The expression given for the lab was

.NOT.  (A .GT. 5 ) .AND. .NOT. (B .GE. 7)

We had to apply De Morgans law to get an equivalent expression. This is how to
go about solving such problems.

Considering expr1=.NOT.(A.GT.5) and
expr2 = .NOT.(B.GE.7)

So the given expression would look like this

expr1 .AND. expr2

If we apply two NOTs to the above eqn we do not
change the value of our expression.Therefore

.NOT..NOT.(expr1 .AND. expr2)

Now by bringing in the first NOT inside the
bracket we can apply DeMorgans law and get

.NOT.(.NOT.(expr1) .OR. .NOT.(expr2))

Now by putting the values of expr1 and expr2 back
we find that we have

.NOT.(.NOT.(.NOT.(A.GT.5)) .OR. .NOT.(.NOT.(B.GE.7)))

Since the two NOT's cancel each other we have our
final expression as

.NOT.((A.GT.5) .OR. (B.GE.7))

Hope this was helpful.