# ECE 108: Discrete Math 1

An **axiom** is a defined core assumption held to be true.

!!! example
    True is not false.

A **theorem** is a true statement derived from axioms via logic or other theorems.

!!! example
    True or false is true.

A **proposition/statement** must be able to have the property that it is exclusively true or false.

!!! example
    The square root of 2 is a rational number.

An **open sentence** becomes a proposition if a value is assigned to the variable.

!!! example
    $x^2-x\geq 0$

## Truth tables

A truth table lists all possible **truth values** of a proposition, containing independent **statement variables**.

!!! example
    | p | q | p and q |
    | --- | --- | --- |
    | T | T | T |
    | T | F | F |
    | F | T | F |
    | F | F | F |

## Logical operators

!!! definition
    - A **compound statement** is composed of **component statements** joined by logical operators AND and OR.

The **negation** operator is equivalent to logical **NOT**.

$$\neg p$$

The **conjunction** operaetor is equivalent to logical **AND**.

$$p\wedge q$$

The **disjunction** operator is equivalent to logical **OR**.

$$p\vee q$$

### Proposation relations

!!! definition
    A **tautology** is a statement that is always true, regardless of its statement variables.

The **implication** sign requires that if $p$ is true, $q$ is true, such that *$p$ implies $q$*. The first symbol is the **hypothesis** and the second symbol is the **conclusion**.

$$p\implies q$$

| $p$ | $q$ | $p\implies q$ |
| --- | --- | --- |
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | F |

The **inference** sign represents the inverse of the implication sign, such that $p$ **is implied by** $q$. It is equivalent to $q\implies p$.

$$p\impliedby q$$

The **if and only if** sign requires that the two propositions imply each other — i.e., that the state of $p$ is the same as the state of $q$. It is equivalent to $(p\implies q)\wedge (p\impliedby q)$.

$$p\iff q$$

The **logical equivalence** sign represents if the truth values for both statements are **the same for all possible variables**, such that the two are **equivalent statements**.

$$p\equiv q$$

$p\equiv q$ can also be defined as true when $p\iff q$ is a tautology.

!!! warning
    $p\equiv q$ is *not a proposition* itself but instead *describes* propositions. $p\iff q$ is the propositional equivalent.

## Common theorems

The **double negation rule** states that if $p$ is a proposition:

$$\neg(\neg p)\equiv p$$

!!! tip "Proof"
    Note that:
    
    | $p$ | $\neg p$ | $\neg(\neg p)$ |
    | --- | --- | --- |
    | T | F | T |
    | F | T | F |
    
    Because the truth values of $p$ and $\neg(\neg p)$ for all possible truth values are equal, by definition, it follows that $p\equiv\neg(\neg p)$.

!!! warning
    Proofs must include the definition of what is being proven, and any relevant evidence must be used to describe why.

The two **De Morgan's Laws** allow distributing the negation operator in a dis/conjunction if the junction is inverted.

$$
\neg(p\vee q)\equiv(\neg p)\wedge(\neg q) \\
\neg(p\wedge q)\equiv(\neg p)\vee(\neg q)
$$

An implication can be expressed as a disjunction. As long as it is stated, it can used as its definition.

$$p\implies \equiv (\neg p)\vee q$$

Two **converse** propositions imply each other:

$$p\implies q\text{ is the converse of }q\implies p$$ 

A **contrapositive** is the negatated converse, and is **logically equivalent to the original implication**. This allows proof by contrapositive.

$$\neg p\implies\neg q\text{ is the contrapositive of }q\implies p$$

### Operator laws

Both **AND** and **OR** are commutative.

$$
p\wedge q\equiv q\wedge p \\
p\vee q\equiv q\vee p
$$

Both **AND** and **OR** are associative.

$$
(p\wedge q)\wedge r\equiv p\wedge(q\wedge r) \\
(p\vee q)\vee r\equiv p\vee(q\vee r)
$$

Both **AND** and **OR** are distributive with one another.

$$
p\wedge(q\vee r)\equiv(p\wedge q)\vee(p\wedge r) \\
p\vee(q\wedge r)\equiv(p\vee q)\wedge(p\vee r)
$$

!!! tip "Proof"
    $$
    \begin{align*}
    (\neg p\vee\neg r)\wedge s\wedge\neg t&\equiv\neg(p\wedge r\vee s\implies t) \\
    \tag*{definition of implication} &\equiv \neg (p\wedge r\vee[\neg s\vee t]) \\
    \tag*{DML} &\equiv\neg(p\wedge r)\wedge\neg[(\neg s)\vee t)] \\
    \tag*{DML} &\equiv(\neg p\vee\neg r)\wedge\neg[(\neg t)\vee t] \\
    \tag*{DML} &\equiv(\neg p\vee\neg r)\wedge\neg(\neg s)\wedge\neg t \\
    \tag*{double negation} &\equiv(\neg p\vee\neg r)\wedge s\wedge\neg t
    \end{align*}
    $$

### Quantifiers

A **quantified statement** includes a **quantifier**, **variable**, **domain**, and **open sentence**.

$$
\underbrace{\text{for all}}_\text{quantifier}\  \underbrace{\text{real numbers}\overbrace{x}^\text{variable}\geq 5}_\text{domain}, \underbrace{x^2-x\geq 0}_\text{open sentence}
$$

The **universal quantifier** $\forall$ indicates "for all".

$$\forall x\in S,P(x)$$

!!! example
    All real numbers greater than or equal to 5, defined as $x$, satisfy the condition $x^2-x\geq 0$.
    
    $$\forall x\in\mathbb R\geq 5,x^2-x\geq 0$$

The **existential quantifier** $\exists$ indicates "there exists at least one".

$$\exists x\in S, P(x)$$

!!! example
    There exists at least one real number greater than or equal to 5, defined as $x$, satisfies the condition $x^2-x\geq 0$.
    
    $$\exists x\in\mathbb R\geq 5,x^2-x\geq 0$$

Quantifiers can also be negated and nested. The opposite of "for each ... that satisfies $P(x)$" is "there exists ... that does **not** satisfy $P(x)$".

$$\neg(\forall x\in S,P(x))\equiv\exists x\in S,\neg P(x)$$

Nested quantifiers are **evaluated in sequence**. If the quantifiers are the same, they can be grouped together per the commutative and/or associative laws.

$$\forall x\in\mathbb R,\forall y\in\mathbb R\equiv \forall x,y\in\mathbb R$$

!!! warning
    This means that the order of the quantifiers is relevant if the quantifiers are different:
    
    $\forall x\in\mathbb R,\exists y\in\mathbb R,x-y=1$ is **true** as setting $y$ to $x-1$ always fulfills the condition.
    
    $\exists y\in\mathbb R,\forall x\in\mathbb R, x-y=1$ is **false** as when $x$ is selected first, it is impossible for every value of $y$ to satisfy the open sentence.