Simplify (using the laws of Boolean algebra)

Question

I have: $(\overline{A} \land \overline{B} \land \overline{C}) \lor (\overline{A} \land \overline{B} \land C) \lor (\overline{A} \land B \land C) \lor (A \land \overline{B} \land \overline{C}) \lor (A \land B \land \overline{C}) \lor (A \land B \land C)$

I found this (step by step) solution:

1. $(\overline{A} \land \overline{B} \land \overline{C}) \lor (\overline{A} \land \overline{B} \land C) \lor (\overline{A} \land B \land C) \lor (A \land \overline{B} \land \overline{C}) \lor (A \land B \land \overline{C}) \lor (A \land B \land C)$

2. $(\overline{A} \land \overline{B}) \lor (\overline{A} \land B \land C) \lor (A \land \overline{B} \land \overline{C}) \lor (A \land B \land \overline{C}) \lor (A \land B \land C)$

3. $(\overline{A} \land \overline{B}) \lor (\overline{A} \land C) \lor (A \land \overline{B} \land \overline{C}) \lor (A \land B \land \overline{C}) \lor (A \land B \land C)$

4. $(\overline{A} \land C) \lor (\overline{B} \land \overline{C}) \lor (A \land B \land \overline{C}) \lor (A \land B \land C)$

5. $(\overline{A} \land C) \lor (A \land \overline{C}) \lor ( \overline{B} \land \overline{C}) \lor (A \land B \land C)$

6. $(A \land \overline{C}) \lor (B \land C) \lor ( \overline{A} \land \overline{B})$

What rule is used here to absorb variables?

1) $(\overline{A} \land \overline{B} \land \overline{C}) \lor (\overline{A} \land \overline{B} \land C) \Leftrightarrow (\overline{A} \land \overline{B})$

2) $(\overline{A} \land \overline{B}) \lor (\overline{A} \land B \land C) \Leftrightarrow (\overline{A} \land \overline{B}) \lor (\overline{A} \land C)$

3) $(\overline{A} \land \overline{B}) \lor (\overline{A} \land C) \lor (A \land \overline{B} \land \overline{C}) \Leftrightarrow (\overline{A} \land C) \lor (\overline{B} \land \overline{C})$

Can you show an example of how to do this?

Answer 1

1) $\def\neg#1{\overline{#1}} (\neg{A} \land \neg{B} \land \neg{C}) \lor (\neg{A} \land \neg{B} \land C) \iff (\neg{A} \land \neg{B})$

$$\begin{split} (\neg{A} \land \neg{B} \land \neg{C}) \lor (\neg{A} \land \neg{B} \land C) \iff & (\neg A\land \neg B)\land(\neg C\lor C) &:\text{distribution, and association}\\ \iff & (\neg A\land \neg B)\land\top &:\text{complementation/tautology}\\ \iff & \neg A\land \neg B &:\text{identity}\end{split}$$

2) $(\neg{A} \land \neg{B}) \lor (\neg{A} \land B \land C) \iff (\neg{A} \land \neg{B}) \lor (\neg{A} \land C)$

Use the above, and idempotence.   Sometimes making things seem more complicated helps to simplify them.

$$\begin{split}(\neg{A} \land \neg{B}) \lor (\neg{A} \land B \land C) & \iff (\neg A\land\neg B\land \neg C)\lor (\neg A\land\neg B\land C)\lor (\neg A\land B\land C) \\ & \iff (\neg A\land\neg B\land \neg C)\lor (\neg A\land\neg B\land C)\lor (\neg A\land\neg B\land C)\lor (\neg A\land B\land C) \\ & \iff (\neg{A} \land \neg{B}) \lor (\neg{A} \land C)\end{split}$$

3) $(\neg{A} \land \neg{B}) \lor (\neg{A} \land C) \lor (A \land \neg{B} \land \neg{C}) \iff (\neg{A} \land C) \lor (\neg{B} \land \neg{C})$

Do you have the idea, now?   Give it a whirl.