# How can you use a truth table to prove that ((~p vv q) ^^ p) vv q is equivalent to q ?

Jun 20, 2016

See explanation...

#### Explanation:

In this truth table, $0$ represents false, $1$ represents true.

((p,color(blue)(q),~p,~p vv q,(~p vv q) ^^ p,color(green)([(~p vv q) ^^ p] vv q)),(0,color(blue)(0),1,1,0,color(green)(0)),(0,color(blue)(1),0,1,0,color(green)(1)),(1,color(blue)(0),0,0,0,color(green)(0)),(1,color(blue)(1),0,1,1,color(green)(1)))

The truth values are evaluated for all possible combinations of $p , q$ true or false.

Notice that the resulting green column is the same as the blue column.

Jun 20, 2016

$\left[\left(\neg p \cup q\right) \cap p\right] \cup q = \left(q \cap p\right) \cup q \equiv q$

#### Explanation:

$\left(\neg p \cup q\right) \cap p = \left(\neg p \cap p\right) \cup \left(q \cap p\right)$

but $\left(\neg p \cap p\right) = \emptyset$ then

$\left(\neg p \cup q\right) \cap p = \left(q \cap p\right)$

so

$\left[\left(\neg p \cup q\right) \cap p\right] \cup q = \left(q \cap p\right) \cup q \equiv q$