Department

Philosophy

Course Code

PHL246H1

Professor

Colin Howson

Exercise Answers.

Chapter One.

1.

i. For the truth table, go to

http://turner.faculty.swau.edu/mathematics/materialslibrary/truth/

and type in “~(A&(B+~C))” for the Boolean expression and hit generate table.

ii.

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.

2.

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.

3.

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.

4.

i. Show that T^A Ù A

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