Main Operator in Formula: Tribar or ~?

  • Thread starter Thread starter steersman
  • Start date Start date
  • Tags Tags
    Formula Operator
Click For Summary

Homework Help Overview

The discussion revolves around identifying the main operator in a logical formula presented in a truth table from an Intro to Logic textbook. The formula in question is represented as (~(p & q) = (~p v ~q)), where the equal sign is intended to be a tribar.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore whether the main operator is the first tilde or the tribar, with some suggesting that the tribar relates the two sides of the equation. There is confusion regarding the role of brackets and the terminology of "main operator." Participants also express uncertainty about the standard definitions and implications of these terms.

Discussion Status

The discussion is active, with participants sharing their thoughts and memories about logical operators. Some have offered tentative confirmations of each other's reasoning, while others express confusion about the terminology and structure of the formula. There is no explicit consensus, but various interpretations are being explored.

Contextual Notes

Some participants mention the redundancy of brackets in the formula and question the standard usage of the term "main operator." There is also a reference to the potential connection to order of operations in logic.

steersman
Messages
46
Reaction score
0
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.
 
Last edited:
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
1K
Replies
25
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 44 ·
2
Replies
44
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K