Sets and logic: worked exercises

End of September 2026: in Year 1 of the liceo classico (the Italian high school focused on Latin and Ancient Greek), sets and logic come into the syllabus now, right after the September-October block on numbers, GCD and LCM, powers, fractions and proportions. With 3 hours of maths a week the topic passes quickly — a week or two, no more — and the test that contains it arrives between mid-October and mid-November, so four to seven weeks from today. The same block also appears in September-October at the liceo scientifico (science-oriented) and, moved to November-December, at the liceo linguistico (languages): if you are at one of these two schools, the part on sets and logic below still applies, only the timing of the test changes.

It is the part of the syllabus that looks like a gift — no sums, no fractions with a sign to check twice — and precisely for that reason it gets studied less than the others. The result is predictable: in the test you lose one or two marks on definitions that, read calmly, take twenty minutes. Here we look at them with the space they deserve: not a box inside a broader post, but the whole topic, with three fully worked exercises.

Sets: the representations

A set is a collection of objects — its elements — identified without ambiguity: given an object, it must always be possible to say whether it belongs to the set or not. The same set can be written in three ways, and they are equivalent: if an exercise asks for a specific form, that is the one you must use.

By listing (tabular form): all the elements are listed between curly brackets.

A = {2, 4, 6, 8}

By defining property: you state the rule an object must satisfy to belong to the set.

A = {x ∣ x is even and 0 < x < 10}

With the Euler-Venn diagram: a closed line containing the element points. It is the most intuitive representation, but on its own it is not enough in a written test: it must always be accompanied by one of the other two.

A point that is often confused: membership and inclusion are not the same relation. An element belongs to a set, $2 \in A$ ; a set is contained in another, that is, it is a subset of it, $B \subseteq A$ . Writing $2 \subseteq A$ or $B \in A$ when $B$ is a set of elements of $A$ is a symbol error, not a reasoning error — but in a test it counts as a full error.

Subsets and the power set

$B$ is a subset of $A$ (written $B \subseteq A$ ) if every element of $B$ is also an element of $A$ . If in addition $B \neq = A$ , it is called a proper subset and written $B \subset A$ . Two cases to remember always, because they often come back as a theory question:

the empty set $\emptyset$ is a subset of every set, including itself;
every set is a subset of itself, $A \subseteq A$ .

The power set of $A$ , written $P (A)$ , is the set of all subsets of $A$ , including $\emptyset$ and $A$ itself. If $A$ has $n$ elements, $P (A)$ has exactly

2^{n}

For example, if $A = {a, b, c}$ (three elements), $P (A)$ has $2^{3} = 8$ elements:

P (A) = {\emptyset, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}

The most common mistake here is to forget $\emptyset$ or $A$ itself in the list: both are subsets in every respect, even if they do not “look like” real subsets.

Operations between sets

Given two sets $A$ and $B$ , inside an ambient set $U$ (the universal set):

Union $A \cup B$ : the elements that are in $A$ or in $B$ (or in both).
Intersection $A \cap B$ : the elements that are in both $A$ and $B$ .
Difference $A ∖ B$ : the elements of $A$ that are not in $B$ .
Complement $A^{c}$ (or $\overline{A}$ ): the elements of $U$ that are not in $A$ , that is $U ∖ A$ .

If $A \cap B = \emptyset$ , the two sets are called disjoint. When a set $U$ is divided into non-empty subsets, pairwise disjoint, whose union rebuilds $U$ in full, we speak of a partition of $U$ : it is the typical situation of a class divided into bands, or of a questionnaire with mutually exclusive answers.

The Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ with $a \in A$ and $b \in B$ . “Ordered” is the key word: $(a, b)$ and $(b, a)$ are different pairs, unless $a = b$ . If $A$ has $m$ elements and $B$ has $n$ , $A \times B$ has $m \cdot n$ .

Logic: propositions and connectives

A logical proposition is a statement to which one can assign with certainty a truth value, true (T) or false (F), and only one. “Rome is the capital of Italy” is a proposition (T); “what time is it?” is not, because it is neither true nor false.

Connectives combine propositions into other propositions:

Connective	Symbol	Read as	True when
Negation	$\neg p$	not $p$	$p$ is false
Conjunction	$p \land q$	$p$ and $q$	both true
Inclusive disjunction	$p \lor q$	$p$ or $q$	at least one true
Implication	$p \Rightarrow q$	if $p$ then $q$	false only if $p$ true and $q$ false
Biconditional	$p \Leftrightarrow q$	$p$ if and only if $q$	same truth value

The point on which almost everyone slips: the “or” of everyday language is often exclusive (“coffee or tea” means one of the two, not both), whereas the “or” of logic, $\lor$ , is inclusive by default: $p \lor q$ is true even when both are true. If an exercise wants the exclusive “or” it says so explicitly; otherwise $\lor$ always admits the “both true” case.

