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Simple logic unification
~ not
-> implies
v OR
^ AND
= Equivalence, should really consist of 3 horizontal lines
1) Reduce the formula to conjunctive normal form
A ^ B -> (B ^ C)
A ^ B -> (B ^ C) is the same as (B ^ C) v~ (A^B).
Using De Morgans this becomes (B ^ C) v (-A v ~B)
Which is the same as (B v (~A v ~B)) ^ (C v (~A v ~B))
Which is the same as C v ~A v ~B
This is in Conjunctive normal form
Your answer is right.
Also expressed as ~(A^B^~C)
I gotta go, so I didn't check you steps thru it though,
I did it with a cube made out of 8 little cubes, where
the bottom was B, and the right was A, and the
back was C, like a right-angled Venn diagram.
Bye.
igloo myrtilles fourmis
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I finally checked your steps.
They are correct!!
Thanks for making me look up what "conjunctive normal form" is.
Since my negated answer is negated, then that
probably makes it not the agreed upon format, like
yours, so I think yours is best, with my new
understanding of conjunctive normal form.
Because yours could be conjuncted with True to make
it look like a conjunction of disjunctions, but
my ~(A^B^~C) is certainly a negated conjunction...
By the way, the digital electronics phrase "Product of Sum" (POS) also
means conjunctive normal form.
And "Sum of Product" (SOP) is the opposite, where you
do the Disjunction of Conjunctions, such as circling the
ones on a Karnaugh map and or'ing the encircled groups
together... (Certain patterns can be circled if using AND
operators, other patterns can be circled if using XOR
operators in checkerboard-ish patterns, sometimes a
checkerboard that is twice as wide as high, where each
1 might be beside one other 1. (Just an example)
Last edited by John E. Franklin (2008-01-31 08:05:34)
igloo myrtilles fourmis
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I have always thought that conjunction is AND (while OR is disjunction).
In this case, you would want
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can someone confirm which one is correct please?
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