i. For the truth table, go to
and type in “~(A&(B+~C))” for the Boolean expression and hit generate table.
A truth table will have 2 to the power of X rows, where X is the number of
letters/variables you are using. In this case, it is 2^5=32 rows.
i. For this truth table, go to the same website as above and paste in “A&(B+C)” as the left
side, and “(A&B)+(A&C)” for the right side.
ii. Use the site again, with “A+(B&C)” for the table for the left side of the equation, and
“(A+B)&(A+C)” for the right.
i. If B1,B2,…,Bn are mutually exclusive, then only one of them can be true at one point
in time. Now, in a conjunction (eg A^B1), if any term is false, then the whole conjunction
is false. So, if only one of B1 through Bn can be true at any point in time, then all but one
are false at the point in time. It follows then that all but one of the conjunctions A^Bx
(where Bx will just be the whole set of B1 through Bn) must be false when any single
sentence in Bx is true. For example, if B1 is true, B2 through Bn is false, and therefore
A^B1 is true and A^B2 through A^Bn are false (you may substitute any number in place
of the 1 and have it still be true).
Long winded, but I hope that gets the point across.
ii. If B1 through Bn are mutually exclusive, and we partition that set into (B1v…vBn-1)
and Bn, then they will also be mutually exclusive.
If Bn is true, then none of B1 through Bn-1 can be true. And so, since there is not a single
truth in the disjunction, the whole disjunction is false. If any of B1 through Bn-1 is true
(granted, only one at a time due to the mutual exclusivity), then Bn must be false. So,
since Bn cannot be true when (B1v…vBn) is, and vice versa, they are mutually exclusive.
i. Show that T^A Ù A