Animated Solution for Mathematics - Sets and Relations: Consider the following two propositions: \\ $P1:\sim (p\to \sim q)$ \\ $P2:(p\wedge \sim q)\wedge ((\sim p)\vee q)$ \\ If the proposition $p\to ((\sim p)\vee q)$ is evaluated as FALSE, then:

Welcome, future engineer! Today, we are going to step into the shoes of a logic detective. We have a mystery: a complex logical proposition that evaluates to FALSE, and we need to uncover the hidden truth values of our variables, $p$ and $q$.

This isn't just about solving a problem; it's about understanding the fundamental architecture of reasoning.

The False Implication

We are given the condition that $p\to ((\sim p)\vee q)$ is FALSE. In the world of logic, an implication $A\to B$ is a very specific structure.

It is only FALSE in one unique, catastrophic scenario: when the premise $A$ is TRUE, and the conclusion $B$ is FALSE. If the premise were FALSE, the implication would be vacuously TRUE.

So, we immediately know two things: 1. $p$ must be TRUE. 2. The entire expression $((\sim p)\vee q)$ must be FALSE.

Now, let's look at our second piece of evidence: $((\sim p)\vee q)=\text{FALSE}$. We already deduced that $p$ is TRUE, which means its negation, $\sim p$, is FALSE.

Now we have an OR statement: $(\text{FALSE}\vee q)=\text{FALSE}$. For an OR operation to be FALSE, both components must be FALSE.

Since $\sim p$ is already FALSE, the only way for this statement to be FALSE is if $q$ is also FALSE. We have cracked the case: $p$ is TRUE and $q$ is FALSE.

Now that we have our suspects, $p=\text{TRUE}$ and $q=\text{FALSE}$, let's test our propositions.

First, consider $P1:\sim (p\to \sim q)$. Substituting our values, we get $\sim q=\sim (\text{FALSE})=\text{TRUE}$.

So, $P1$ becomes $\sim (\text{TRUE}\to \text{TRUE})$. Since $\text{TRUE}\to \text{TRUE}$ is TRUE, the negation makes $P1$ FALSE.

Finally, let's look at $P2:(p\wedge \sim q)\wedge ((\sim p)\vee q)$. Substituting our values: $(p\wedge \sim q)$ becomes $(\text{TRUE}\wedge \text{TRUE})$, which is TRUE.

The second part, $((\sim p)\vee q)$, becomes $(\text{FALSE}\vee \text{FALSE})$, which is FALSE. So, $P2$ is $\text{TRUE}\wedge \text{FALSE}$.

As any good detective knows, an AND operation requires both sides to be TRUE to succeed. Since one side is FALSE, $P2$ is also FALSE.

We have systematically dismantled the problem. Both $P1$ and $P2$ evaluate to FALSE.

Logic isn't just about symbols; it's about the clarity of thought. Keep practicing, and you'll find that even the most complex logical structures become simple, elegant puzzles.