Logic Practice

1. This is not a proposition because the truth value depends on the unspecified variable $x$.

2. This is a proposition, and it is true.

3. This is not a proposition because it is a command.

4. This is a proposition, and it is true.

Since $Q$ is false, $\neg Q$ is true, so

Also, $\neg P$ is false, so

Therefore,

An implication is false only when the antecedent is true and the conclusion is false. Since $P$ is false,

A biconditional is true when both statements have the same truth value. Since both $P$ and $Q$ are false,

Let $P$ be "the number is divisible by $6$" and $Q$ be "the number is divisible by $3$."

The contrapositive of $P \to Q$ is

So the contrapositive is:

If a number is not divisible by $3$, then it is not divisible by $6$.

Apply De Morgan's laws:

Then negate the disjunction:

So an equivalent formula is

A biconditional means both implications hold:

This matches the definition of "if and only if."

The negation of a universal statement is existential:

In words: there exists at least one $x$ for which $P(x)$ is false.

The negation of an existential statement is universal:

Now apply De Morgan's law inside the predicate:

No. The order of quantifiers matters.

The first statement says that for every $x$, you may choose a possibly different $y$.

The second statement says there is one single $y$ that works for every $x$.

These are different claims, so they are not logically equivalent.

1. $P \lor \neg P$ is a tautology because it is true for every truth value of $P$.

2. $P \land \neg P$ is a contradiction because it is never true.

First apply De Morgan's law:

So the formula becomes

Distribute $P$ across the disjunction:

Since $\neg P \land P \equiv F$, this simplifies to

First flip the universal quantifier:

Then use the implication equivalence:

So

Therefore the negation is

Use hypothetical syllogism on the first two premises:

Then apply modus ponens with $P$:

So the conclusion is

First use disjunctive syllogism:

Then apply modus ponens to $Q \to R$ and $Q$:

So the conclusion is

Use proof by cases on $P \lor Q$.

If $P$ is true, then $P \to R$ gives $R$.

If $Q$ is true, then $Q \to R$ gives $R$.

Since one of the two cases must hold, $R$ follows in either case. Therefore,

is a valid conclusion.

Use the distributive law:

This is in conjunctive normal form because it is an AND of OR-clauses.

Let $x$ be arbitrary and assume $x \in A \cap B$.

By the definition of intersection, this means $x \in A$ and $x \in B$.

In particular, $x \in A$.

Since every element of $A \cap B$ is in $A$, we conclude

Apply resolution to the first two premises:

Now use disjunctive syllogism with $\neg R$:

So the conclusion is

It is enough to give one counterexample.

Take the integer $1$. It is not even, so the claim "every integer is even" is false.

Thus a single counterexample disproof is sufficient.

No. This is the fallacy of affirming the consequent.

The premise $P \to Q$ does not allow you to conclude $P$ from $Q$.

In this example, many shapes are rectangles without being squares, so the conclusion does not follow.

If being licensed is necessary for driving, then driving cannot happen without being licensed:

If being licensed is sufficient for driving, then licensed implies driving:

Since $\neg P$ must be true, choose

Then $P \lor Q$ becomes true only if $Q$ is true, so choose

One satisfying assignment is $P = F$ and $Q = T$.

The formula requires both $P \to Q$ and $P$ to be true, which forces $Q$ to be true by modus ponens.

But the formula also requires $\neg Q$.

So the formula cannot be true under any assignment. It is a contradiction.

First replace each implication:

Now negate the conjunction:

Use De Morgan's law:

Negate each disjunction:

This is an equivalent formula using only $\neg$, $\land$, and $\lor$.

For $\forall x\, \exists y\, P(x,y)$, every $x$ can choose itself as $y$. So this statement is true.

For $\exists y\, \forall x\, P(x,y)$, we would need one single $y$ that equals both $1$ and $2$. That is impossible, so this statement is false.

Thus the first statement is true and the second is false.

Let $x$ be arbitrary. Then

means

That is equivalent to saying $x$ is not in both $A$ and $B$, so

This is the same as

which means

Since the two sides contain exactly the same elements, the sets are equal.

No, the set is inconsistent.

From $\exists x\, P(x)$, choose a witness $c$ such that $P(c)$ is true.

From $\forall x(P(x) \to Q(x))$, we get $P(c) \to Q(c)$, so $Q(c)$ is true.

But $\forall x\, \neg Q(x)$ gives $\neg Q(c)$.

So we obtain both $Q(c)$ and $\neg Q(c)$, which is impossible. Therefore the set is not consistent.