To find the logically equivalent statement to ∼[(𝑝 ∨ 𝑟) → ∼(𝑝 ∨ 𝑞)], let's break down the expression step by step.

Step 1: Recall the logical implication (→)

An implication $A \to B$ is logically equivalent to $\sim A \vee B$, meaning "if $A$, then $B$" is the same as "not $A$ or $B$".

So, $(p \vee r) \to \sim(p \vee q)$ becomes:

$\sim(p \vee r) \vee \sim(p \vee q)$

Step 2: Negating the entire expression (outer negation)

The original expression has a negation in front: $\sim[(p \vee r) \to \sim(p \vee q)]$.

Using De Morgan's law, $\sim(A \vee B)$ is equivalent to $\sim A \wedge \sim B$. So we apply De Morgan's law to $\sim[\sim(p \vee r) \vee \sim(p \vee q)]$, which gives:

$(p \vee r) \wedge (p \vee q)$

Step 3: Simplifying the expression

The expression $(p \vee r) \wedge (p \vee q)$ says that both conditions must be true. This is a conjunction of two disjunctions.

Now, let's distribute the terms:

$(p \wedge p) \vee (p \wedge q) \vee (r \wedge p) \vee (r \wedge q)$

Simplifying:

$p \vee (p \wedge q) \vee (r \wedge p) \vee (r \wedge q)$

Since $p \vee (p \wedge q)$ simplifies to just $p$, we can further reduce the expression to:

$p \vee (r \wedge q)$

Step 4: Match the result with the options

The expression $p \vee (r \wedge q)$ is logically equivalent to $p \vee (\sim q \wedge r)$, which corresponds to the choice:

$\boxed{p \vee (\sim q \wedge r)}$

Answer: $p \vee (\sim q \wedge r)$

Would you like further details on any step?

Related Questions:

1. What is the meaning of logical equivalence in propositional logic?
2. How do De Morgan's laws apply to negating complex expressions?
3. How does the distributive property work in logical conjunction and disjunction?
4. What are the rules for simplifying logical expressions?
5. How can truth tables help verify logical equivalence?

Tip:

Always simplify expressions by breaking them down step by step, especially when dealing with negations and implications.