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**.
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:
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 $P(x)$".
Nested quantifiers are **evaluated in sequence**. If the quantifiers are the same, they can be grouped together per the commutative and/or associative laws.
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.
There are a variety of methods to prove or disprove statements.
- **Deduction**: a chain of logical inferences from a starting assumption to a conclusion
- **Case analysis**: exhausting all possible cases (e.g., truth table)
- **Contradiction**: assuming the conclusion is false, which follows that a core assumption is false, therefore the conclusion must be true
- **Contrapositive**: is equivalent to the original statement
- **Counterexample**: disproves things
- **Induction**: Prove for a small case, then prove that that applies for all cases
Implications can be proven in two simple steps:
1. It is assumed that the hypothesis is true (the implication is always true when it is false)
2. Proving that it follows that the conclusion is true
!!! example "Proving implications"
Prove that if $n+7$ is even, $n+2$ is odd.
$\text{Proof:}$
$\text{Assume }n+7\text{ is an even number. It follows that for some }k\in\mathbb Z$
$$
\begin{align*}
n+7&=2k \\
\text{s.t.} n+2&=2k-5 \\
&=2(k-3)+1
\end{align*}
$$
$\text{which is of the form }2z+1,z\in\mathbb Z,\text{ thus } n+2\text{ is odd.}$
!!! example "Proof by contradiction"
Prove that there is no greatest integer.
$\text{Proof:}$
$\text{ Let }n\in\mathbb Z\text{ be given and assume }\overbrace{\text{for the sake of contradiction}^\text{FTSOC}}\text{ that }n\text{ is the largest integer. Note that }n+1\in\mathbb Z\text{ and }n+1>n.\text{ This contradicts the initial assumption that }n\text{ is the largest integer, therefore there is no largest integer.}$
- A proof **without loss of generality** (WLOG) indicates that the roles of variables do not matter — so long as the symbols CTRL-H'd, the proof remains exactly the same. For example, "WLOG, let $x,y\in\mathbb Z$ st. $x<y$."
Induction is a proof technique that can be used if the open sentence $P(n)$ depends on the parameter $n\in\mathbb N$. Because induction works in discrete steps, it generally cannot be applied domains of all real numbers.
To do so, the following must be proven:
- $P(1)$ must be true (the base case)
- $P(k+1)$ must be true for all $P(k)$, assuming $P(k)$ is true (the inductive case)
!!! warning
The statement **cannot** be assumed to be true, so one side must be derived into the other side.
!!! tip "Proof"
This should more or less be exactly followed. For the statement $\forall n\in\mathbb Z,n!>2^n$:
> We use mathematical induction on $n$, where $P(n)$ is the statement $n!>2^n$.
>
> **Base case**: Our base case is $P(4)$. Note that $4!=24>16=2^4$, so the base case holds.
>
> **Inductive step**: Let $k\geq 4$ for an arbitrary natural number and assume that $k!>2^k$. Multiplying by $k+1$ gives
>
> $$(k+1)k^2>(k+1)2^k$$
>
> By definition $(K=1)k!=(k+1)!$. Since $k\geq 4$, $k+1>2$ and thus $(k+1)2^k>2\cdot 2^k=2^{k+1}$. Putting this together gives
>
> $$(k+1)!>2^{k+1}$$
>
> Thus $P(k+1)$ is true and by the Principle of Mathematical Induction (POMI), $P(n)$ is true for all $n\geq 4$.
Induction can be applied to the whole set of integers by proving the following:
- $P(0)$
- if $i\geq 0, P(i)\implies P(i+1)$
- if $i\leq 0, P(i)\implies P(i-1)$
Alternatively, some steps can be skipped in **strong induction** by proving that if for $k\in\mathbb N$, $P(i)$ holds for all $i\leq k$, so $P(k+1)$ holds. In other words, by assuming that the statement is true for all values before $k$. If strong induction is true, regular induction must also be true, but not vice versa.
- A **set** is an unordered collection of distinct objects.
- An **element/member** of a set is an object in that set.
- A **multiset** is an unordered collection of objects.
