#### Question

Without using truth table show that P↔q ≡(P ∧ q) ∨ (~ p ∧ ~ q)

#### Solution

`L.H.S= pq-=(p ->q) ^^ (q->p)`

`-=(~p vv q) ^^ (~q vv p)`

`-=[~p^^(~qvvp)]vv[q^^(~qvvp)] " by distributive law" `

`-=[(~p^^~q)vv(~p^^p)]vv[(q^^~q)vv(q^^p)] " by distributive law"`

`-=[(~p^^q)vvF]vv[Fvv(q^^p)] "by complement law" `

`-=(~p^^~q)vv(q^^p) " by identity law"`

`-=(p^^q)vv(~p^^~q) " by commutative law"`

=R.H.S

Is there an error in this question or solution?

#### APPEARS IN

Solution Without using truth table show that P↔q ≡(P ∧ q) ∨ (~ p ∧ ~ q) Concept: Mathematical Logic - Algebra of Statements.