Selected items from the Introduction to set theory and to methodology and philosophy of mathematics and computer programming.

1. Introduction to set theory and to methodology and philosophy of
mathematics and computer programming
Properties of union, intersection and complement
An overview
by Jan Plaza
c 2017 Jan Plaza
Use under the Creative Commons Attribution 4.0 International License
Version of February 5, 2017

2. Proposition. Let X, Y , Z ⊆ U. Then:
1. (Identity Laws)
∅ ∪ X = X
U ∩ X = X
2. (Dominance Laws or Annihilation Laws)
∅ ∩ X = ∅
U ∪ X = U
3. (Idempotence Laws)
X ∪ X = X
X ∩ X = X
4. (Associativity Laws)
(X ∪ Y ) ∪ Z = X ∪ (Y ∪ Z)
(X ∩ Y ) ∩ Z = X ∩ (Y ∩ Z)
5. (Commutativity Laws)
X ∪ Y = Y ∪ X
X ∩ Y = Y ∩ X

3. Proposition. Let X, Y , Z ⊆ U. Then:
1. (Identity Laws)
∅ ∪ X = X analogous to 0 + x = x
U ∩ X = X analogous to 1 · x = x
2. (Dominance Laws or Annihilation Laws)
∅ ∩ X = ∅ analogous to 0 · x = 0
U ∪ X = U
3. (Idempotence Laws)
X ∪ X = X
X ∩ X = X
4. (Associativity Laws)
(X ∪ Y ) ∪ Z = X ∪ (Y ∪ Z) analogous to (x + y) + z = x + (y + z)
(X ∩ Y ) ∩ Z = X ∩ (Y ∩ Z) analogous to (x · y) · z = x · (y · z)
5. (Commutativity Laws)
X ∪ Y = Y ∪ X analogous to x + y = y + x
X ∩ Y = Y ∩ X analogous to x · y = y · x

4. 6. (Distributivity Laws)
X ∩ (Y ∪ Z) = (X ∩ Y ) ∪ (X ∩ Z)
X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z)
7. (Absorption Laws)
X ∩ (X ∪ Y ) = X
X ∪ (X ∩ Y ) = X
8. (De Morgan Laws)
(X ∪ Y )c
= Xc ∩ Y c
(X ∩ Y )c
= Xc ∪ Y c
9. (Complement Laws)
X ∪ Xc = U
X ∩ Xc = ∅
10. (Double Complement Law)
(Xc)c
= X

5. 6. (Distributivity Laws)
X ∩ (Y ∪ Z) = (X ∩ Y ) ∪ (X ∩ Z) analogous to x · (y + z) = x · y + x · z
X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z)
7. (Absorption Laws)
X ∩ (X ∪ Y ) = X
X ∪ (X ∩ Y ) = X
8. (De Morgan Laws)
(X ∪ Y )c
= Xc ∩ Y c
(X ∩ Y )c
= Xc ∪ Y c
9. (Complement Laws)
X ∪ Xc = U
X ∩ Xc = ∅
10. (Double Complement Law)
(Xc)c
= X analogous to −(−x) = x

6. We will prove the distributivity of union with respect to intersection:
X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z).
To prepare for the proof, recall that,
1. z ∈ X ∪ Y iff z ∈ X or z ∈ Y ,
2. z ∈ X ∩ Y iff z ∈ X and z ∈ Y ,

7. Proof of X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z).

8. Proof of X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z).
By the Axiom of Extensionality it is enough to prove that:
for every u, u ∈ X ∪ (Y ∩ Z) is equivalent to

9. Proof of X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z).
By the Axiom of Extensionality it is enough to prove that:
for every u, u ∈ X ∪ (Y ∩ Z) is equivalent to u ∈ (X ∪ Y ) ∩ (X ∪ Z).

10. Proof of X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z).
By the Axiom of Extensionality it is enough to prove that:
for every u, u ∈ X ∪ (Y ∩ Z) is equivalent to u ∈ (X ∪ Y ) ∩ (X ∪ Z).
In order to prove it, consider an arbitrary u.

11. Proof of X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z).
By the Axiom of Extensionality it is enough to prove that:
for every u, u ∈ X ∪ (Y ∩ Z) is equivalent to u ∈ (X ∪ Y ) ∩ (X ∪ Z).
In order to prove it, consider an arbitrary u.
We have the following equivalences.
u ∈ X ∪ (Y ∩ Z)
iff
iff
iff
u ∈ (X ∪ Y ) ∩ (X ∪ Z).

