# Thread: Why is ((¬⊥)) not a formula? (Why isn't it T)

1. ## Why is ((¬⊥)) not a formula? (Why isn't it T)

I stumbled across this in my homework problems, but why is ((¬⊥)) not a valid formula? My perception would be that it is T, but I must admit I am terrible at logic-- illogical even .

2. Originally Posted by Nirvana
I stumbled across this in my homework problems, but why is ((¬⊥)) not a valid formula? My perception would be that it is T, but I must admit I am terrible at logic-- illogical even .
¬⊥must be a typo.

¬ is the symbol for negation.
$\vdash$ is the symbol for "therefore the conclusion is".

A letter representing a premise must follow ¬; example ¬Q or ¬S or ¬X. ¬Q= not a queen or ¬S= not silly, or ¬X = not X-Ray.

⊥ is not a logic operator, neither ⊥ is T. In math ⊥ means perpendicular.

Don't feel bad. You will be an expert in no time.

3. In mathematical logic, $\bot$ often denotes falsehood (see the table of logic symbols in Wikipedia).

I stumbled across this in my homework problems, but why is ((¬⊥)) not a valid formula? My perception would be that it is T, but I must admit I am terrible at logic-- illogical even .
When you say "valid formula", do you mean "syntactically well-formed formula" or "formula that is true in every interpretation"?

4. Originally Posted by emakarov
In mathematical logic, $\bot$ often denotes falsehood (see the table of logic symbols in Wikipedia).

When you say "valid formula", do you mean "syntactically well-formed formula" or "formula that is true in every interpretation"?
Nice table.
My $\bot$ hurt.

5. I meant well formed (syntactically correct) formula, not one that is always true

6. Don't think most people use that symbol anyway. It's more common to use 1 or true for the tautology and 0 or false for the contradiction.
I don't know about the logic that you are learning, but the negation of 0 is 1 in my book. so it's T, yeh >.>

Even with other definitions:
not p is the same as p -> 0
So not 0 is 0 -> 0, which is always true, so 1.

7. Originally Posted by TiRune
I don't know about the logic that you are learning, but the negation of 0 is 1 in my book. so it's T, yeh >.>
@TiRune: Your perception is the same as mine! However this is not the correct answer according to the book and the prof. Apparently negating the falsum symbol is not a syntactically correct formula, and I'm stumped as to why. BTW: the class is Discrete Mathematics For Computer Science

8. Apparently negating the falsum symbol is not a syntactically correct formula
Aha, so we are talking about syntax here, not truth. Everything depends on the definitions you've been given, but I can make several remarks.

If $\bot$ was defined as a contraction, for example, for $A\land\neg A$ or $0 = 1$, then $\bot$ is not literally a formula; it's a notation that we, humans, use, but which ultimately has to be expanded into the original syntax.

The same can be said about $\neg$. One often defined $\neg A$ as $A\to\bot$; thus $\neg\bot$ is $\bot\to\bot$. However, these two options seem unlikely to me.

In the thread title you wrote ((¬⊥)) with two pairs of parentheses. Maybe your book is very strict on these things and, for example, $(A\land B)$ is considered a formula when A and B are formulas, but $((A))$ is not a formula even if $(A)$ is.

One important remark: you wrote that you thought $\neg\bot$ is $\top$. These two formulas may be semantically equivalent (true in the same interpretations), but this is not related to whether they are well-formed syntactically. We do not even mention interpretations, truth, and similar things when we talk about formula syntax.

There may be other weird things about this particular presentation of formula syntax, but this is not an important point. To me, $((\neg\bot))$ is a formula, and if this is not so in your course, the reason is most likely in some unusual or extremely pedantic definition.