Show that p → q and q → p are not equivalent.

Truth Table for p → q & q → p p q p → q q → p T T T T T F F T F T T F F F T T Truth tables and all the interpretations of the statements p → q & q → p are not same Therefore p → q and q → p are not equivalent Hence proved.

Prove that a - 1 - 1 = a for every a ∈ G , group.

Given that G is a group ∀ a ∈ G ⇒ a - 1 ∈ G ⇒ a - 1 - 1 ∈ G By definition we have a * a - 1 = a - 1 * a = e and a - 1 * a - 1 - 1 = a - 1 - 1 * a - 1 = e where e is an identity element of G Consider a * a - 1 = e Post operating a - 1 - 1 on both sides we get a * a - 1 * a - 1 - 1 = e * a - 1 - 1 = a - 1 - 1 ⇒ a * a - 1 * a - 1 - 1 = a - 1 - 1 ∵ * is associative ⇒ a * e = a - 1 - 1 ⇒ a = a - 1 - 1 Hence, proved.

Prove that a - 1 - 1 = a for every a ∈ G , group.

Given that G is a group ∀ a ∈ G ⇒ a - 1 ∈ G ⇒ a - 1 - 1 ∈ G By definition we have a * a - 1 = a - 1 * a = e and a - 1 * a - 1 - 1 = a - 1 - 1 * a - 1 = e where e is an identity element of G Consider a * a - 1 = e Post operating a - 1 - 1 on both sides we get a * a - 1 * a - 1 - 1 = e * a - 1 - 1 = a - 1 - 1 ⇒ a * a - 1 * a - 1 - 1 = a - 1 - 1 ∵ * is associative ⇒ a * e = a - 1 - 1 ⇒ a = a - 1 - 1 Hence, proved.

Show that p → q and q → p are not equivalent.

Truth Table for p → q & q → p p q p → q q → p T T T T T F F T F T T F F F T T Truth tables and all the interpretations of the statements p → q & q → p are not same Therefore p → q and q → p are not equivalent Hence proved.

Show that ( p ∧ q ) → ( p ∨ q ) is a tautology.

Given; To prove ( p ∧ q → ( p ∨ q ) ) is tautology Formulating the table p q p&and;q p&or;q (p&and;q)→(p&or;q) T T T T T T F F T T F T F T T F F F F T ∵ All true ∴ Tautology Hence, proved.