Presentation: Math Review

# CS-3110: Formal Languages and Automata

## Math Review

### Chapter 1 + 2

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# Example 1: Direct Proof

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# Problem: De Morgan's laws for sets

Given:

* The definitions of the set operations from last lecture, and the laws of predicate logic

* Two sets A and B

Show that:

$$
\overline{A \cup B} = \overline{A} \cap \overline{B}\\\\
\overline{A \cap B} = \overline{A} \cup \overline{B}
$$

---

# Some Notation

Remember: Two sets are equal if they have the same elements

$$
A = B \equiv x \in A \leftrightarrow x \in B
$$

Additionally, our set-builder notation is really just shorthand:

$$
y \in \\{x | P(x)\\} \equiv P(y)
$$

Also recall De Morgan's laws for logic:

$$
\neg (a \wedge b) \leftrightarrow (\neg a) \vee (\neg b)\\\\
\neg (a \vee b) \leftrightarrow (\neg a) \wedge (\neg b)
$$

---

# With our new notation

What we want to show is:

$$
\overline{A \cup B} =  \overline{A} \cap \overline{B}
$$

Which is exactly the same as:

$$
\forall y: y\in \overline{A \cup B} \leftrightarrow y\in (\overline{A} \cap \overline{B})
$$

For a direct proof, we start with the left side and then show that we can get to the right side by applying equivalence 
transformations (logical equivalences and notational equalities)

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# A direct proof

For all y:

$$\begin{aligned}
y \in \overline{A \cup B}\:\leftrightarrow&\: y \in \\{x | \neg x \in (A \cup B) \\}\\\\
                          \leftrightarrow&\: \neg y \in (A \cup B) \\\\
                          \leftrightarrow&\: \neg y \in \\{x | x\in A \vee x \in B\\}\\\\
                          \leftrightarrow&\: \neg (y \in A \vee y \in B)\\\\
                          \leftrightarrow&\: (\neg y \in A) \wedge (\neg y \in B)\\\\
                          \leftrightarrow&\: (y \in \\{x | \neg x\in A\\}) \wedge (y \in \\{x | \neg x \in B\\}\\\\
                          \leftrightarrow&\: y \in \overline{A} \wedge y \in \overline{B}\\\\
                          \leftrightarrow&\: y \in \\{x | x \in \overline{A} \wedge \overline{B}\\}\\\\
                          \leftrightarrow&\: y \in (\overline{A} \cap \overline{B})
\end{aligned}
$$

---

# A direct proof

We usually skip some of the more tedious rewriting.

For all y:

$$\begin{aligned}
y \in \overline{A \cup B}\:\leftrightarrow&\: \neg y \in (A \cup B) \\\\
                          \leftrightarrow&\: \neg (y \in A \vee y \in B)\\\\
                          \leftrightarrow&\: (\neg y \in A) \wedge (\neg y \in B)\\\\
                          \leftrightarrow&\: y \in \overline{A} \wedge y \in \overline{B}\\\\
                          \leftrightarrow&\: y \in (\overline{A} \cap \overline{B})
\end{aligned}
$$

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# Example 2: Induction Proof

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# Problem: Generalized DeMorgan's law

Given:

* Set operations, as before, plus the proof from example 1

* n (but at least 2) sets `$A_1, A_2, \ldots A_n$`

Show that:

$$
\overline{A_1 \cup A_2 \cup \cdots \cup A_n} = \overline{A_1} \cap \overline{A_2} \cap \cdots \cap \overline{A_n}\\\\
\overline{A_1 \cap A_2 \cap \cdots \cap A_n} = \overline{A_1} \cup \overline{A_2} \cup \cdots \cup \overline{A_n}
$$

---

# Induction Proof

An induction proof needs 2 "ingredients":

* A base case: Our statement holds for some small/fixed number 
 
 * A step: We can show that if our statement holds for (any) n, it also holds for n+1 
 
Once we have shown these two things, we can conclude that our statement holds for any n: For a given n, start at the base case, and then apply the step as often as is necessary to reach that n.

---

# Induction Proof

Base case: We have two sets

$$
\overline{A_1 \cup A_2} = \overline{A_1} \cap \overline{A_2}
$$

How do we show this?

We just did, in example 1. If, say, the homework does not have such an "example 1", you'd insert a proof here, using your favorite technique (and yes, nested induction proofs are a thing).

---

# Induction Step

Now we need to show that for **any** n we can take

$$
\overline{A_1 \cup A_2 \cup \cdots \cup A_n} = \overline{A_1} \cap \overline{A_2} \cap \cdots \cap \overline{A_n}
$$

to conclude

This means: We can **use** the case for n **as an additional premise**!

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# Induction Step

Now we actually show the step:

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# Induction Proof Application

Say we have 37 sets. We can show that DeMorgan's laws hold for these 37 sets:

* We know (from example 1) that it works with 2 sets

* We apply the induction step proof once: We know it also works with 3 sets

* We apply the induction step proof another time: We know it works with 4 sets

* ...

* We apply the induction step proof another time: We know it works with 37 sets

---

# Induction Proof

* But what if we have more (or fewer?) than 37 sets?

* We could still use the same procedure!

* The induction proof **shows** that the step works

* And that the base case holds

* In other words: The proof itself needs to work for **any** n, and must not be a simple enumeration of steps (which may break!)

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# Never trust patterns!

