Show that (p↔q)≡(p∧q)∨(¬p∧¬q) using equivalence laws.

Question

Can you give the answer step by step? Need answer ASAP in one hour

Answer 1

I am assuming we are working in a boolean algebra. If you don't know what this means then the answer is almost certainly yes.

first use the definition of ↔

(p ↔ q) ≡ (p → q) ∧ (q → p)

next use the definition of → on each of the factors

(p ↔ q) ≡ (¬p ∨ q) ∧ (¬q ∨ p)

next use the distributive property, treating (¬q ∨ p) as one big term

(p ↔ q) ≡ (¬p ∧ (¬q ∨ p)) ∨ (q ∧ (¬q ∨ p))

next use the distributive property inside each set of big parentheses

(p ↔ q) ≡ ((¬p ∧ ¬q) ∨ (¬p ∧ p)) ∨ ((q ∧ ¬q) ∨ (q ∧ p))

(notice how this looks like foil with polynomials. that actually works here)

since (¬p ∧ p) as well as (q ∧ ¬q) are both false, we have

(p ↔ q) ≡ ((¬p ∧ ¬q) ∨ false) ∨ (false ∨ (q ∧ p))

it is always true that a ∨ false ≡ false ∨ a ≡ a.

(p ↔ q) ≡ (¬p ∧ ¬q) ∨ (q ∧ p)

by commutativity, we have

(p ↔ q) ≡ (p ∧ q) ∨ (¬p ∧ ¬q)