Section 1.4: Nested Quantifiers

1. Section 1.4: Nested Quantifiers We will now look more closely at how quantifiers can be nested in a proposition and how to interpret more complicated logical statements involving nested quantifiers.

2. Ex: xy[x + y = y + x] - This is the commutative law for addition xy[x + y = 0] - This says every integer has an add. inv. xyz[x + (y + z) = (x + y) + z] Ex: xy[(x > 0)  (y > 0)  (xy > 0)] [True] For any x and for any y, if x and y are both positive then xy is positive. Ex: xy[(x > 0)  (y > 0)  (xy > 0)] [False] For any x and for any y, x and y are both positive and xy is positive. Ex: x[C(x)  y[C(y)  F(x, y)]] where C(x) is “x has a computer” and F(x, y) is “x and y are friends”. U.d. for x and y is all students. “For any student x, x has a computer or there is some student y where y has a computer and x and y are friends.” Any student either has a computer or has a friend that has a computer.

3. Ex: xyz[F(x, y)  F(x, z)  (y  z)  F(y, z)] Consider the implication F(x, y)  F(x, z)  (y  z)  F(y, z). This says that if (x is friends with y and y is friends with z and y and z are different people) then (y is not friends with z). Now looking at the quantifiers, we are dealing with “There exists some person x, such that for all persons y and all persons z, if x is friends with y and with z and y and z are different people then y is not friends with z.” So it says that someone exists satisfying the condition that no two of his friends are friends with one another. The question of when an  is appropriate versus using  confuses many people. Note that we are using an implication because we are saying that for all y and all z that satisfy the hypothesis, they satisfy the consequence. If the  were replaced with  we would be saying that all x and all y do satisfy both the hypothesis and consequent.

4. Interpreting a logical statement with nested quantifiers is a tricky business. So is producing a quantified statement to represent a given English statement (which we are about to turn to). This is by far the material that students last semester in this class had the most trouble with (test-wise). I urge you to spend time reading through the examples in the text and even more working through the exercises. This material is important because it is the basis for mathematical reasoning and we will be using it (more implicitly than explicitly) throughout this course. It may be that students have the most trouble with it when it is formalized because they did much better at implicitly using these techniques when constructing proofs later in the class. Nonetheless, the more comfortable you get at working with quantified logic statements, the more prepared you will be to work with and manipulate mathematical statements later.

5. Ex: “If a person is female and is a parent, then that person is someone’s mother.” Let’s use all people as our universe of discourse and translate this sentence into a quantified statement. F(x): “x is female” P(x): “x is a parent” M(x, y): “x is the mother of y” So we have “If F(x) and P(x) then M(x, someone)”. That is “F(x)  P(x)  M(x, someone)”. [How to quantify x?] x should be universally quantified because we are making a statement for all people x who satisfy the hypothesis: x[F(x)  P(x)  M(x, someone)] x[F(x)  P(x)  M(x, y)] [How to quantify y?][Where to put it?] x[F(x)  P(x)  yM(x, y)] or xy[F(x)  P(x)  M(x, y)]

6. Ex: “Everyone has exactly one best friend.” x[“x has exactly one best friend”] B(x, y): “x’s best friend is y” Now we want to be able to say “x has exactly one best friend”. This statement has two parts really. It is saying that x does have a best friend and furthermore that x has at most one best friend. We can think of this as at least one and at most one. The at least one part is easy. xyB(x,y). This says for every person x, there exists a best friend y. This covers at least one. Now we will add some stuff for at most one. xy[B(x,y)  z[(z  y)  B(x, z)]] xyz[B(x,y)  ((z  y)  B(x, z))]

7. Ex: “There is a person who has taken a flight on every airline.” T(p, f): “person p has taken flight f” Q(f, a): “f is a flight on airline a” Note that we need to have separate universes of discourse here. Let the u.d. for p be all people, for f be all flights, and for a be all airlines. paf[T(p,f)  Q(f, a)] Ex: “The sum of two odd integers is an even integer.” O(x): “x is odd” E(x): “x is even” xy[O(x)  O(y)  z[(x + y = z)  E(z)]] xyz[O(x)  O(y)  (x + y = z)  E(z)] Let u.d. for o be the odd integers and u.d. for e be the even integers. o1o2e[o1 + o2 = e]

8. Negating Nested Quantifiers If we have a quantified statement, then just as with statements in propositional logic, we indicate the negation of the statement by placing the negation operator() in front of it. This negation should apply to the entire statement so we may need to group the entire statement into parenthesis to force precedence of operations. Then we can successively apply the rules we have seen for negation of quantified statements to simplify it.

9. Ex: The negation of xy[x + y = 0] is xy[x + y = 0] [simplify] xy[x + y = 0]  x[y[x + y = 0]]  xy[(x + y = 0)]  xy[x + y  0] Ex: The negation of xyz[B(x,y)  ((z  y)  B(x, z))] is xyz[B(x,y)  ((z  y)  B(x, z))] xyz[B(x,y)  ((z  y)  B(x, z))]  xyz[(B(x,y)  ((z  y)  B(x, z)))]  xyz[B(x,y)  ((z  y)  B(x, z))]  xyz[B(x,y)  ((z  y)  B(x, z))]  xyz[B(x,y)  (z  y)  B(x, z)]  xy[B(x,y)  z[(z  y)  B(x, z)]] There is some x who doesn’t have any best friend or has more than 1.

10. Homework problems from Section 1.4 Problems 3, 8, 10, 15, 19, 26, 32, 37 from this section will be included on the next homework assignment.