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# Never trust patterns!

$$
\int_0^\infty 2\cos(x)dx = \pi\\\\
\int_0^\infty 2\cos(x) \frac{\sin(x)}{x} dx = \pi\\\\
\int_0^\infty 2\cos(x) \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3} dx = \pi\\\\
\int_0^\infty 2\cos(x) \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3} \frac{\sin(x/5)}{x/5}dx = \pi\\\\
\vdots\\\\
\int_0^\infty 2\cos(x) \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3} \frac{\sin(x/5)}{x/5}\cdots  \frac{\sin(x/113)}{x/113} dx < \pi\\
$$

[Borwein Integrals](https://www.youtube.com/watch?v=851U557j6HE)

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# Example 3: Proof By Contradiction

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# Problem: Infinitely Many Primes

Given (premises):
 
 * The usual laws for integer arithmetic and sets
 
 * A prime number is a number p with exactly two proper divisors: 1 and p
 
 * There exists at least one prime number, and the smallest prime number is 2
 
Show that:

* There are infinitely many prime numbers
 
 * Use a proof by contradiction
 
---

# Negation of the Conclusion

To prove by contradiction, we first start with negating the conclusion:

Let us assume that there are only finitely many prime numbers.

In fact, let us name and order them all:

$$
p_1 \lt p_2 \lt p_3 \lt \ldots \lt p_n 
$$

Where

$$
p_1 = 2
$$

---

# A new number

Let us look at this number:

$$
N = p_1 \cdot p_2 \cdot p_3 \cdots p_n + 1
$$

Some things we "know":

* N is larger than all of our prime numbers, because the smallest number in our product is 2, and we add 1 to the result 
 
 * N is not a prime number (because we said we have named **all** prime numbers)
 
 * Therefore, N must be divisible by one of our prime numbers 
 
(You may have seen this proof before, with some "handwaving" here that obviously N can't be divisible by any of the primes we have)

---

# Dividing N

First, observe that this product is, by definition, also divisible by each of our prime numbers:

$$
p_1 \cdot p_2 \cdot p_3 \cdots p_n
$$

With the knowledge that N must be divisible by some `$p_i$`:

$$
p_i | N \equiv N = k \cdot p_i\\\\
p_i | p_1 \cdot p_2 \cdot p_3 \cdots p_n \equiv p_1 \cdot p_2 \cdot p_3 \cdots p_n = j \cdot p_i
$$

$$
N - p_1 \cdot p_2 \cdot p_3 \cdots p_n = k \cdot p_i - j \cdot p_i\\\\
p_1 \cdot p_2 \cdot p_3 \cdots p_n + 1 - p_1 \cdot p_2 \cdot p_3 \cdots p_n = k \cdot p_i - j \cdot p_i\\\\
1 = (k-j) \cdot p_i \equiv p_i | 1 
$$

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# Dividing 1

* We have concluded that the same prime number that divides N also divides 1

* The smallest prime number is 2

* This means we would have some number greater or equal to 2 that divides 1

* But such a number does not exist ....

We have reached a contradiction

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# Evaluating our assumption

* We started with assuming that there only finitely many prime numbers

* After this assumption we have only used our given premises

* But we reached a contradiction

* This means that our assumption must be wrong, and there must indeed be **infinitely many** prime numbers

QED

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# Example 4: Proof By Contrapositive

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# Proof By Contrapositive

Given (premises):
 
 * The usual laws for integer arithmetic and sets
 
 * We call an integer n **even** iff there is an integer k, such that n=2k
 
 * We call an integer n **not even** (or "odd") iff there is an integer k such that n=2k+1
 
Show that:

* If `$n^2$` is even, then n is even 
 
 * Use a proof by contrapositive 
 
---

# Proof by Contrapositive

* A proof by contrapositive is a weird beast

* We can only use it to show something of the form "if p then q"

* And we do this by showing "If not q, then not p"

* And we (usually) do **that** by using "not q" as an additional premise, and showing "not p" using a direct proof

---

# Proof by Contrapositive

To show:

### If `$n^2$` is even, then n is even

The contrapositive is:

### If n is not even, then `$n^2$` is not even

We show this by starting with an n that is not even and manipulating it.

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# The Proof

$$
\begin{aligned}
n =&\: 2k+1\\\\
n^2 = n\cdot n =&\: (2k+1)\cdot(2k +1)\\\\
               =&\: 4k^2 + 4k + 1\\\\
               =&\: 2(2k^2 + 2k) + 1\\\\
               =&\: 2k_1 + 1
\end{aligned}
$$

This new `$k_1$` surely is an integer (we got it from adding and multiplying integers), so there is a "k" such that our `$n^2 = 2k_1 + 1$` which means that it is not even.

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# Proofs and Implications

* Note: We started with "n = 2k+1" **as an additional premise**

* Anything we proved from there is a **consequence** of this premise

* This is how we got the implication we needed ("if n is not even, then `$n^2$` is not even")

* In fact, *every* proof is an "implication": From the premises follow the conclusions

* If we **only** use equivalences and equalities, the "implication" goes in both directions, and becomes another equivalence (see example 1)

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# Proofs

* Proofs require practice and patience

* You may not always see what the path to the solution may be

* Sometimes it helps to re-examine all the givens, and see which ones have yet to be used

* The proof method also sometimes gives you a clue what to do (for induction proofs, you have to use the statement for the n-case somewhere)