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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?

Wind
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  • The reference "here" in "What rule is used..." is not clear. Are you asking about one of the steps or all of them? When you ask for "an example of how to do this", the reference to "this" is similarly unclear. – hardmath Jun 13 '18 at 20:26
  • For example, about: $(\overline{A} \land \overline{B} \land \overline{C}) \lor (\overline{A} \land \overline{B} \land C) \Leftrightarrow (\overline{A} \land \overline{B})$ – Wind Jun 13 '18 at 20:28
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    In that case, we have, by using the distributive property, $\Big((\overline{A} \land \overline{B})\land (\overline C \lor C)\Big) \iff (\overline A \land \overline B)$. Now what do you know about $(\overline C \lor C)$. – amWhy Jun 13 '18 at 20:32
  • @amWhy $(\overline C \lor C) \equiv 1$ – Wind Jun 13 '18 at 20:35
  • Also, $\Big((\overline{A} \land \overline{B}) \land 1\Big) \iff (\overline A \land \overline B)$ – Wind Jun 13 '18 at 20:36
  • Exactly. So on the left-hand side we have $$\overline A \land \overline B \land 1 \equiv \overline A \land \overline B$$ $P \equiv P\lor 1$ – amWhy Jun 13 '18 at 20:37
  • Can I use this same for: $(\overline{A} \land \overline{B}) \lor (\overline{A} \land B \land C) \Leftrightarrow (\overline{A} \land \overline{B}) \lor (\overline{A} \land C)$

    ?

     – Wind Jun 13 '18 at 20:39

1 Answers1

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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.

Graham Kemp
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