ece108: truth tables part 2

docs/1b/ece108.md

@@ -49,6 +49,11 @@ 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**.
 
 $$p\implies q$$
@@ -60,4 +65,48 @@ $$p\implies q$$
 | 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.
+
+$$p\implies \equiv (\neg p)\vee q$$