Recall the operator from the lecture on fast arithmetic hardware:

(g1,p1) (g2,p2) = (g1 | (p1& g2), p1&p2)

In other words, the operator takes two (g,p) pairs and produces a new (g,p) pair. Written another way, the result of (g1,p1) (g2,p2) is the pair (g,p) where

g = g1 | (p1& g2)

p = p1& p2

Show that the operator is associative. That is, using the definition of the operator and the rules of Boolean Algebra, show that the following equality holds:((g1,p1) (g2,p2)) (g3,p3) = (g1,p1) ((g2,p2) (g3,p3)) for any Boolean values g1, p1, g2, p2, g3, p3. To do this, use the definition of the operator on the above equation until the resulting Boolean expressions only involve the operators & and |. Then apply the laws of Boolean Algebra to both sides until both sides have the same form.

You need to be very careful, and you need to determine which of the laws to apply, and when. If done properly, the problem requires just 5 or 6 steps (maybe less, depending upon how you combine steps). Be sure to use parentheses to group expressions appropriately. Include explanations for each step