Question

Without using truth table prove that ( p V q) Ʌ (p V ~q) = p​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: (p\lor q)\land (p\lor \sim q)$

We know,

$\boxed{ \rm{ \:(p\lor q)\land (p\lor r) \: = \: p\lor (q\land r) \: }}$

So, using this distributive property, we have

$\rm \: = \: p\lor (q\land \sim q)$

We know,

$\boxed{ \rm{ \:p\land \sim p= c \: }}$

So, using this complement law, we get

$\rm \: = \: p\lor c$

$\rm \: = \: p$

Hence,

$\rm\implies \:\rm \: (p\lor q)\land (p\lor \sim q) \: = \: p$

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Additional Information :-

1. Commutative Law :-

$\boxed{ \rm{ \:p\lor q \: = \: q\lor p \: }}$

$\boxed{ \rm{ \:p\land q \: = \: q\land p \: }}$

2. Associative Law

$\boxed{ \rm{ \:p\lor (q\lor r) = (p\lor q)\lor r \: }}$

$\boxed{ \rm{ \:p\land (q \land \: r) = (p\land q)\land r \: }}$

3. Idempotent Law

$\boxed{ \rm{ \:p\lor p = p \: }}$

$\boxed{ \rm{ \:p\land p = p \: }}$

4. Distributive Law

$\boxed{ \rm{ \:(p\lor q)\land (p\lor r) \: = \: p\lor (q\land r) \: }}$

$\boxed{ \rm{ \:(p\land q)\lor (p\land r) \: = \: p\land (q\lor r) \: }}$

5. De Morgan's Law

$\boxed{ \rm{ \:\sim (p\lor q) = \sim p\land \sim q \: }}$

$\boxed{ \rm{ \:\sim (p\land q) = \sim p\lor \sim q \: }}$

6. Complement Law

$\boxed{ \rm{ \:p\land \sim p= c \: }}$

$\boxed{ \rm{ \:p\lor \sim p= t \: }}$

$\boxed{ \rm{ \:\sim t = c \: }}$

$\boxed{ \rm{ \:\sim c = t \: }}$