Let's go through each of the questions based on the image you've uploaded.

1. For each pair of propositions $P$ and $Q$ below, state whether or not $P \equiv Q$.

• a) $P = p, Q = p \lor q$

We are asked whether $p \equiv p \lor q$. This is false in general because $p \lor q$ could be true even when $p$ is false, depending on $q$.

• b) $P = p \land q, Q = \neg p \lor \neg q$

The expression $P$ is $p \land q$, which is true if both $p$ and $q$ are true. The expression $Q$ is $\neg p \lor \neg q$, which is true if at least one of $p$ or $q$ is false. These are negations of each other, so $P \not\equiv Q$.

• c) $P = p \to q, Q = \neg p \lor q$

This is true because $p \to q$ is logically equivalent to $\neg p \lor q$ by the definition of implication.

• d) $P = p \land (\neg q \lor r), Q = p \lor (q \land \neg r)$

We need to check whether these two compound propositions are logically equivalent. In this case, $P$ and $Q$ represent two different logical expressions, so they are generally not equivalent.

2. Determine the truth value of each statement below (domain of discourse is $\mathbb{R}$).

• a) $\forall x (x^2 > x)$

This is false, because for $x = 1$, we have $x^2 = x$, and for $0 < x < 1$, $x^2 < x$.

• b) $\exists x (x^2 > x)$

This is true because there exist values of $x$, such as $x = 2$, for which $x^2 > x$.

• c) $\forall x (x > 1 \to x^2 > x)$

This is true, because for any $x > 1$, we have $x^2 > x$.

• d) $\exists x (x > 1 \land x^2 > x)$

This is true, because there exist values of $x$, such as $x = 2$, for which $x > 1$ and $x^2 > x$.

3. Determine the truth value of each statement below (domain of discourse is $\mathbb{R} \times \mathbb{R}$).

• a) $\forall x \forall y (x^2 < y + 1)$

This is false. For example, take $x = 2$ and $y = 1$. Then $x^2 = 4$, which is not less than $y + 1 = 2$.

• b) $\forall x \exists y (x^2 < y + 1)$

This is true, because for any $x$, you can always find a $y$ such that $x^2 < y + 1$ by choosing a sufficiently large $y$.

• c) $\exists x \forall y (x^2 < y + 1)$

This is false because for any fixed $x$, you cannot make $x^2$ less than $y + 1$ for all $y$. For large values of $y$, $x^2 \geq y + 1$.

• d) $\exists x \exists y (x^2 < y + 1)$

This is true because you can always find $x$ and $y$ such that $x^2 < y + 1$, for example, $x = 1$ and $y = 1$.

Do you have any questions or would you like more details on any of these solutions?

5 Related Questions:

1. What does it mean for two propositions to be logically equivalent?
2. How can implication ($\to$) be transformed into disjunction ($\lor$) in logical statements?
3. Why does $x^2 > x$ not hold for all real numbers $x$?
4. How can quantifiers affect the truth value of a logical statement in mathematics?
5. What strategies can be used to determine whether statements involving multiple quantifiers are true?

Tip:

When working with logical equivalence, remember that transforming the expressions using standard logical identities (e.g., De Morgan's laws, distributive properties) can help clarify their structure.