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steersman
Sep17-04, 10:45 AM
In my Intro to Logic textbook there is a truth-table with the formula:
(~(p & q) = (~p v ~q))
The equal sign is meant to be a tribar

What is the main operator in this formula?
I think it is the first tilde but the textbook indicates that it is the tribar. Does anyone know the correct answer? Thanks.

Norman
Sep17-04, 10:51 AM
the tilde on the left hand side only operates on the (p & q) part. So I would say (if 6 years after taking these classes I can still remember) that the equal sign (tribar) is the main operator, since it relates the two sides. I wouldn't bet too much on me, my memory is bad. But hopefully someone will confirm or disprove me.
Hope I helped a little.

steersman
Sep17-04, 11:08 AM
I think you're right but I'm still confused with these brackets. It seems that the first bracket encompasses the whole formula, including (p & q) and the = (tribar). Argh! I was never good at algebra.

Gokul43201
Sep17-04, 11:59 AM
I've never heard of the usage "main operator" and am not sure is it's standard terminology, or how it is defined.

The first bracket does encompass the whole formula, and so, is redundant.

Norman
Sep17-04, 12:02 PM
Those brackets are definately redundant so just forget about them. I have never heard of this main operator, but my guess is that it has something to do with order of operations stuff.

zefram_c
Sep18-04, 02:12 AM
I've not heard of this "main operator" before, but it probably entails this. If I could denote the entire formula by one single operator and compound statement(s), what's that operator? In your case, you have

(~(p & q) = (~p v ~q))

So define the following
r = p&q
s = ~p
t = ~q
u = ~r
z = s v t

Then you can rewrite the formula as
(~(p & q) = (~p v ~q))
(~r = (s v t))
(u = z)
Hence '=' is the main operator there. Of course, you're expected to do this mentally - it's quite easy once you get the knack of it.