Natural deduction - finishing the proof 
Give a natural deduction proof of ($\neg A \lor B) \Leftrightarrow C$
from hypotheses $\neg A \to C$ and $B \Leftrightarrow C$

My attempt so far:

*

*$\neg A \to C$

*$B \Leftrightarrow C$

*C

*$\neg A$ (1,3, $\to e)$

*$\neg A \lor B$ (4, $\lor i)$
How can I finish the proof?
 A: Are you familiar with deduction trees?
Showing $\neg A \lor B \rightarrow C$:
$$
  \dfrac
  {
    \dfrac
    {
        [\neg A \lor B]
        \quad
        \dfrac{[\neg A] \quad \neg A \rightarrow C}{C}
        \quad
        \dfrac{[B] \quad {\dfrac{B \leftrightarrow C}{B \rightarrow C}}}{C}
    }
    {
      C
    }
  }
  {
    \neg A \lor B \rightarrow C
  }
$$
Showing $C \rightarrow \neg A \lor B$:
$$
\dfrac
{
  \dfrac
  {
    \dfrac
    {
      [C]
      \quad
      \dfrac{B \leftrightarrow C}{C \rightarrow B}
    }
    {
      B
    }
  }
  {
    \neg A \lor B
  }
}
{
  C \rightarrow \neg A \lor B
}
$$
A: To derive $(\neg A \lor B) \Leftrightarrow C$ from hypotheses $\lnot A \to C$ and $B \leftrightarrow C$, you need to obtain $C$ from the assumption of $(\neg A \lor B)$ and $(\neg A \lor B)$ from the assumption of $C$.
$
\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\def\Ae#1{\qquad\mathbf{\forall\,Elim}\colon #1 \\}
\def\Ai#1{\qquad\mathbf{\forall\,Intro}\colon #1 \\}
\def\Ee#1{\qquad\mathbf{\exists\,E}\colon #1 \\}
\def\Ei#1{\qquad\mathbf{\exists\,Intro}\colon #1 \\}
\def\R#1{\qquad\mathbf{R}\colon #1 \\}
\def\ci#1{\qquad\mathbf{\land\,I}\colon #1 \\}
\def\ce#1{\qquad\mathbf{\land\,E}\colon #1 \\}
\def\oe#1{\qquad\mathbf{\lor\,E}\colon #1 \\}
\def\ii#1{\qquad\mathbf{\to I}\colon #1 \\}
\def\ie#1{\qquad\mathbf{\to E}\colon #1 \\}
\def\be#1{\qquad\mathbf{\leftrightarrow E}\colon #1 \\}
\def\bi#1{\qquad\mathbf{\leftrightarrow I}\colon #1 \\}
\def\qi#1{\qquad\mathbf{=I}\\}
\def\qe#1{\qquad\mathbf{=E}\colon #1 \\}
\def\ne#1{\qquad\mathbf{\neg E}\colon #1 \\}
\def\ni#1{\qquad\mathbf{\neg I}\colon #1 \\}
\def\IP#1{\qquad\mathbf{IP}\colon #1 \\}
\def\X#1{\qquad\mathbf{\bot\,Elim}\colon #1 \\}
\def\DNE#1{\qquad\mathbf{DNE}\colon #1 \\}
$
$
\fitch{
\lnot A \to C\\
B \leftrightarrow C
}{
 \fitch{\lnot A \lor B}{
 \vdots\\
 C
}\\
\fitch{C}{
  \vdots\\
 \lnot A \lor B
}
}
$
A couple of points about your proof:

*

*Remember to clearly mark your assumptions. Sometimes, indentation is used to accomplish that task.

*Your line 4 is incorrect: you cannot derive $\lnot A$ from $\lnot A \to C$ and C. The rule of modus ponens states that, for any propositions $P, Q$ $$P \to Q, P \vdash Q$$
