(PQ)
is a tautology.There are eight (8) questions listed below, you may choose to do any five (5) of them. Each question will count 20 points. If you complete more than 5 questions, you must clearly indicate which 5 you wish to be graded. There will be no extra credit - don't bother asking for it. Good Luck!!.
1. Use a truth table to show that
~P (PQ)
is a tautology.
2. Use logical symbols to rewrite these statements and determine if the argument is valid (that is, the premise [first three statements] implies the conclusion [last statement]).
Everyone who is sane can do Logic.
No lunatics are fit to
serve on a jury.
None of your sons can do Logic.
Therefore,
none of your sons are fit to serve on a jury.
Use the predicates, S(x) = "x is sane", L(x) = "x can do Logic", J(x) = "x is fit to serve on a jury" and N(x) = "x is one of your sons".[Also note that a lunatic is a person who is not sane.]
3. If B and C are sets, either prove the following statement or show that it is false (i.e., provide a counterexample).
If x 
B and
B is not a subset of C then x 
C.
4. Prove that for any sets A and B,
B where Xc is the complement of the set X.
5. Prove or disprove (that is, provide a counterexample):
x)(P(x)
Q(x)) ((x)P(x) (x)Q(x) ).
6. Let A , B and C be subsets of a universal set U. Prove that:
(B
C) = (AB) (AC).7. Let there be 9 points in 3-space with integer coefficients. Prove that there is a pair of these points so that the line segment determined by them contains an interior point whose coordinates are integers.
8. Let A and B be subsets of a universal set U. Prove that:
(A - B)(B - A) = (AB) - (AB)
| P | Q | ~P | PQ | ~P(PQ) |
|---|---|---|---|---|
| T | T | F | T | T |
| T | F | F | F | T |
| F | T | T | T | T |
| F | F | T | T | T |
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Let S(x) = "x is sane", L(x) = "x can do Logic", J(x) = "x is fit to serve on a jury" and N(x) = "x is one of your sons". Symbolically we have:
x)(S(x)L(x)),x)(~S(x)~J(x)) 
~(
x)(~S(x)
J(x)),x)(N(x)~L(x)) ~(x)(N(x)L(x)).~L(x) ~S(x) (by the contrapositive of the first statement)~J(x). Thus, (x)(N(x)~J(x)) ~(x)(N(x)J(x)).Return to Questions
The statement is false. Consider the sets B = {a, b} and C = {a, c}. Clearly, a B and B is not a subset of C, but aC.
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(A-B)c = (ABc)c (Definition of A - B)
= AcBcc (DeMorgan's Law for sets)
= AcB (Double complement)
Return to questions.
The statement is false. A counterexample is given by: P(x) = "x is an even integer" and Q(x) = "x is an odd integer". In the universe of integers, every x is either even or odd, but it is clearly not true that every x is even or every x is odd.
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For any element x, we have:
xA(BC) 
(xA) ((xB) (xC))
((xA)(xB)) ((xA)(xC)) [Distributive Law in Logic]
(xAB)(xAC) [Definition of union]
x(AB)(AC) [Definition of intersection]
Thus, the two sets are equal.
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Special Homework Assignment #1.
Return to questions.
x(A - B)(B - A)((xA)(xB))((xB)(xA))
(((xA)(xB))(xB))(((xA)(xB))(xA))
(((xA)(xB))((xB)(xB)))(((xA)(xA))((xB)(xA)))
(((xA)(xB))((xB)(xA)))
(xAB)(xBA)
x(AB) - (BA)
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