a) Show De Morgan's law $\sim (X\blacktriangle Y) \leftrightarrow (\sim X\blacktriangledown \sim Y)$

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Let the functions $\blacktriangle $ and $\blacktriangledown$ be defined as $X\blacktriangle Y = \sim (X\wedge Y) $ and $X\blacktriangledown Y = \sim (X\vee Y)$

a) Show De Morgan's law $\sim (X\blacktriangle Y) \leftrightarrow (\sim X\blacktriangledown \sim Y)$

My solution:

De Morgan's laws:

$\sim (P\vee Q)\leftrightarrow (\sim P)\wedge (\sim Q)$ and

$\sim (P\wedge Q) \leftrightarrow (\sim P) \vee (\sim Q)$

Functions:

$\sim (X\blacktriangle Y) \leftrightarrow (\sim X\blacktriangledown \sim Y)$ =

$\sim( \sim (X\wedge Y)) \leftrightarrow $ $ \sim (\sim X\vee \sim Y)$ (Is this correct?)

Question:

So I don’t have much of a solution to show here. But did I correctly translate the functions $\blacktriangle $ and $\blacktriangledown$ to boolean algebra? Sorry I have a hard time understanding this problem. :/