### Negation

Truth Table

A B A $$\land$$ B A $$\lor$$ B
True True True True
True False False True
False True False True
False False False False

The $$\land$$ acts as “AND” and it is False when there is at least one False, and True otherwise.
The $$\lor$$ acts as “OR” and it is True when there is at least one True, and False otherwise.

$$\lnot$$ denotes a “NOT” sign

A B $$\lnot$$A
True True False
True False False
False True True
False False True

## De Morgan’s Laws:

### Law 1:

$$\lnot$$ ($$\lnot$$ A) = A

### Law 2:

($$\lnot$$A)∨($$\lnot$$B) ≡ $$\lnot$$(A ∧ B)
($$\lnot$$A)∧($$\lnot$$B) ≡ $$\lnot$$(A ∨ B)

### Law 3:

(A ∧ B) ∧ C ≡ A ∧ (B ∧ C)
(A ∨ B) ∨ C ≡ A ∨ (B ∨ C)
associativity law

Proving Law 2

Start out with: $$\lnot$$(($$\lnot$$A)∧($$\lnot$$B))
$$\lnot$$(($$\lnot$$A)∧($$\lnot$$B)) ≡ ($$\lnot$$($$\lnot$$A))∨($$\lnot$$($$\lnot$$B))
The ($$\lnot$$A)∧($$\lnot$$B) undergoes negation
≡ (A) ∨ (B)
This yields just (A) ∨ (B)

Knowing Law 1
($$\lnot$$A)∧($$\lnot$$B) ≡ $$\lnot$$($$\lnot$$($$\lnot$$A)∧($$\lnot$$B))) ≡ $$\lnot$$(A ∨ B)

By negating (A) ∨ (B), we can get: ($$\lnot$$A)∧($$\lnot$$B) ≡ $$\lnot$$(A ∨ B)

## Implication

Hypothesis: If … then

Conclusion: then…

A B A $$\Rightarrow$$ B
True True True
True False False
False True True
False False True

If A, then B

Think of this like “If you give me $100, I will give you$200 tomorrow.” The implication is similar to if I keep my promise in this situation.

If the hypothesis is correct (I gave you $100 to begin with), then the implication is true if I give you back$200 tomorrow, or false I accept the money and do not give you back the money.

The hypothesis is false if you do not give me $100. I may give you$200 regardless tomorrow, making the implication true.

The hypothesis can also be false if you do not give me money, and I give you no money. In this case, the implication is also true.

Here are four implications. Only one is false. Identify the false implication.

1. If x is a real number and $$x^2$$ < 0, then $$x^4$$ ≥ 0.
2. If x is a real number and $$x^2$$ < 0 then $$x^4$$ < 0.
3. If n is a positive integer, then $$n^2$$ +13 is not a perfect square.
4. If n is an integer and $$2^{2n}$$ is an odd integer, then $$2^{−2n}$$ is an odd integer.

Only the implication “If n is a positive integer, then $$n^2$$ + 13 is not a perfect square.” is false.