From 89e8b634e8456da177ba1b167bf415ef19d9b911 Mon Sep 17 00:00:00 2001
From: Price Hiller <price@orion-technologies.io>
Date: Fri, 26 Jan 2024 23:18:31 -0600
Subject: [PATCH] style: format with typstfmt

---
 Spring-2023/CS-2233/Assignment-1/Solution.typ | 24 +++++++++----------
 1 file changed, 12 insertions(+), 12 deletions(-)

diff --git a/Spring-2023/CS-2233/Assignment-1/Solution.typ b/Spring-2023/CS-2233/Assignment-1/Solution.typ
index 7ab2bdb..24471cb 100644
--- a/Spring-2023/CS-2233/Assignment-1/Solution.typ
+++ b/Spring-2023/CS-2233/Assignment-1/Solution.typ
@@ -49,15 +49,16 @@ Section 001
     radius: 4pt,
     stroke: luma(50%) + .5pt,
     fill: luma(90%),
-  )[If you are interested in viewing the source code of this document, you can do so by clicking
+  )[If you are interested in viewing the source code of this document, you can do so
+    by clicking
     #text(
       blue,
       link(
         "https://gitlab.orion-technologies.io/philler/college/-/blob/Development/Spring-2023/CS-2233/Assignment-1/Solution.typ?ref_type=heads",
         "here",
       ),
-    ). This document was written in Typst and a bit of infinite _withering_ pain in Neovim, a Vim
-    derivative. Here's to hoping everything below is correct.],
+    ). This document was written in Typst and a bit of infinite _withering_ pain in
+    Neovim, a Vim derivative. Here's to hoping everything below is correct.],
 )
 = Problems
 
@@ -125,17 +126,17 @@ Section 001
 
     Expressing "You understand propositional logic if you passed CS 2233" logically
     would be:
-    #m[s → q]
+    #m[s → p]
 
     Now slightly rewriting the statement with our logic embedded, "If you can
-    register for CS 3343, then you have passed CS 2233 and $(s → q)$".
+    register for CS 3343, then you have passed CS 2233 and $(s → p)$".
 
-    Then further refining the statement: "If you can register for CS 3343 then $p ∧ (s → q)$.
+    Then further refining the statement: "If you can register for CS 3343 then $p ∧ (s → p)$.
 
-    And with a final refinement: $r → p ∧ (s → q)$
+    And with a final refinement: $r → p ∧ (s → p)$
 
     Thus the final expression logically would be:
-  ][$r → p ∧ (s → p)$]
+  ][$r → (p ∧ (s → p))$]
 ]
 #problem(
   3,
@@ -379,8 +380,8 @@ Section 001
   d. [10 points] By creating a sequence of logical equivalences and annotating
   each step
   #solve[Notice by step three, both logical sequences are equivalent.][
-    #table(columns: 2, [
-      *$¬q → (p ∧ r)$*
+    #table(columns: 2, align: left, [
+      #align(center)[*$¬q → (p ∧ r)$*]
       #grid(columns: 2, row-gutter: .5em, column-gutter: 1em)[
         1. $¬¬q ∨ (p ∧ r)$
       ][Conditional Identities][
@@ -388,8 +389,7 @@ Section 001
       ][Double Negation Law][
         3. $(q ∨ p) ∧ (q ∨ r)$
       ][Distributive Laws]
-    ], [
-      #align(left)[*$(¬q → r) ∧ (q ∨ p)$*]
+    ], [#align(center)[*$(¬q → r) ∧ (q ∨ p)$*]
       #grid(columns: 2, row-gutter: .5em, column-gutter: 1em)[
         1. $(¬¬q ∨ r) ∧ (q ∨ p)$
       ][Conditional Identities][