---
title: "De Morgan; negative OR equals positive AND"
subject: "Digital Logic"
chapter: 2
key: "semester_3/digital_logic:demorgan-negative-or-positive-and"
source: https://www.bctengineer.com/q/semester_3/digital_logic/demorgan-negative-or-positive-and
site: BCT Engineering
---

# De Morgan; negative OR equals positive AND

Digital Logic · Chapter 2: Digital Logic

- Asked in: 2074 Chaitra, 4+2 marks
- Video: [Lesson 2 - Negative Logic and DeMorgan's Theorem](https://www.youtube.com/watch?v=TyB1jFj5hQ8) — LBEbooks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/digital_logic/demorgan-negative-or-positive-and
- Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-2

## Question

State and prove De-Morgan's theorems with necessary diagrams. Prove that negative logic OR Gate is equivalent to positive logic AND Gate.

## Short answer

De Morgan's theorems: (A+B)' = A'B' and (AB)' = A'+B', proven by truth tables. A gate with truth table 0,0,0,1 in positive logic (H=1) acts as AND; in negative logic (H=0) the complemented levels give OR, so negative-logic OR equals positive-logic AND.

## Answer

#### De Morgan's theorems
**First theorem:** the complement of a sum (OR) of variables equals the product (AND) of their complements.
$$
\overline{A + B} = \bar{A} \cdot \bar{B}
$$

**Second theorem:** the complement of a product (AND) of variables equals the sum (OR) of their complements.
$$
\overline{A \cdot B} = \bar{A} + \bar{B}
$$

In words: "break the bar and change the sign".

**Proof by truth table**

| A | B | $\overline{A+B}$ | $\bar A \cdot \bar B$ | $\overline{AB}$ | $\bar A + \bar B$ |
|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 |

For all four input combinations, column 3 equals column 4 and column 5 equals column 6, so both theorems are proved.

**Algebraic proof (Theorem 1):** let $X = A + B$, $Y = \bar A \bar B$. Then $X + Y = A + B + \bar A\bar B = A + B + \bar B = 1$ and $XY = A\bar A\bar B + B\bar A\bar B = 0$, so $Y = \bar X$. Theorem 2 follows by duality.

**Diagrams**

```text
 Theorem 1: NOR = bubbled AND
 A --|‾‾‾\             A --o|‾‾‾\
     | OR  )o-- Y   =       | AND )-- Y
 B --|___/             B --o|___/

 Theorem 2: NAND = bubbled OR
 A --|‾‾‾\             A --o|‾‾‾\
     | AND )o-- Y   =       | OR  )-- Y
 B --|___/             B --o|___/

 (o = inversion bubble)
```

#### Negative-logic OR = positive-logic AND
Take a gate whose output is HIGH only when both inputs are HIGH:

| A | B | Y |
|---|---|---|
| L | L | L |
| L | H | L |
| H | L | L |
| H | H | H |

**Positive logic** (H = 1): Y = 1 only for A = B = 1 → **AND**.

**Negative logic** (H = 0, L = 1):

| A | B | Y |
|---|---|---|
| 1 | 1 | 1 |
| 1 | 0 | 1 |
| 0 | 1 | 1 |
| 0 | 0 | 0 |

Y = 0 only when both inputs are 0 → **OR**.

**Algebraically:** a negative-logic OR, seen in positive logic, has all inputs and the output complemented:

$$
\begin{aligned} Y &= \overline{\bar A + \bar B} \\ &= \bar{\bar A} \cdot \bar{\bar B} \quad (\text{De Morgan}) \\ &= A \cdot B \end{aligned}
$$

So the negative-logic OR gate is the same circuit as the positive-logic AND gate.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In negative logic HIGH =
   A. 1
   B. Z
   C. unknown
   D. 0
   Answer: D. LOW = 1. (Key point: Negative logic)

2. Gate with Y high only when A and B are high is?
   A. OR in positive logic
   B. NAND
   C. XOR
   D. AND in positive logic
   Answer: D. Y = 1 only for A = B = 1. (Key point: Table)

3. In negative logic same gate acts as?
   A. AND
   B. NOR
   C. OR
   D. NAND
   Answer: C. Y = 0 only when both 0. (Key point: Negative view)

4. Y = (A'+B')' =
   A. A+B
   B. AB
   C. A'B'
   D. A'+B
   Answer: B. De Morgan. (Key point: Algebra)

## Related in Digital Logic

- [De Morgan; positive NAND equals negative NOR](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Ademorgan-positive-nand-negative-nor) — Digital Logic, chapter 2
- [Positive/negative logic; verify 3-variable De Morgan](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Ademorgan-three-variable-verify) — Digital Logic, chapter 2
- [Prove De Morgan's theorems; XOR using NAND](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Ademorgan-prove-xor-nand) — Digital Logic, chapter 2, asked 3×
- [Prove positive XOR equals negative XNOR](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Apositive-xor-negative-xnor-prove) — Digital Logic, chapter 2, asked 3×
- [De Morgan; X-NOR using a universal gate](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Ademorgan-xnor-universal-gate) — Digital Logic, chapter 2
- [Positive XOR equals negative XNOR](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Apositive-xor-negative-xnor-define) — Digital Logic, chapter 2, asked 3×

## Related in other subjects

- [State and prove De Morgan's theorems](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Ade-morgan-state-prove) — Basic Electronics Engineering, chapter 6, asked 4×