Let $\mathcal{L}$ be the formal system of statement calculus. Let $A$ be a formula of $\mathcal{L}$. Then each of the three formulas \begin{gather*} A \vee (\sim A) \\ A \implies [(\sim A) \implies A] \\ (\sim A) \implies [A \implies (\sim A)] \\ \end{gather*} is a theorem of $\mathcal{L}$. By case analysis, $$[(\sim A) \implies A] \vee [A \implies (\sim A)]$$ is a theorem of $\mathcal{L}$. But this seems intuitively incorrect.
For example, each of the statements \begin{gather*} (\forall x \in \mathbb{N})[(x < 3) \implies (x \ge 3)]\\ (\forall x \in \mathbb{N})[(x \ge 3) \implies (x < 3)] \end{gather*} is false. However, the disjunction of these two statements is true, by the argument above.
Edit. I guess the example was not very good.
Suppose you flip a coin. It is a fair coin and it cannot stand up on its side. It lands at your feet on a solid, flat surface.
Let $A$ be the statement that the coin shows heads. Then $(\sim A)$ is the statement that the coin shows tails. The statement \begin{gather*} \text{"if the coin shows heads, then it shows tails,}\\ \text{or}\\ \text{if the coin shows tails, then it shows heads"} \end{gather*} is true, even though both disjuncts are false.
Maybe this question no longer belongs on math.stackexchange.com
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