Wow,
I must be confusing this with any/all is/isn't. In various software
(mail scripts, iTunes smart playlists, etc):
any IS IS IS IS
equals
all NOT NOT NOT NOT
Dan
On May 2, 2006, at 15:11, Matt Kettler wrote:
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
