Discrete Structures

# Lecture Notes, 27 Sep 2010

 Last Update: 27 September 2010 Note: or material is highlighted

1. §1.4: Nested Quantifiers (cont'd):

1. Uniqueness:

"There is exactly one dog" can be represented as:

x[Dog(x) ∧ ∀y[Dog(y) → y = x]]

1. I.e., There is (at least) one thing in the domain that is a dog,
and, any (other) dog in the domain is identical to that one.

• Sometimes, logicians abbreviate this (with the "E-shriek!" quantifier) as:

∃!xDog(x)

(read "There is a unique x in the domain such that x is a dog".)

2. Some analogies with computer programs:

1. The first "∃" is an existential quantifier.

• It can be read "there is something (call it x) in the domain such that…"

2. The first "x" is an existentially quantified (or "bound") variable.

• Think of it as being like a variable-declaration in a programming language.

3. The scope of "∃x" is [Dog(x) ∧ ∀y[Dog(y) → y = x]]

• Think of as being like the body of a procedure.

4. The "∀" is a nested universal quantifier.

• Think of it as being like the beginning of a nested procedure.

5. The "y" is a universally quantified (or "bound") variable.

• Think of it as being like a local variable-declaration in the nested procedure.

6. The scope of "∀y" is [Dog(y) → y = x]

• Think of it as being like the body of the nested procedure.

7. All 3 occurrences of "x" are said to be bound occurrences of x in "∃x[…]"

• Think of them as being variables that are "local" to the "procedure" that is the scope of the existential quantifier.

8. All 3 occurrences of y are bound occurrences of y in "∀y[…]"

• Think of them as being variables that are "local" to the "nested procedure" that is the scope of the universal quantifier.

9. The rightmost occurrence of x (i.e., the one in the equality proposition "y=x")
is said to be a free occurrence of x in "∀y[…]"

• Think of it as being a variable that is global to the "nested procedure" that is the scope of the universal quantifier.

• In general, a variable is:
• "bound" if it's "local"
• "free" if it's "global".

And now for something apparently completely different…

I didn't get a chance to cover this in today's lecture,
and I think I won't spend any lecture time on it (in order to move on with other topics),
but I thought you might find this interesting.

3. Is the present King of France bald?

2. It's a trick question: There is no present King of France!
3. OK…so…is "The present King of France is bald" true, or is it false?
Surely, it's got to be one or the other!

4. Here's how Bertrand Russell analyzed it:

• x[IsPresentlyKingOfFrance(x) ∧ ∀y[IsPresentlyKingOfFrance(y) → y = x] ∧ IsBald(x)]

• I.e., there is someone who is presently King of France,
and there's only one (see IA, above!)
and he's bald.

• On this analysis, the sentence is false,
because there is no present King of France.

• But if it's false, how do we say that the present King of France is not bald?

• There are two ways to negate the English sentence:

1. It's not the case that the present King of France is bald.
2. The present King of France is not bald.

• On the first way, we just put "¬" in front of the whole existentially quantified proposition
• That negation will be true, because the existentially quantified proposition was false.

• On the second way, we put the "¬" in front of "IsBald(x)"
• Now the negation is also false!
—again, because there's no present King of France!

2. Peano's Axioms:

1. More examples of nested quantifiers, here used to axiomatically characterize ("define") the natural numbers.

2. For Giuseppe Peano's biography, see here and here.

3. For more on Peano's axioms, see this course's webpages on:

4. Peano's Axioms:

1. Syntax and semantics of our representation:

Let N(x) = "x is a natural number".

(i.e., x ∈ {0,1,2,3,…}
or: x is one of: 0 ⇒ 1 ⇒ 2 ⇒ 3 ⇒ …)

Let S(x,y) = "x is succeeded by y" or "y is a successor of x".

(i.e., y is a very next number after x)

2. Axiom 1: 0 is a natural number
(or: There is a natural number; call it "0")

In FOL: N(0)

3. Axiom 2: Every natural number has a unique successor that is a natural number.

In FOL: ∀x[N(x) → ∃!y[N(y) ∧ S(x,y)]]

4. Axiom 3: No natural number is its own successor.

In FOL: ∀xy[(N(x) ∧ N(y) ∧ S(x,y)) → xy]

5. Axiom 4: 0 is not the successor of any natural number.
(i.e., no natural number has 0 as its successor;
i.e., it all starts with 0)

In FOL: ¬∃x[N(x) ∧ S(x,0)]

6. Axiom 5: If (what appear to be) "two" natural numbers have the same successor, then "they" are the same.

In FOL: ∀xy[(N(x) ∧ N(y)) → (∃z[N(z) ∧ S(x,z) ∧ S(y,z)] → x = y)]

7. Axiom 6: The Principle of Mathematical Induction (later…)