14 lines
338 B
Markdown
14 lines
338 B
Markdown
|
---
|
||
|
|
tags:
|
||
|
|
- logic
|
||
|
|
---
|
||
|
|
|
||
|
|
# Biconditional introduction
|
||
|
|
|
||
|
|
The biconditional means if $P$ is the case, $Q$ must be the case and if $Q$ is
|
||
|
|
the case, $P$ must be the case. Thus to introduce this operator we must
|
||
|
|
demonstrate both that $Q$ follows from $P$ and that $P$ follows from $Q$. We do
|
||
|
|
this via two sub-proofs.
|
||
|
|
|
||
|
|
|