How would I show that X is equivalent to ((¬X ↔ X ) ∨ X )?

Question

I use the https://en.wikipedia.org/wiki/Fitch_notation, or fitch notation, for logical deduction systems.

I don't know how to derive a contradiction in the other half of the biconditional where $X \to \lnot X$. In this proof, the goal is show that one could derive $((\lnot X \leftrightarrow X ) \lor X )$ from $X$.

1. $X$ (premise)

2. | $\lnot X$ (subproof 1)

3. | $X$ Reiterate line 1

4. | $X$ (subproof 2)
5. . . . (need to reach a contradiction for negated $X$)

If I were to indirectly prove this, it would ultimately result in the same path as trying to find a contradiction for the biconditional.

Answer 1

Here is a Fitch proof that proves the equivalence:

Answer 2

I am not familiar with the Fitch notation, but you should mimic this proof:

"->": If $X$, then clearly $(¬X↔X) \vee X$.

"<-": Let $(¬X↔X) \vee X$. We either have $X$ or $¬X$.
If we have $X$ we are done.
If we have $¬X$, then by assumption we still have $¬X↔X$, which implies $X$.