[p or(~p and q)] = ?

[p or(~p and q)] = ?

$[p \text{ or } (~p \text{ and }q)] = p + \bar{p}\cdot q$

$\text{true }=1 \quad \text{false }=0 \qquad \text{ and } = "\cdot" \quad \text{ or } = "+"\\ \begin{array}{|r|r|r|r||r|} \hline p & q & \bar{p} & \bar{p} \cdot q & p+ \bar{p} \cdot q & p \text{ and } q \\ \hline 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0 & 1 & 1\\ \hline \end{array}$

[p or(~p and q)] =

p v (~p ^ q )]  =

(p v ~p)  ^  (p v q)  =   [this is the universal set intersected with (p v q)  ]  =

p v q

Can some other mathematician check this???

Let's see....

Truth table

Minimal Forms

Logic Circuit

New Edited:

[p or(~p and q)] = ?

$[p \text{ or }( \text{~} p \text{ and } q)] = p + \bar{p}\cdot q = p+q= p \text{ or } q$

$\text{true }=1 \quad \text{false }=0 \qquad \text{ and } = "\cdot" \quad \text{ or } = "+"\\ \begin{array}{|r|r|r|r||r|} \hline p & q & \bar{p} & \bar{p} \cdot q & p+ \bar{p} \cdot q & p+q = p \text{ or } q \\ \hline 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0 & 1 & 1\\ \hline \end{array}$

Here's a way of visualizing it:

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