Algebraic proof of double negation inclusion?

[jk22]

I search for a vector space based proof of the following :

The logic on values implies ~~v(A)=v(A)

If the value of A is v(A)$\in\{0,1\}$ then it is simply $1-(1-v(A))=v(A)$

But if we suppose A="the sky is red"
Then as on operator acting on A, ~A is not defined since for example ~A="the sky is not red" is a set in which by using the axiom of choice we could choose ~A="the sky is green" and then ~~A="the sky is not green" which does not imply that the sky is red as in the starting point.

So we have A$\subset$~~A.

Where could the proof of this based on orthogonal spaces be found ? Does it work only for dimension 4 and higher ?

 

[Mark44]

Then as on operator acting on A, ~A is not defined since for example ~A="the sky is not red" is a set in which by using the axiom of choice we could choose ~A="the sky is green" and then ~~A="the sky is not green"

This doesn't make sense, logically. If A is defined as A = "the sky is red," then ~A is the negation of that sentence, which as you said, is "the sky is not red."

I don't see how it could possibly be valid to choose a meaning for that the last sentence of "the sky is green." Further, given the sentence A, there are only two possible choices: that it is true or that it is false. You can't attach additional meaning to the sentence "the sky is not red."

 

[mfb]

~A="the sky is not red" is a set in which by using the axiom of choice we could choose ~A="the sky is green"

An element of a set (or a subset of a set) is not the same as the set. This should be clear in this sentence, because obviously "the sky is not red" and "the sky is green" are not the same, they cannot both be ~A.

 

[Stephen Tashi]

But if we suppose A="the sky is red"

by using the axiom of choice we could choose ~A="the sky is green"

You are confusing sets with propositions about sets. If you use "A" to denote the set of things that are red then you need different notation to denote the proposition "the sky is an element of the set A". If you wish to use "A" to denote the proposition "The sky is red" then the you need different notation to indicate "the set of things that are red".

The intuitive idea that sets are intimately related to propositions is correct. For example the statement about sets "A is a subset of B" is defined to be the statement: "For each element x, if x is an member of A then x is a member of B". As another example, the set "the complement of set A" can be defined by the proposition "For each element x, x is a member of the complement of A if and only if x is not an element of A".

Note that those examples involve "propositonal functions" using the variable "x" and the "quantifier "for each" that turns the propositional functions into propositions. We can also formulate examples using the quantifier "there exists". We an also formulate examples without using a variable, such as "The sky on Mars is green", where "The sky on Mars" is understood to be a single specific element.

You must make a distinction between propositions like "The sky on Mars is green" versus sets such as "The set of things that are not red". A proposition such as "The sky on Mars is green" is a statement so it has a "truth value " (True or False). A set such as "The set of things that are not red" isn't a statement. In terms of grammer, it is a noun. ( A set may be the empty set, but we don't say the empty set has the property of being "False".)