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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.