Sets are expressed with curly brackets:
$$\{s_1, s_2,\dots\}$$
Numbers are defined as sets of recursively empty sets:
$$
\begin{align*}
0&:=\empty \\
1&:=\{\empty\} \\
2&:=\{\empty,\{\empty\}\}
\end{align*}
$$
### Special sets
- $\mathbb N$ is the set of **natural numbers** $\{1, 2, 3,\dots\}$
- $\mathbb W$ is the set of **whole numbers** $\{0, 1, 2,\dots\}$
- $\mathbb Z$ is the set of **integers** $\{\dots, -1, 0, 1, \dots\}$
- $\mathbb Z^+_0$ is the set of **positive integers, including zero** — these modifiers can be applied to the set of negative integers and real numbers as well
- $2\mathbb Z$ is the set of **even integers**
- $2\mathbb Z + 1$ is the set of **odd integers**
- $\mathbb Q$ is the set of **rational numbers**
- $\mathbb R$ is the set of **real numbers**
- $\empty$ or $\{\}$ is the **empty set** with no elements
### Set builder notation
!!! definition
- The **domain of discourse** is the context of the current problem, which may limit the universal set (e.g., if only integers are discussed, the domain is integers only)
$x$ is an element if $x$ is in $\mathcal U$ and $P(x)$ is true.
$$\{x\in\mathcal U|P(x)\}$$
!!! example
All even numbers: $A=\{n\in\mathbb Z,\exists k\in\mathbb Z,n=2k\}$
$f(x)$ is an element if $x$ is in $\mathcal U$, and $P(x)$ is true:
$$\{f(x)|\underbrace{x\in\mathcal U, P(x)}_\text{swappable, omittable}\}$$
!!! example
- All even numbers: $A=\{2k|k\in\mathbb Z\}$
- All rational numbers: $\mathbb Q=\{\frac a b | a,b\in\mathbb Z,b\neq 0\}$
The **complement** of a set is the set containing every element **not** in the set.
$$\overline S$$
The **universal set** is the set containing everything, and is the complement of the empty set.
- An **index set** $I$ is the set containing all relevant indices.
A **partition** of a set $S$ is a set of **disjoint** sets that create the original set when unioned.
$$S=\bigcup_{i\in I}A_i$$
!!! example
$\{\{1\},\{2,3\},\{4,\dots\}\}$ is a partition of $\mathbb N$.
A **powerset** of a set $A$ is the set of all possible subsets of that set.
$$\mathcal P(A)=\{X|X\subseteq A\}$$
The empty set is the subset of every set so is part of each powerset. The number of elements in a subset is equal to the the number of elements in the original set as a power of two.
To prove $A\subseteq B\implies \mathcalP(A)\subseteq \mathcal P(B)$:
**Proof:** Let $A\subseteq B$ and $X\in\mathcal P(A)$. By definition, since $X\in\mathcal P(A), X\subseteq A$. Since $A\subseteq B$, it follows that $X\subseteq B$. Thus by the definition of the powerset, $X\in\mathcal P(B)$.
## Functions
!!! definition
- A **surjective** function has an equal codomain and range.
A **function** a relation between two sets $f:X\to Y$ such that each $x\in X$ **maps to** a unique $f(x)\in Y$.
$$
\begin{align*}
f:\ &X\to Y \\
&x\longmapsto f(x)
\end{align*}
$$
!!! example
Sample function with multiple cases and indices:
$$
\begin{align*}
f:\ &X\to Y \\
&x_i\longmapsto \begin{cases}
y_1 & i\in\{1,2\} \\
y_3 & i\in\{3,4,5\}
\end{cases}
\end{align*}
$$
The **domain** $\text{dom}(f)$ is the input set.
$$X=\text{dom}(f)$$
The **codomain** $\text{cod}(f)$ is the output set.
$$Y=\text{cod}(f)$$
The **range** $\text{rang}(f)$ is the subset of $Y$ that is actually mapped to by the domain.
A **surjective function**, **surjection**, or **onto** is a function that has its codomain equal to its range. A surjection $g:Y\to X$ exists if and only if an injection $f:X\to Y$ exists.
