin
Dec 13 '18 at 18:04
$begingroup$
@Fabio, yes, that is the answer - thank you!
$endgroup$
– andreikashin
Dec 13 '18 at 18:04
$begingroup$
math.stackexchange.com/a/60724/599981
$endgroup$
– andreikashin
Dec 14 '18 at 9:44
$begingroup$
math.stackexchange.com/a/60724/599981
$endgroup$
– andreikashin
Dec 14 '18 at 9:44
add a comment |
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$begingroup$
What do you mean "by combinatorics"? The Karnaugh map itself may be considered as a concept based on combinatorics.
$endgroup$
– Vincenzo
Dec 12 '18 at 21:50
$begingroup$
I mean using Boolean Algebra simplification rules
$endgroup$
– andreikashin
Dec 13 '18 at 17:39
$begingroup$
Look up the consensus theorem.
$endgroup$
– Fabio Somenzi
Dec 13 '18 at 17:46
$begingroup$
@Fabio, yes, that is the answer - thank you!
$endgroup$
– andreikashin
Dec 13 '18 at 18:04
$begingroup$
math.stackexchange.com/a/60724/599981
$endgroup$
– andreikashin
Dec 14 '18 at 9:44