Little Frank
10/26/2023 · Senior High School
Without using truth table prove that : \( (p \wedge q) \vee(-p \wedge q) \vee(p \wedge-q)=p \vee q \)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To prove the logical equivalence \( (p \wedge q) \vee (-p \wedge q) \vee (p \wedge -q) = p \vee q \) without using a truth table, we can use logical identities and simplifications.
1. **Start with the left-hand side**:
\[
(p \wedge q) \vee (-p \wedge q) \vee (p \wedge -q)
\]
2. **Factor out common terms**:
Notice that the first two terms share \( q \):
\[
= q \wedge (p \vee -p) \vee (p \wedge -q)
\]
Here, \( p \vee -p \) is a tautology (always true), so we can simplify:
\[
= q \vee (p \wedge -q)
\]
3. **Distribute \( q \) over \( (p \wedge -q) \)**:
We can rewrite \( q \vee (p \wedge -q) \) using the distributive law:
\[
= (q \vee p) \wedge (q \vee -q)
\]
Again, \( q \vee -q \) is a tautology, so we simplify:
\[
= q \vee p
\]
4. **Rearranging**:
Since \( q \vee p \) is the same as \( p \vee q \) (due to the commutative property of disjunction), we have:
\[
= p \vee q
\]
5. **Conclusion**:
Thus, we have shown that:
\[
(p \wedge q) \vee (-p \wedge q) \vee (p \wedge -q) = p \vee q
\]
This completes the proof.
Quick Answer
We can simplify \( (p \wedge q) \vee (-p \wedge q) \vee (p \wedge -q) \) to \( p \vee q \) by factoring out common terms and using logical identities.
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