Simplifying Boolean Algebra - factorising

Question

How is:

$P \lor ¬Q \lor P \land ¬Q$​

Factorise​d to get:

$P \lor ¬Q \land (1 \lor P)$

I've been staring at it all weekend. I understand what factorisation is and can do it in math.

I tried expanding it to get:

P v ¬Q ^ v P v ¬Q ^ P

It makes no sense.

Answer 1

I would recommend to use $+$ in place of $\vee$, a product with no symbol instead of $\wedge$ and $\overline{Q}$ instead of $\neg Q$. Now, using commutativity, associativity, the rule $1 + x = 1$ and the distributivity law $a(b + c) = ab + ac$, your two expressions become \begin{align} P + \overline{Q} + P\overline{Q} &= P + P\overline{Q} + \overline{Q} = P(1 + \overline{Q}) + \overline{Q} = P + \overline{Q}\\ P + \overline{Q}(1 + P) &= P + \overline{Q} \end{align}

Answer 2

First expression $(1)$ can be expanded using distributivity :

$$P\lor \lnot Q \lor (P\land \lnot Q) \tag{1}$$

$$\iff (P\lor \lnot Q \lor P)\land (P\lor \lnot Q \lor \lnot Q) \tag{1}$$

Since $P\lor P = P$ and $\lnot Q \lor \lnot Q = \lnot Q$, we have the first expression equivalent to:

$$ (P\lor \lnot Q) \land (P\lor \lnot Q) \tag{1}$$ Which is the same as:

$$ (P\lor \lnot Q) \tag{1}$$

As for the second expression $(2)$, you have: $$ P\lor \lnot Q \land (1\lor P) \tag{2}$$

$$\iff P\lor \lnot Q \land 1 \tag{2}$$

$$\iff (P\lor \lnot Q) \land 1 \tag{2}$$

$$\iff P\lor \lnot Q \tag{2}$$

EDIT:

Since OP wants actually to get from his first expression to his second expression, here is a way to do it:

We previously saw that first expression is equivalent to:

$$(P\lor \lnot Q)$$

Next we can "artificially" add $1$ on the right:

$$\iff (P\lor \lnot Q)\land 1$$

Then, since $1 = 1\lor P$ (as OP correctly pointed out in one of his comments):

$$\iff (P\lor \lnot Q)\land (1 \lor P)$$

And that's it.