Page 76 - HOW TO PROVE IT: A Structured Approach, Second Edition

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62 Quantificational Logic
∃yL(x, y) could be written as “x likes someone,” so the original statement
means that for every person x, x likes someone. In other words, everyone likes
someone. On the other hand, ∃y∀xL(x, y) means that there is some person y
such that ∀xL(x, y) is true. As we saw in Example 2.1.1, ∀xL(x, y) means
“Everyone likes y,” so ∃y∀xL(x, y) means that there is some person y such
that everyone likes y. In other words, there is someone who is universally liked.
These statements don’t mean the same thing. It might be the case that everyone
likes someone, but no one is universally liked.

Example 2.1.4. What do the following statements mean? Are they true or
false? The universe of discourse in each case is N, the set of all natural numbers.

1. ∀x∃y(x < y).
2. ∃y∀x(x < y).
3. ∃x∀y(x < y).
4. ∀y∃x(x < y).
5. ∃x∃y(x < y).
6. ∀x∀y(x < y).

Solutions
1. This means that for every natural number x, the statement ∃y(x < y) is true.
In other words, for every natural number x, there is a natural number bigger
than x. This is true. For example, x + 1 is always bigger than x.
2. This means that there is some natural number y such that the statement
∀x(x < y) is true. In other words, there is some natural number y such that
all natural numbers are smaller than y. This is false. No matter what natural
number y we pick, there will always be larger natural numbers.
3. Thismeansthatthereisanaturalnumberxsuchthatthestatement∀y(x < y)
is true. You might be tempted to say that this statement will be true if x = 0,
but this isn’t right. Since 0 is the smallest natural number, the statement
0 < y is true for all values of y except y = 0, but if y = 0, then the statement
0 < y is false, and therefore ∀y(0 < y) is false. Similar reasoning shows
that for every value of x the statement ∀y(x < y) is false, so ∃x∀y(x < y)
is false.
4. This means that for every natural number y, there is a natural number smaller
than y. This is true for every natural number y except y = 0, but there is no
natural number smaller than 0. Therefore this statement is false.
5. This means that there is a natural number x such that ∃y(x < y) is true.
But as we saw in the first statement, this is actually true for every natural
number x, so it is certainly true for at least one. Thus, ∃x∃y(x < y) is true.