Russell Miller wrote: > On Tuesday 02 May 2006 14:59, Dan wrote: >> Is: >> >> A && (B || C || D || E || F) >> >> equivalent to?: >> >> A && (!B && !C && !D && !E && !F)
No the two are NOT equivalent. The first statement will be true if A and any one of B-F is true. The second statement will be true if A and all of B-F are false. It would be true if you modified the second rule to be: A && !(!B && !C && !D && !E && !F) >> > I believe that's a fundamental logic rule, so yes. > > A && B == ~A || ~B No, that is not a fundamental logic rule. It is not even true, they are in fact exact opposites of each other. I'm going to change the use of ~ to !, just to match the OP's syntax of the NOT operation. ( A && B == !A || !B ) Truth tables: A B (A && B) ( !A || !B) 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 The correct rule is: !( A && B) == !A || !B It's called DeMorgan's theorem. Note the difference being that there's negation on both sides. A B !(A && B) ( !A || !B) 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 0