12. Proof of X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z).
By the Axiom of Extensionality it is enough to prove that:
for every u, u ∈ X ∪ (Y ∩ Z) is equivalent to u ∈ (X ∪ Y ) ∩ (X ∪ Z).
In order to prove it, consider an arbitrary u.
We have the following equivalences.
u ∈ X ∪ (Y ∩ Z)
iff
u ∈ X ∨ u ∈ (Y ∩ Z)
iff
iff
u ∈ (X ∪ Y ) ∩ (X ∪ Z).

13. Proof of X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z).
By the Axiom of Extensionality it is enough to prove that:
for every u, u ∈ X ∪ (Y ∩ Z) is equivalent to u ∈ (X ∪ Y ) ∩ (X ∪ Z).
In order to prove it, consider an arbitrary u.
We have the following equivalences.
u ∈ X ∪ (Y ∩ Z)
iff
u ∈ X ∨ u ∈ (Y ∩ Z)
iff
u ∈ X ∨ (u ∈ Y ∧ u ∈ Z)
iff
u ∈ (X ∪ Y ) ∩ (X ∪ Z).

14. Proof of X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z).
By the Axiom of Extensionality it is enough to prove that:
for every u, u ∈ X ∪ (Y ∩ Z) is equivalent to u ∈ (X ∪ Y ) ∩ (X ∪ Z).
In order to prove it, consider an arbitrary u.
We have the following equivalences.
u ∈ X ∪ (Y ∩ Z)
iff
u ∈ X ∨ u ∈ (Y ∩ Z)
iff
u ∈ X ∨ (u ∈ Y ∧ u ∈ Z)
iff
(u ∈ X ∪ Y ) ∧ (u ∈ X ∪ Z)
iff
u ∈ (X ∪ Y ) ∩ (X ∪ Z).

15. Proof of X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z).
By the Axiom of Extensionality it is enough to prove that:
for every u, u ∈ X ∪ (Y ∩ Z) is equivalent to u ∈ (X ∪ Y ) ∩ (X ∪ Z).
In order to prove it, consider an arbitrary u.
We have the following equivalences.
u ∈ X ∪ (Y ∩ Z)
iff
u ∈ X ∨ u ∈ (Y ∩ Z)
iff
u ∈ X ∨ (u ∈ Y ∧ u ∈ Z)
(u ∈ X ∨ u ∈ Y ) ∧ (u ∈ X ∨ u ∈ Z)
iff
(u ∈ X ∪ Y ) ∧ (u ∈ X ∪ Z)
iff
u ∈ (X ∪ Y ) ∩ (X ∪ Z).

16. Proof of X ∪ (Y ∩ Z) = (X ∪ Y ) ∩ (X ∪ Z).
By the Axiom of Extensionality it is enough to prove that:
for every u, u ∈ X ∪ (Y ∩ Z) is equivalent to u ∈ (X ∪ Y ) ∩ (X ∪ Z).
In order to prove it, consider an arbitrary u.
We have the following equivalences.
u ∈ X ∪ (Y ∩ Z)
iff
u ∈ X ∨ u ∈ (Y ∩ Z)
iff
u ∈ X ∨ (u ∈ Y ∧ u ∈ Z)
iff (because x ∨ (y ∧ z) ↔ (x ∨ y) ∧ (x ∨ z) is a tautology)
(u ∈ X ∨ u ∈ Y ) ∧ (u ∈ X ∨ u ∈ Z)
iff
(u ∈ X ∪ Y ) ∧ (u ∈ X ∪ Z)
iff
u ∈ (X ∪ Y ) ∩ (X ∪ Z).

17. Exercise: To complete the proof above, verify that x ∨ (y ∧ z) ↔ (x ∨ y) ∧ (x ∨ z) is a
tautology.
Exercise: Prove the remaining points of the Proposition above.

18. Proposition. Uc = ∅
Proof. Version 1 - A proof by reduction to the propositional calculus.
By exensionality, it is enough to take an arbitrary x ∈ U and show that x ∈ Uc iff
x ∈ ∅.
Notice that:
x ∈ Uc
iff
x ∈ U
iff
⊥
iff
x ∈ ∅.

19. Proposition. Uc = ∅
Proof. Version 2 - by rewriting expressions.
Uc
= (by absorption)
Uc ∩ (Uc ∪ U)
= (by commutativity)
Uc ∩ (U ∪ Uc)
= (by commutativity)
(U ∪ Uc) ∩ Uc
= (by complement laws)
U ∩ Uc
= (by complement laws)
∅.

20. Exercise
Prove ∅c = U
Exercise
Prove (X ∪ Y ) ∩ Xc = (X ∪ Y c)c
.
Proofs by reduction for propositional calculus are easier.