What guarantee do we have that in a proof by contradiction, $¬¬A$ does not cause an absurd? This could be something truly stupid. But let me describe it: Let's take a proof by contradiction, we suppose $¬ A$, and when an absurdity comes from this, we deduce $¬¬A$ must be true. 


*

*In this case, we verified what happens when something is false and found an absurdity, and hence it must be true, but we actually didn't verify if it being true would imply some problem.
Now take - for example - Russell's paradox. We verify if it is true and an absurdity happens and then, we verify if it is false and then another absurdity happens.


*

*Notice that in this case, we verified what happens if it is truth or false. 


So in general what guarantee do we have that $¬A$ causes an absurdity, $¬¬A$ is true and $¬¬A$ does not cause another absurd just as in Russell's paradox?
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 A: You did not understand Russell's paradox correctly. Here it is formally (using natural deduction):
$
\def\imp{\Rightarrow}
\def\eq{\Leftrightarrow}
$

If $\exists S\ \forall x\ ( x \in S \eq \neg x \in x )$:
  Let $R$ be such that $\forall x\ ( x \in R \eq \neg x \in x )$.   [by $\exists$-elim]
  $R \in R \eq \neg R \in R$.   [by $\forall$-elim]
  $R \in R \imp \neg R \in R$.   [by $\eq$-elim]
  $\neg R \in R \imp R \in R$.   [by $\eq$-elim]
  If $R \in R$:
    $\neg R \in R$.   [by $\imp$-elim]
    Contradiction.
  $\neg R \in R$.   [by $\imp$-intro and $\neg$-intro]
  $R \in R$.   [by $\imp$-elim]
  Contradiction.
$\neg \exists S\ \forall x\ ( x \in S \eq \neg x \in x )$   [by $\imp$-intro and $\neg$-intro]

In particular, we do not have a sentence that is proven true and false at the same time, except under the false assumption of the existence of a Russell collection. The above proof is valid in ZF set theory, and also in many other systems, as long as they support the rules used. It thus implies that such systems cannot have unrestricted specification of collections.
A: This is my understanding:
When we work with proofs by contradiction, We take as an axiom that every statement is either true or false, but not both.
Thus, If $\neg A$ is false, then (by another axiom) $A$ is true.
However, when it comes to Russell's paradox, you encounter a statement that is both true and false. Thus, the above "setting" is invalid (since some of the axioms are not true) and we cannot apply the "proof by contradiction" logic.
