Simplify logical formulas using various logical equivalence laws | Step-by-Step Solution
Problem
Use the laws stated in the text to find simpler formulas equivalent to these formulas: a) ¬(¬P ∨ Q) ∨ (P ∧ ¬R), b) ¬(¬P ∧ Q) ∨ (P ∧ ¬R), c) (P ∧ R) ∨ [¬R ∧ (P ∨ Q)]
🎯 What You'll Learn
- Apply logical equivalence laws
- Simplify complex logical expressions
- Understand logical transformations
Prerequisites: Propositional logic basics, Logical operators, Truth tables
💡 Quick Summary
Hi there! I can see you're working with logical formula simplification, which is really just like simplifying algebraic expressions but with logical operators instead of numbers. The key is to systematically apply Boolean algebra laws like De Morgan's laws, double negation, and distributive properties. When you look at these formulas, what do you notice about the negation symbols and parentheses - can you think about which laws might help you "break into" those parentheses first? I'd suggest starting from the innermost parts and working outward, just like you would with nested algebraic expressions. You already know these logical equivalence laws, so trust your instincts about which ones to apply when you see patterns like ¬(something ∨ something else) or double negations!
Step-by-Step Explanation
1. What We're Solving:
We need to simplify three logical formulas using Boolean algebra laws like De Morgan's laws, distributive laws, and absorption laws. Think of this like simplifying algebraic expressions, but with logical operators instead of arithmetic ones!2. The Approach:
The key strategy is to apply logical equivalence laws systematically:- De Morgan's Laws: ¬(P ∨ Q) ≡ ¬P ∧ ¬Q and ¬(P ∧ Q) ≡ ¬P ∨ ¬Q
- Double Negation: ¬¬P ≡ P
- Distributive Laws: P ∧ (Q ∨ R) ≡ (P ∧ Q) ∨ (P ∧ R)
- Absorption Laws: P ∨ (P ∧ Q) ≡ P
3. Step-by-Step Solution:
Part a) ¬(¬P ∨ Q) ∨ (P ∧ ¬R)
Step 1: Apply De Morgan's Law to ¬(¬P ∨ Q)
- ¬(¬P ∨ Q) becomes ¬(¬P) ∧ ¬Q
- ¬(¬P) becomes P
- So we have: P ∧ ¬Q
- (P ∧ ¬Q) ∨ (P ∧ ¬R)
- P ∧ (¬Q ∨ ¬R)
Part b) ¬(¬P ∧ Q) ∨ (P ∧ ¬R)
Step 1: Apply De Morgan's Law to ¬(¬P ∧ Q)
- ¬(¬P ∧ Q) becomes ¬(¬P) ∨ ¬Q
- ¬(¬P) becomes P
- So we have: P ∨ ¬Q
- (P ∨ ¬Q) ∨ (P ∧ ¬R)
- P ∨ ¬Q ∨ (P ∧ ¬R)
- P ∨ ¬Q
Part c) (P ∧ R) ∨ [¬R ∧ (P ∨ Q)]
Step 1: Distribute ¬R into (P ∨ Q)
- ¬R ∧ (P ∨ Q) becomes (¬R ∧ P) ∨ (¬R ∧ Q)
- (P ∧ R) ∨ (¬R ∧ P) ∨ (¬R ∧ Q)
- P ∧ (R ∨ ¬R) ∨ (¬R ∧ Q)
- P ∧ True ∨ (¬R ∧ Q) = P ∨ (¬R ∧ Q)
4. The Answers:
- a) P ∧ (¬Q ∨ ¬R)
- b) P ∨ ¬Q
- c) P ∨ (¬R ∧ Q)
5. Memory Tip:
Remember "DUDA" - De Morgan's first, Undo double negations, Distribute when possible, Absorb redundant terms. Always work from the innermost parentheses outward, just like solving nested algebraic expressions!Great job working through these! Boolean algebra becomes much easier once you recognize these common patterns. Keep practicing, and you'll start seeing these simplifications almost automatically! 🎉
⚠️ Common Mistakes to Avoid
- Misapplying logical laws
- Incorrectly canceling out negations
- Not fully distributing logical operators
This explanation was generated by AI. While we work hard to be accurate, mistakes can happen! Always double-check important answers with your teacher or textbook.
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📷 Problem detected:
Solve: 2x + 5 = 13
Step 1:
Subtract 5 from both sides...
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