Main Operator in Formula: Tribar or ~?

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The discussion centers on identifying the main operator in the formula ~(p & q) = (~p v ~q), where the equal sign is represented as a tribar. Participants agree that the tribar serves as the main operator since it connects the two sides of the equation. The tilde on the left only applies to the conjunction (p & q), while the brackets are deemed redundant. Some express confusion over the term "main operator," suggesting it relates to order of operations. Ultimately, the consensus is that the tribar is the correct designation for the main operator in this logical expression.
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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.
 
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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.
 
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.
 
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.
 
Those brackets are definitely 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.
 
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.
 
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