The problem asks to determine whether the given logical statement is true or false:

$\forall x \in \mathbb{R}, \text{IF } |x + 2| < 5 \text{ AND } |x - 1| < 3, \text{ THEN } x < 1.$

Step-by-Step Analysis:

1. First Inequality: $|x + 2| < 5$

   • This absolute value inequality can be rewritten as: $-5 < x + 2 < 5$ Subtracting 2 from all parts: $-7 < x < 3$

2. Second Inequality: $|x - 1| < 3$

   • This absolute value inequality can be rewritten as: $-3 < x - 1 < 3$ Adding 1 to all parts: $-2 < x < 4$

3. Combined Inequalities:

   • Now, we combine the two ranges:
     • $-7 < x < 3$ from the first inequality.
     • $-2 < x < 4$ from the second inequality.
   • The intersection of these intervals is: $-2 < x < 3$

4. The Conclusion: $x < 1$

   • We now check whether $x < 1$ holds for all values in the interval $-2 < x < 3$.
   • Clearly, this is false for values in the interval $1 \leq x < 3$, where $x$ is greater than or equal to 1.

Conclusion:

The statement is false because the conclusion $x < 1$ does not hold for all values of $x$ in the range $-2 < x < 3$.

Would you like more details or have any questions?

Here are five related questions for practice:

1. What is the solution set for $|x + 3| < 4$?
2. How would the inequality change if the conclusion was $x \leq 1$?
3. Can you find a counterexample for this logical statement in a different domain (e.g., integers)?
4. How does the solution change if the inequalities use "greater than or equal to" instead of strict inequalities?
5. How would the problem change if we had $|x - 1| > 3$?

Tip: When solving absolute value inequalities, always consider splitting the absolute value into two separate inequalities to handle both the positive and negative cases.