Question

State Demorgan’s theorems and prove both theorems using a truth table.

Answer 1

De Morgan's Theorems are fundamental rules in logic and Boolean algebra that relate conjunctions (AND operations) and disjunctions (OR operations) through negation. The theorems are stated as follows:

1. **De Morgan's First Theorem:**
   \[
   \neg (A \land B) = \neg A \lor \neg B
   \]

2. **De Morgan's Second Theorem:**
   \[
   \neg (A \lor B) = \neg A \land \neg B
   \]

### Proof Using Truth Tables

Let's create truth tables to prove both theorems.

#### Proof of De Morgan's First Theorem:
\[
\neg (A \land B) = \neg A \lor \neg B
\]

| A | B | \( A \land B \) | \( \neg (A \land B) \) | \( \neg A \) | \( \neg B \) | \( \neg A \lor \neg B \) |
|---|---|-----------------|------------------------|--------------|--------------|--------------------------|
| T | T | T               | F                      | F            | F            | F                        |
| T | F | F               | T                      | F            | T            | T                        |
| F | T | F               | T                      | T            | F            | T                        |
| F | F | F               | T                      | T            | T            | T                        |

**Conclusion:** The columns for \( \neg (A \land B) \) and \( \neg A \lor \neg B \) are identical, proving that:
\[
\neg (A \land B) = \neg A \lor \neg B
\]

#### Proof of De Morgan's Second Theorem:
\[
\neg (A \lor B) = \neg A \land \neg B
\]

| A | B | \( A \lor B \)  | \( \neg (A \lor B) \) | \( \neg A \) | \( \neg B \) | \( \neg A \land \neg B \) |
|---|---|-----------------|------------------------|--------------|--------------|--------------------------|
| T | T | T               | F                      | F            | F            | F                        |
| T | F | T               | F                      | F            | T            | F                        |
| F | T | T               | F                      | T            | F            | F                        |
| F | F | F               | T                      | T            | T            | T                        |

**Conclusion:** The columns for \( \neg (A \lor B) \) and \( \neg A \land \neg B \) are identical, proving that:
\[
\neg (A \lor B) = \neg A \land \neg B
\]

### Summary
Both of De Morgan's Theorems have been verified through truth tables, demonstrating the relationship between conjunctions and disjunctions under negation.

Answer 2

De Morgan's Theorems are fundamental rules in Boolean algebra and digital logic design. They provide a way to simplify expressions involving AND and OR operations. There are two primary theorems:

1. **De Morgan’s First Theorem**:
   \[
   \overline{A \cdot B} = \overline{A} + \overline{B}
   \]
   This theorem states that the complement of a product (AND) of two variables is equal to the sum (OR) of their complements.

2. **De Morgan’s Second Theorem**:
   \[
   \overline{A + B} = \overline{A} \cdot \overline{B}
   \]
   This theorem states that the complement of a sum (OR) of two variables is equal to the product (AND) of their complements.

Let's prove both theorems using truth tables.

### Proof of De Morgan’s First Theorem

**Theorem**: \(\overline{A \cdot B} = \overline{A} + \overline{B}\)

1. **Construct the Truth Table for \(A \cdot B\) and \(\overline{A \cdot B}\):**

   | \(A\) | \(B\) | \(A \cdot B\) | \(\overline{A \cdot B}\) |
   |-------|-------|--------------|-------------------------|
   | 0     | 0     | 0            | 1                       |
   | 0     | 1     | 0            | 1                       |
   | 1     | 0     | 0            | 1                       |
   | 1     | 1     | 1            | 0                       |

2. **Construct the Truth Table for \(\overline{A}\) and \(\overline{B}\), and then \(\overline{A} + \overline{B}\):**

   | \(A\) | \(B\) | \(\overline{A}\) | \(\overline{B}\) | \(\overline{A} + \overline{B}\) |
   |-------|-------|-----------------|-----------------|------------------------------|
   | 0     | 0     | 1               | 1               | 1                            |
   | 0     | 1     | 1               | 0               | 1                            |
   | 1     | 0     | 0               | 1               | 1                            |
   | 1     | 1     | 0               | 0               | 0                            |

3. **Comparison**:
   By comparing the columns \(\overline{A \cdot B}\) and \(\overline{A} + \overline{B}\), we see that the results are identical for all combinations of \(A\) and \(B\). Thus, \(\overline{A \cdot B} = \overline{A} + \overline{B}\).

### Proof of De Morgan’s Second Theorem

**Theorem**: \(\overline{A + B} = \overline{A} \cdot \overline{B}\)

1. **Construct the Truth Table for \(A + B\) and \(\overline{A + B}\):**

   | \(A\) | \(B\) | \(A + B\) | \(\overline{A + B}\) |
   |-------|-------|-----------|----------------------|
   | 0     | 0     | 0         | 1                    |
   | 0     | 1     | 1         | 0                    |
   | 1     | 0     | 1         | 0                    |
   | 1     | 1     | 1         | 0                    |

2. **Construct the Truth Table for \(\overline{A}\) and \(\overline{B}\), and then \(\overline{A} \cdot \overline{B}\):**

   | \(A\) | \(B\) | \(\overline{A}\) | \(\overline{B}\) | \(\overline{A} \cdot \overline{B}\) |
   |-------|-------|-----------------|-----------------|----------------------------------|
   | 0     | 0     | 1               | 1               | 1                                |
   | 0     | 1     | 1               | 0               | 0                                |
   | 1     | 0     | 0               | 1               | 0                                |
   | 1     | 1     | 0               | 0               | 0                                |

3. **Comparison**:
   By comparing the columns \(\overline{A + B}\) and \(\overline{A} \cdot \overline{B}\), we see that the results are identical for all combinations of \(A\) and \(B\). Thus, \(\overline{A + B} = \overline{A} \cdot \overline{B}\).

### Conclusion
Both De Morgan's Theorems are verified through truth tables. The tables show that for any combination of input values for \(A\) and \(B\), the outputs of the expressions on both sides of the equations are identical, confirming the validity of De Morgan's Theorems.