Pls Help. A math question stuck me for days.

Question

Prove that (p if and only if p and q) are logically equivalent to (not p or q) using laws of logic.

Sentence1: Ray goes to work if and only if Ray goes to work and John goes to school.

Sentence2: Ray does not go to work or John goes to school.

Write both sentences as logical expressions. Then, by simplifying one expression to the other one using laws of logic (Cannot use a Truth Table), prove both expressions are logically equivalent.

Answer 1

Notation of logical operations (could vary in different textbooks):

negation (not): ∼

conjunction (and): ∧

disjunction (or): ∨

implication (if... then...): →

equivalency (if and only if): ↔

Priority of operations: ∼, ∧, ∨, →, ↔

Notation of logical equivalency: ≡

Sentence 1 as logical expression: p ↔ (p ∧ q)

Sentence 2 as logical expression: ∼p ∨ q

We need to prove that p ↔ (p ∧ q) ≡ ∼p ∨ q

Laws of logical equivalency that we will use.

Commutative law: a ∨ b ≡ b ∨ a

Associative law: (a ∨ b) ∨ c ≡ a ∨ (b ∨ c)

Distributive law: a ∨ (b ∧ c) ≡ (a ∨ b) ∧ (a ∨ c)

De Morgan law: ∼(a ∧ b) ≡ ∼a ∨ ∼b

Equivalency law: a ↔ b ≡ (a → b) ∧ (b → a)

Implication as a disjunction: a → b ≡ ∼a ∨ b

Inverse law: a ∨ ∼a ≡ T

Identity law: a ∧ T ≡ a

Domination law: a ∨ T ≡ T

Let work with the expression for the Sentence 1:

p ↔ (p ∧ q) equivalency law

≡ (p → (p ∧ q)) ∧ ((p ∧ q) → p)

Work with p → (p ∧ q) implication as a disjunction

≡ ∼p ∨ (p ∧ q) distributive law

≡ (∼p ∨ p) ∧ (∼p ∨ q) inverse law

≡ T ∧ (∼p ∨ q) identity law

≡ ∼p ∨ q

Work with (p ∧ q) → p implication as a disjunction

≡ ∼(p ∧ q) ∨ p De Morgan law

≡ (∼p ∨ ∼q) ∨ p commutative and associative laws

≡ (∼p ∨ p) ∨ ∼q inverse law

≡ T ∨ ∼q domination law

≡ T

So, p ↔ (p ∧ q) ≡ (p → (p ∧ q)) ∧ ((p ∧ q) → p) ≡ (∼p ∨ q) ∧ T ≡ ∼p ∨ q

QED