Let's create a truth table for the expression (p → q) ∧ ¬p

Method:

1. List all possible combinations of truth values for the variables.
2. Build the truth table column by column, working from the simplest components.
3. Use the previously computed columns to determine the values for more complex expressions.

Example: Simplify ¬(p ∧ (q ∨ ¬r)) using logical equivalences.

Solution:

1. ¬(p ∧ (q ∨ ¬r))
2. Apply De Morgan's Law: ¬p ∨ ¬(q ∨ ¬r)
3. Apply De Morgan's Law again: ¬p ∨ (¬q ∧ ¬(¬r))
4. Apply Double Negation: ¬p ∨ (¬q ∧ r)
5. The simplified expression is: ¬p ∨ (¬q ∧ r)

Example: Determine if the following argument is valid:

Premises:

• If I study hard, I will pass the exam. (p → q)
• I did not pass the exam. (¬q)

Conclusion: I did not study hard. (¬p)

Analysis:

Let's check if the conclusion logically follows from the premises:

1. p → q (Premise 1)
2. ¬q (Premise 2)
3. We want to determine if ¬p follows logically

This is an example of Modus Tollens (p → q, ¬q ⊢ ¬p), which is a valid argument form.

Therefore, the argument is valid.

For A = {a, b} and B = {1, 2, 3}:

A × B = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}

B × A = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}

Note that A × B ≠ B × A unless A = B.

|A × B| = |A| × |B| = 2 × 3 = 6

Example: Prove that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

We need to show that every element in the left-hand side is also in the right-hand side, and vice versa.

Part 1: Show x ∈ A ∪ (B ∩ C) → x ∈ (A ∪ B) ∩ (A ∪ C)

If x ∈ A ∪ (B ∩ C), then either x ∈ A or x ∈ B ∩ C.

• If x ∈ A, then x ∈ A ∪ B and x ∈ A ∪ C, so x ∈ (A ∪ B) ∩ (A ∪ C).
• If x ∈ B ∩ C, then x ∈ B and x ∈ C, which means x ∈ A ∪ B and x ∈ A ∪ C, so x ∈ (A ∪ B) ∩ (A ∪ C).

Part 2: Show x ∈ (A ∪ B) ∩ (A ∪ C) → x ∈ A ∪ (B ∩ C)

If x ∈ (A ∪ B) ∩ (A ∪ C), then x ∈ A ∪ B and x ∈ A ∪ C.

There are three cases to consider:

• If x ∈ A, then x ∈ A ∪ (B ∩ C).
• If x ∉ A but x ∈ B and x ∈ C, then x ∈ B ∩ C, so x ∈ A ∪ (B ∩ C).
• If x ∉ A and either x ∉ B or x ∉ C, then x cannot be in both A ∪ B and A ∪ C, which contradicts our assumption.

Therefore, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

For three sets, the Venn diagram becomes more complex with 8 distinct regions:

1. Elements in A only
2. Elements in B only
3. Elements in C only
4. Elements in A and B, but not C
5. Elements in A and C, but not B
6. Elements in B and C, but not A
7. Elements in A, B, and C
8. Elements in none of A, B, or C (but in U)

This allows us to visualize complex set operations like (A ∩ B) ∪ (B ∩ C) ∪ (A ∩ C).

Example: Let's examine the "less than or equal to" relation (≤) on the set of integers:

• Reflexive? Yes, because a ≤ a for all integers a.
• Symmetric? No, because 3 ≤ 5 but 5 ≰ 3.
• Antisymmetric? Yes, because if a ≤ b and b ≤ a, then a = b.
• Transitive? Yes, because if a ≤ b and b ≤ c, then a ≤ c.

Function Composition

Given functions f: A → B and g: B → C, the composition of g and f, denoted g ∘ f: A → C, is defined as:

(g ∘ f)(a) = g(f(a)) for all a ∈ A

Example:

Let f(x) = x² and g(x) = x + 3

(g ∘ f)(x) = g(f(x)) = g(x²) = x² + 3

(f ∘ g)(x) = f(g(x)) = f(x + 3) = (x + 3)²

Note that function composition is not commutative: g ∘ f ≠ f ∘ g in general.

Inverse Functions

If f: A → B is a bijective function, then there exists an inverse function f⁻¹: B → A such that:

• f⁻¹(f(a)) = a for all a ∈ A
• f(f⁻¹(b)) = b for all b ∈ B

Example:

Let f(x) = 2x + 3

To find f⁻¹, solve for x:

y = 2x + 3

x = (y - 3)/2

Therefore, f⁻¹(y) = (y - 3)/2

We can verify that:

f⁻¹(f(x)) = f⁻¹(2x + 3) = ((2x + 3) - 3)/2 = x

f(f⁻¹(y)) = f((y - 3)/2) = 2((y - 3)/2) + 3 = y - 3 + 3 = y