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.
$\text{Proof:}$
$\text{ Let }n\in\mathbb Z\text{ be given and assume }\overbrace{\text{for the sake of contradiction}^\text{FTSOC}}\text{ that }n\text{ is the largest integer. Note that }n+1\in\mathbb Z\text{ and }n+1>n.\text{ This contradicts the initial assumption that }n\text{ is the largest integer, therefore there is no largest integer.}$