Truth tables and De Morgan’s laws

The truth table lists all the possible combinations of the values of $p$ and $q$ and the resulting value of the compound proposition. For $p \land q$ and $p \lor q$ :

$p$	$q$	$p \land q$	$p \lor q$
T	T	T	T
T	F	F	T
F	T	F	T
F	F	F	F

De Morgan’s laws link negation to the two connectives, and it is the same pattern you see in the complement operation between sets:

\neg (p \land q) = \neg p \lor \neg q

\neg (p \lor q) = \neg p \land \neg q

In words: negating ” $p$ and $q$ ” does not give “not $p$ and not $q$ ”, it gives “not $p$ or not $q$ ” — the connective changes when you negate. It is exactly the same reason why, between sets, $\overline{A \cap B} = \overline{A} \cup \overline{B}$ : the two laws are the same idea written in two different languages, and connecting them is what distinguishes someone who has understood from someone who has learned a formula by heart.

Quantifiers and their negation

The universal quantifier $\forall$ is read “for all”; the existential quantifier $\exists$ is read “there exists at least one”. Negating them swaps one for the other:

\neg (\forall x, p (x)) = \exists x, \neg p (x)

\neg (\exists x, p (x)) = \forall x, \neg p (x)

Translated without symbols: the negation of “all the students in the class have the book” is not “no student has the book” — it is “there is at least one student who does not have the book”. Confusing the negation of “all” with “none” is the most frequent mistake on this topic, and it happens because in spoken language the two are often used as if they were equivalent: in logic they are not.

Three worked exercises

1. Sets with a Venn diagram and counting

In a class of 25 students, 15 take the Greek course and 12 take the Latin course as an optional afternoon subject; 6 students take both courses. How many students take neither course?

Let $G$ be the set of those who take Greek and $L$ the set of those who take Latin. We know:

∣ G ∣ = 15 ∣ L ∣ = 12 ∣ G \cap L ∣ = 6

To count how many students take at least one of the two courses, you do not simply add 15 and 12: those who take both would be counted twice. You use the union formula:

∣ G \cup L ∣ = ∣ G ∣ + ∣ L ∣ - ∣ G \cap L ∣ = 15 + 12 - 6 = 21

The students who take neither course are those of the class minus those in the union:

25 - 21 = 4

4 students take neither Greek nor Latin. The step never to skip is the subtraction of $∣ G \cap L ∣$ : that is where you go wrong when you try to “add by eye” without going through the formula, and it is exactly the same pattern as, when drawing the Venn diagram, writing first the 6 of the intersection, then $15 - 6 = 9$ in $G$ alone, then $12 - 6 = 6$ in $L$ alone, and finally $9 + 6 + 6 = 21$ inside the two circles.

2. Truth table of a compound proposition

Build the truth table of $(p \lor q) \land \neg q$ .

You proceed column by column, from the innermost. First $\neg q$ , then $p \lor q$ , finally the end result as the conjunction of the two intermediate columns:

$p$	$q$	$p \lor q$	$\neg q$	$(p \lor q) \land \neg q$
T	T	T	F	F
T	F	T	T	T
F	T	T	F	F
F	F	F	T	F

The proposition turns out to be true in only one case, when $p$ is true and $q$ is false. It is worth noting: $(p \lor q) \land \neg q$ is logically equivalent to $p \land \neg q$ — the final column coincides with the one you would get by calculating $p \land \neg q$ directly. Verifying it by building that table too is a good checking exercise, because two logically equivalent propositions always have the same column of values.

3. Negation of a proposition with quantifiers

Negate the proposition: “All the students in the class handed in the assignment and nobody arrived late.”

The proposition, in symbols, is a conjunction of two universal statements:

(\forall x, submitted (x)) \land (\forall x, \neg late (x))

To negate it you first apply De Morgan to the conjunction — negating " $P \land Q$ " gives " $\neg P \lor \neg Q$ " — then you negate each universal quantifier, turning it into an existential one:

\neg [(\forall x, submitted (x)) \land (\forall x, \neg late (x))] = (\exists x, \neg submitted (x)) \lor (\exists x, late (x))

Translated: “there is at least one student who did not hand in the assignment, or there is at least one student who arrived late.” Not “nobody handed it in” and not “everyone arrived late” — the negation of “all” always remains “there exists at least one who does not”, on each of the two pieces separately.

What to do about a practice test

We do not yet have a dedicated practice test on sets and logic: it is a topic that lends itself more to applied theory questions than to a closed multiple-choice test, and we prefer to build it properly rather than publish it incomplete. What there is, and covers the rest of the same test, is the simulated test on numbers, fractions and powers: 15 questions in 50 minutes on the September-October block — GCD, LCM, powers, fractions, proportions — that is, everything in your test that is not sets or logic. The correct answer to every question is free; the guided step-by-step solution is unlocked with the first lesson.

If you want the part on the rest of the first test — the typical mistakes with signs, powers and order of operations — I have written it in full in the first maths test in Year 1 of high school, where sets and logic are only a box; here, instead, they are the whole topic. If you are at the liceo artistico (art high school), the October syllabus is similar but with different timetables: numbers, fractions and powers in Year 1 of the art high school. And if you want a review of the whole year in a single document, there is the Year 1 maths formula sheet. The full syllabus for maths tutoring is here, year by year. Where the “if… then” structure becomes the hypothesis and conclusion of a geometry proof is in triangle congruence criteria: the proofs.

If after reading this guide sets are still confusing, or it is propositional logic that does not click, the free introductory call is for exactly this: you tell me what you have done in class and I tell you honestly where to concentrate the first lessons before the test arrives.

Sets and logic in Year 1 of high school: worked exercises