Math Problem Statement

without using truth table prove (~pvq)^(p^(P^q))

Solution

To prove the logical equivalence of the expression (¬pq)(p(¬pq))(\neg p \vee q) \wedge (p \wedge (\neg p \vee q)), we can simplify the expression using logical identities. Here's the step-by-step proof:

Step 1: Simplify the expression

Consider the given expression:

(¬pq)(p(¬pq))(\neg p \vee q) \wedge (p \wedge (\neg p \vee q))

This expression is a conjunction (\wedge) of two components: (¬pq)(\neg p \vee q) and (p(¬pq))(p \wedge (\neg p \vee q)).

Step 2: Distribute (¬pq)(\neg p \vee q) over the conjunction

The expression can be expanded by distributing (¬pq)(\neg p \vee q) over the conjunction:

[ [(\neg p \vee q) \wedge p] \wedge [(\neg p \vee q) \wedge (\neg p \vee q)] ]

Since (¬pq)(¬pq)(\neg p \vee q) \wedge (\neg p \vee q) simplifies to (¬pq)(\neg p \vee q), the expression becomes:

[(¬pq)p](¬pq)[(\neg p \vee q) \wedge p] \wedge (\neg p \vee q)

Step 3: Simplify the conjunction (¬pq)p(\neg p \vee q) \wedge p

Now, consider the conjunction (¬pq)p(\neg p \vee q) \wedge p. This can be rewritten using the distributive law:

[(¬pp)(qp)][(\neg p \wedge p) \vee (q \wedge p)]

Since ¬pp\neg p \wedge p is a contradiction (i.e., always false), it simplifies to:

false(qp)\text{false} \vee (q \wedge p)

This further simplifies to:

qpq \wedge p

Thus, the original expression simplifies to:

(qp)(¬pq)(q \wedge p) \wedge (\neg p \vee q)

Step 4: Distribute qpq \wedge p over the conjunction

Distribute qpq \wedge p over the conjunction:

[ [q \wedge p \wedge \neg p] \vee [q \wedge p \wedge q] ]

Step 5: Simplify each term

  1. The term qp¬pq \wedge p \wedge \neg p is a contradiction, as pp and ¬p\neg p cannot both be true, so this term simplifies to false.

  2. The term qpqq \wedge p \wedge q simplifies to qpq \wedge p, because qqq \wedge q is just qq.

So the entire expression simplifies to:

qpq \wedge p

Conclusion:

Thus, the original expression simplifies to qpq \wedge p. Therefore, (¬pq)(p(¬pq))(\neg p \vee q) \wedge (p \wedge (\neg p \vee q)) is logically equivalent to qpq \wedge p.

Would you like further details or have any questions?

Here are some related questions:

  1. What is the distributive law in logic?
  2. How does the distributive law apply to conjunctions and disjunctions?
  3. What is the significance of contradictions like p¬pp \wedge \neg p in logic?
  4. How can logical identities be used to simplify complex expressions?
  5. What are common pitfalls when simplifying logical expressions?

Tip: When simplifying logical expressions, always look for contradictions like p¬pp \wedge \neg p or tautologies like p¬pp \vee \neg p, as they can drastically simplify the expression.