A statement following from the law of excluded middle

Question

Does the statement ~~$A\equiv A$ follow from the law of excluded middle? According to my book which is not on logic it does, but I do not know how to use the law of excluded middle for this simple tautology.

Answer 1

Assume $A$ and $\neg A$. Then $A\land \neg A$, which is absurd, hence $\neg\neg A$. So $A\implies \neg\neg A$: this holds without the law of excluded middle.

Now assume $\neg\neg A$. If $A$, then $A$. If $\neg A$, then $\neg A\land \neg\neg A$ which is absurd, hence $A$.

So whether $A$ holds or $\neg A$ holds, we always get that $A$ holds : thus [it's this "thus" that uses the law of excluded middle] $\neg\neg A\implies A$

Answer 2

An informal argument showing that $\lnot \lnot A \equiv A$ follows from the law of excluded middle is the following: by the law of excluded middle, either $A$ or $\lnot A$ holds, there is no third option. So, $\lnot \lnot A$ (whose truth value is determined by the truth value of $A$) should be equivalent to either $A$ or $\lnot A$. Since $\lnot \lnot A$ is the negation of $\lnot A$, it is impossible that $\lnot \lnot A$ and $\lnot A$ hold at the same time and hence they are not equivalent. Therefore, $\lnot \lnot A$ is equivalent to $A$.

Answer 3

This question isn't clear. Is the question "Can (~~A≡A) follow from the law of the excluded middle" the answer is 'yes'. That follows immediately from every single tautology implying every other tautology. Or equivalently, "all tautologies imply every other tautology."

However, if the question is "does the law of the excluded middle necessarily imply (~~A≡A)", where "imply" gets understood to mean that the law of the excluded middle will appear in any proof of (~~A≡A) where (~~A≡A) is not an axiom, then the answer is 'no'. As a simple example, (~~A -> A) and (A -> ~~A) along with modus ponens and ((A -> B) -> ((B -> A) -> (A ≡ B))) will imply that (~~A ≡ A). There is no law of the excluded middle there, and (~~A≡A) is not an axiom.