Without using truth table prove that: p ∧ q ∨ ~ p ∧ q ∨ p ∧ ~ q ≡ p ∨ q

Use algebra of statements to prove that p ∧ q ∨ ~ p ∧ q ∨ p ∧ ~ q ≡ p ∨ q . So take LHS of the statement, p ∧ q ∨ ~ p ∧ q ∨ p ∧ ~ q ≡ p ∨ ~ p ∧ q ∨ p ∧ ~ q (Distributive Law) ≡ T ∧ q ∨ p ∧ ~ q (Complement Law) ≡ q ∨ p ∧ ~ q (Identity law) ≡ q ∨ p ∧ q ∨ ~ q (Distributive Law) ≡ q ∨ p ∧ T (Complement Law) ≡ p ∨ q (Identity Law) Hence Proved.

Without using truth table prove that: p ∧ q ∨ ~ p ∧ q ∨ p ∧ ~ q ≡ p ∨ q

Use algebra of statements to prove that p ∧ q ∨ ~ p ∧ q ∨ p ∧ ~ q ≡ p ∨ q . So take LHS of the statement, p ∧ q ∨ ~ p ∧ q ∨ p ∧ ~ q ≡ p ∨ ~ p ∧ q ∨ p ∧ ~ q (Distributive Law) ≡ T ∧ q ∨ p ∧ ~ q (Complement Law) ≡ q ∨ p ∧ ~ q (Identity law) ≡ q ∨ p ∧ q ∨ ~ q (Distributive Law) ≡ q ∨ p ∧ T (Complement Law) ≡ p ∨ q (Identity Law) Hence Proved.

Without using truth table prove that: ~ p ∨ ~ q → p ∧ ~ q ≡ p ∨ ~ q ∧ ~ p ∨ q

Use algebra of statements to prove that ~ p ∨ ~ q → p ∧ ~ q ≡ p ∨ ~ q ∧ ~ p ∨ q . So take LHS of the statement, Recall that ~ p → q ≡ p ∧ ~ q ~ p ∨ ~ q → p ∧ ~ q ≡ p ∨ ~ q ∧ ~ p ∧ ~ q (Conditional Law) ≡ p ∨ ~ q ∧ ~ p ∨ ~ ~ q (De Morgans Law) ≡ p ∨ ~ q ∧ ~ p ∨ q Hence Proved.