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how can i prove that a k-map for n variables gives the most simplified representation of a Boolean function ?

(by simplified i mean we cannot eliminate another variable)

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  • \$\begingroup\$ They give the most complete representation. It's up to the person interpreting it to find the most simplified expression. \$\endgroup\$ Commented Mar 7, 2015 at 14:53
  • \$\begingroup\$ OK then how to prove that "They give the most complete representation". \$\endgroup\$
    – KFkf
    Commented Mar 7, 2015 at 14:55
  • \$\begingroup\$ All combinations of variables shown in the map is the power set of those variables, since all possible representations of those variables are included. And since the power set is the definition of all possible representations of a boolean expression, it directly follows that a map gives all possible representations. \$\endgroup\$ Commented Mar 7, 2015 at 15:02

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How well a karnaugh map can simplify is downto the creator of the map with regards to drawing optimal loops.

How to determine whether what has been drawn (post further boolean reduction) is not really possible with k-maps on their own

One option is the use of Espresso Heuristic Logic reduction via a fantastic piece of software called: Logic friday ( http://www.sontrak.com/ ) Quite often this has reduced some logic of mine even further as I usually forget the odd one or two don't care states

The real question however is ... is this check on reduction for personal/professional use (in which case carry on) or homework? because if it is homework and logicFriday realises a further simplification, you had better be able to realise it via k-maps for the marks

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