- [[biconditional implication]], [[logical implication]]
# Idea
For any [[logical implication]] $p \rightarrow q$, we can translate by saying the following.
$p$ is a *sufficient* condition for $q$.
- For $p$, it is *sufficient* that $q$
- $p$ being true guarantees that $q$ is true
$q$ is a *necessary* condition for $p$.
- For $p$, it is *necessary* that $q$, which can be translated to $q \rightarrow p$.
- For $q$ to be true, $p$ must also/always be true. $q$ cannot be true unless $p$ is true.
*Necessity* and *sufficiency* describe a conditional or implicational relationship between two statements.
*Sufficiency* is represented by the *direct implication*, $p \rightarrow q$.
*Necessity* is represented by the *converse implication*, $q \rightarrow p$.
## Why is $q$ necessary for $p$ in the statement $p \rightarrow q$?
![[20231230110837.png]]
Interpreting the [[truth table]]:
- When $p$ is true and $q$ is true, $p \rightarrow q$ is true.
- When $p$ is true and $q$ is false, $p \rightarrow q$ is false.
- When $p$ is false (regardless of $q$), $p \rightarrow q$ is true.
From the table, for $p$ to be true (column 1, rows 1 and 2), it is *necessary* for $q$ to be true (row 1) if $p \rightarrow q$ is also true (row 1). In other words, for $p \rightarrow q$ to hold true, whenever $p$ is true, $q$ must necessarily be true. Thus, when $p \rightarrow q$, for $q$ must (necessarily) be true if $p$ is true.
**Important point/distinction**: It also doesn't assert that $q$ alone makes $p$ true, *but rather that $q