P logically implies Q is written as P ⇒ Q, which means any of the following

1. Every assignment of truth values to the atomic proposition p, q, r that makes P T also makes Q T
2. P → Q is always T, regardless of the truth values assigned to the atomic propositions
3. P → Q is a tautology
4. From a truth table

p	q	r	.....	P	Q	P→Q
T	T	T		F	T	T
T	T	F		T	T	T
T	F	T		F	F	T
T	F	F		F	F	T
F	T	T		F	F	T
F	T	F		T	T	T
F	F	F		T	T	T

P is logically equivalent to Q written P ≡ Q ∨ P <=⇒ Q means P ⇒ Q ∧ Q ⇒ P

1. Every assignment of truth values on atomic proposition p, q, r, ... makes both P ∧ Q T ∨ F
2. P <-→ Q is a tautology
3. P & Q Truth table is identical

De Margon's Law

1) LHS = ¬(p ∧ q) ≡ (¬ p) ∨ (¬ q) = RHS

p	q	¬ p	¬ q	p ∧ q	¬(p ∧ q)	(¬ p) ∨ (¬ q)	LHS <→ RHS
T	T	F	F	T	F	F	T
T	F	F	T	F	T	T	T
F	T	T	T	F	T	T	T
F	F	T	T	F	T	T	T

																Columns Are Identical

2. ¬ (p ∨ q) ≡ (¬ p) ∧ (¬ q)

"Direct Proof of LHS ⇒ RHS"

Assume LHS: ¬(p ∨ q) is T then p ∨ q is F, meaning both p is F ∧ so is q. Then both ¬ q is T ∧ ¬ p T. Theref∨e, ¬ p ∧ ¬ q is T. Hence RHS is T.

So, LHS ⇒ RHS

Distributive Laws

$p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)$ $p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)$

Algebraically: $a·(b+c) = ab + ac$ $a + (b·c) != (a+b)(a+c)$

Direct Proof: $p ∧ (¬ q ∨ r) ⇒ (p ∧ q) ∨ (p ∧ r)$

• Assume LHS is T
• Therefore both p is T ∧ q ∨ r is T
• Then either q is T ∨ r is T
• In case (1), p ∧ q is T which means (p ∧ q) ∨ (p ∧ r) is T so RHS is T
• In case (2), p ∧ r is T which means (p ∧ r) is T so RHS is T

So, LHS ⇒ RHS