Logical expressions

Question

How can I transform the logical expressions $(\ref{eq:first})$ into $(\ref{eq:second})$ or $(\ref{eq:second})$ into $(\ref{eq:first})$?
Please show me in a didactic way for a better understanding.

$$ (a \land b \land c) \lor (a \land \lnot b \land \lnot c) \lor (\lnot a \land b \land \lnot c) \lor (\lnot a \land \lnot b \land c)\label{eq:first}\tag{1} $$

$$ \{[(a \land b) \lor (\lnot a \land \lnot b)] \land c\} \lor \{\lnot[(a \land b) \lor (\lnot a \land \lnot b)] \land \lnot c\}\label{eq:second}\tag{2} $$

Answer 1

HINT

Start doing the same thing for the rest: take out the $\neg c$ using Distribution.

Then use Distribution on $a \land \neg b$ and $\neg a \land b$: you get $4$ terms:

$(a \lor \neg a) \land (a \lor b) \land (\neg b \lor \neg a) \land (\neg b \lor b)$

$a \lor \neg a$ and $\neg b \lor b$ can be removed by Complement.

Finally, a couple of DeMorgan's and you're done.

Answer 2

Hint:

$$ (a \land b \land c) \lor (a \land \lnot b \land \lnot c) \lor (\lnot a \land b \land \lnot c) \lor (\lnot a \land(\lnot b \land c)) \\\equiv[a\land(b= c)]\lor[\lnot a\land(b\ne c)] \\\equiv a=(b=c)$$

$$ \{[(a \land b) \lor (\lnot a \land \lnot b)] \land c\} \lor \{\lnot[(a \land b) \lor (\lnot a \land \lnot b)] \land \lnot c\} \\\equiv[(a=b)\land c]\lor[(a\ne b)\land \lnot c] \\\equiv(a=b)=c$$