How do you prove A ↔ B ∴ (A ∧ B) ∨ (¬A ∧ ¬B) using a fitch diagram?

Question

Furthermore, how can I prove A ↔ B ∴ (A ∧ B) ∨ (¬A ∧ ¬B) without using Modus Tollens and De Morgan's using the fitch diagram?

Answer 1

That is a good start. If you are allowed to use the Law of the Excluded Middle, then you may next do a Proof by Cases: $A$ or $\lnot A$.   Else you may go the proof by contradiction route. $$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{A\leftrightarrow B}{A\to B\\B\to A\\A\vee\neg A\\\fitch{A}{~~\vdots\\(A\wedge B)\vee(\neg A\wedge\neg B)}\\\fitch{\neg A}{~~\vdots\\(A\wedge B)\vee(\neg A\wedge\neg B)}\\(A\wedge B)\vee(\neg A\wedge\neg B)}\quad\fitch{A\leftrightarrow B}{A\to B\\B\to A\\\fitch{\neg((A\wedge B)\vee(\neg A\wedge\neg B))}{\fitch{A}{~~\vdots\\(A\wedge B)\vee(\neg A\wedge\neg B)\\\bot}\\\neg A\\~~\vdots\\(A\wedge B)\vee(\neg A\wedge\neg B) \\\bot}\\\neg\neg((A\wedge B)\vee(\neg A\wedge\neg B))\\(A\wedge B)\vee(\neg A\wedge\neg B)}$$

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without using Modus Tollens

You may derive $P\to Q, \neg Q\vdash \neg P$ as follows:

$$\fitch{~~1.~~P\to Q\\~~2.~~\neg Q}{\fitch{~~3.~~P}{~~4.~~Q\quad{\to}\mathsf E~1,3\\~~5.~~\bot\quad\neg\mathsf E~2,3}\\~~6.~~\neg P\qquad\neg\mathsf I~3{-}5}$$