Are \alpha and \beta logically equivalent if ~\alpha and ~\beta are?

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The discussion centers on the logical equivalence of two formulas, α and β, in propositional calculus, specifically exploring whether they are logically equivalent if their negations, ~α and ~β, are also logically equivalent. Participants emphasize the need to define "logically equivalent" and suggest that the question may be a homework query, recommending it be posted in a more appropriate section. There is a call for the original poster to demonstrate their attempt at solving the problem to facilitate better assistance. The conversation highlights the importance of understanding logical equivalence in formal logic. Clarity in definitions and context is crucial for addressing such mathematical inquiries effectively.
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Be \alpha and \beta two formulas of the propositional calculus, show that \alpha and \beta are logically equivalent if and only if ~\alpha and ~\beta are logically equivalent.
 
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cristina89 said:
Be \alpha and \beta two formulas of the propositional calculus, show that \alpha and \beta are logically equivalent if and only if ~\alpha and ~\beta are logically equivalent.

Is this a homework question? If so, it should be posted in the homework section: precalculus mathematics. In any case, you need to show your attempt at a solution.
 
What is your definition of "logically equivalent"?
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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