Negation of Equivalence

Four students, Simona, Laura, Ingrid, and Bianka, were given the task of finding the negation of the equivalence: $$(4>5)\Leftrightarrow(5\in \mathbb{Q}).$$ Here are their solutions:

Simona remembered that the negation of an equivalence $A\Leftrightarrow B$ is: $$A\Leftrightarrow B'$$

And she also remembered that $\mathbb{I}\cup \mathbb{Q}=\mathbb{R}$ ($\mathbb{Q}$ denotes the set of rational numbers) so the negation of $5\in \mathbb{Q}$ is $5\in \mathbb{I}$.

She decided the negation of the given statement is: $$(4>5)⇔(5∈\mathbb{I})$$

Laura recalled that the negation of an equivalence is: $$A'\Leftrightarrow B$$

She proposed the negation of the given statement as: $$(4\leq5)\Leftrightarrow (5\in \mathbb{Q})$$

Ingrid remembered that an equivalence is an implication going both ways: $$(A\Rightarrow B)\wedge(B\Rightarrow A)$$ Therefore, she split the statement into two implications:

\begin{array}{lc} 1) & (4>5)\,\Rightarrow(5\in \mathbb{Q})\cr 2) & (5\in \mathbb{Q})\Rightarrow(4>5) \end{array} She proceeded to make negations of both implications: \begin{array}{lcc} 1) &(4>5)\wedge(5\in \mathbb{I})\cr 2) &\,\,(4\leq5)\wedge(5\in \mathbb{Q}) \end{array}

Finally, she combined these statements using a disjunction: $$((4>5)\wedge(5\in \mathbb{I}))\vee((4\leq5)\wedge(5\in \mathbb{Q}))$$

Bianka, similarly to Ingrid, found in her notes that equivalence is an implication going both ways and she split the statement into two implications: \begin{array}{lc} 1) &(4>5)\,\Rightarrow(5\in \mathbb{Q})\cr 2) &(5\in \mathbb{Q})\Rightarrow(4>5) \end{array}

Then, she found negations of both implications: \begin{array}{lc} 1) &(5\in \mathbb{I})\wedge(4>5)\cr 2) &(4\leq5)\wedge(5\in \mathbb{Q}) \end{array} And finally, she composed the statement using a disjunction: $$((5\in \mathbb{I})\wedge(4>5))\vee((4\leq5)\wedge(5\in \mathbb{Q}))$$

Who solved the problem correctly?