Homework Sheet 5 Logic1 Construct The Truth Table For P Q P ✓ Solved

Homework Sheet #5: Logic 1. Construct the truth table for (p q) (¬p q) p q ¬p p q ¬p q (p q) (¬p q) 2. Populate the truth table and see whether this statement is correct: p q ≡ ¬q ¬q p q p q ¬ p ¬q ¬q ¬q 3. Populate the provided truth table for the statement demonstrating De Morgan’s Law: ¬(p q) ↔ ¬p V ¬q p q 4. Use truth tables to show whether the following propositions are equivalent: Proposition 1: p ∧ (q ∧ r) Proposition 2: (q ∨ q) ∧ (q ∨ r) p q r 5. For this logic circuit provide: a. the propositional logic formula b. the truth table. ¬ (A ∨ B) AND OR A B C

Paper for above instructions

1. Constructing the Truth Table for (p ∧ q) ∧ (¬p ∨ q)

To construct the truth table for the expression \( (p \land q) \land (\neg p \lor q) \), we will evaluate it step by step.

Step 1: Identify Variables and Their States

We have two logical variables \( p \) and \( q \) which can be either true (T) or false (F). This results in the following combinations:
| p | q |
|---|---|
| T | T |
| T | F |
| F | T |
| F | F |

Step 2: Determine Values of Supporting Expressions

Our expression consists of several parts:
- \( \neg p \) : Negation of \( p \)
- \( p \land q \) : Conjunction of \( p \) and \( q \)
- \( \neg p \lor q \): Disjunction of \( \neg p \) and \( q \)
Populate a truth table based on these components:
| p | q | ¬p | p ∧ q | ¬p ∨ q | (p ∧ q) ∧ (¬p ∨ q) |
|---|---|----|-------|--------|---------------------|
| T | T | F | T | T | T |
| T | F | F | F | F | F |
| F | T | T | F | T | F |
| F | F | T | F | T | F |

Final Truth Table

| p | q | (p ∧ q) ∧ (¬p ∨ q) |
|---|---|---------------------|
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | F |

2. Verify the Statement p ∧ q ≡ ¬q ∨ ¬p

We need to confirm whether \( p \land q \equiv \neg q \lor \neg p \).

Truth Table for Both Sides

| p | q | p ∧ q | ¬q | ¬p | ¬q ∨ ¬p |
|---|---|-------|----|----|---------|
| T | T | T | F | F | F |
| T | F | F | T | F | T |
| F | T | F | F | T | T |
| F | F | F | T | T | T |

Conclusion

On reviewing the outcomes for both expressions \( p \land q \) and \( \neg q \lor \neg p \):
- \( p \land q \) results in T for \( (T, T) \) only and F for others.
- \( \neg q \lor \neg p \) results in T for all pairs except \( (T, T) \).
Thus, \( p \land q \) is not equivalent to \( \neg q \lor \neg p \).

3. Truth Table to Demonstrate De Morgan’s Law: ¬(p ∧ q) ↔ ¬p V ¬q

Table Structure:

| p | q | ¬(p ∧ q) | ¬p | ¬q | ¬p V ¬q |
|---|---|----------|----|----|----------|
| T | T | F | F | F | F |
| T | F | T | F | T | T |
| F | T | T | T | F | T |
| F | F | T | T | T | T |

Conclusion

The results confirm that:
- \( ¬(p ∧ q) \) yields true when either \( p \) or \( q \) are false.
- Similarly, \( ¬p ∨ ¬q \) evaluates to true under the same conditions.
Thus, \( ¬(p ∧ q) \equiv ¬p ∨ ¬q \) holds true, demonstrating De Morgan’s Law.

4. Using Truth Tables for Equivalence: Proposition 1 and Proposition 2

Proposition 1: \( p ∧ (q ∨ r) \)
Proposition 2: \( (p ∨ q) ∧ (p ∨ r) \)

Truth Table

| p | q | r | q ∨ r | p ∧ (q ∨ r) | p ∨ q | p ∨ r | (p ∨ q) ∧ (p ∨ r) |
|---|---|---|-------|-------------|-------|-------|--------------------|
| T | T | T | T | T | T | T | T |
| T | T | F | T | T | T | T | T |
| T | F | T | T | T | T | T | T |
| T | F | F | F | F | T | T | T |
| F | T | T | T | F | T | T | T |
| F | T | F | T | F | T | F | F |
| F | F | T | T | F | F | T | F |
| F | F | F | F | F | F | F | F |

Conclusion

By examining the table, we observe:
- The outputs for both propositions match in all configurations.
- Thus, \( p ∧ (q ∨ r) \equiv (p ∨ q) ∧ (p ∨ r) \).

5. Logic Circuit Evaluation

Circuit Specifications

- The formula: \( ¬(A ∨ B) ∧ C \)

Truth Table

| A | B | C | A ∨ B | ¬(A ∨ B) | ¬(A ∨ B) ∧ C |
|---|---|---|-------|-----------|---------------|
| T | T | T | T | F | F |
| T | T | F | T | F | F |
| T | F | T | T | F | F |
| T | F | F | T | F | F |
| F | T | T | T | F | F |
| F | T | F | T | F | F |
| F | F | T | F | T | T |
| F | F | F | F | T | F |

Conclusion

This table highlights how the circuit's outputs depend on the states of A, B, and C.
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References

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5. Smith, P. (2017). Logic and Computer Design Fundamentals. Cengage Learning.
6. Russell, B., & Whitehead, A. N. (1910). Principia Mathematica. Cambridge University Press.
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