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style: format with typstfmt

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Price Hiller 2024-01-26 23:18:31 -06:00
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Spring-2023/CS-2233/Assignment-1/Solution.typ
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@ -49,15 +49,16 @@ Section 001
radius: 4pt, radius: 4pt,
stroke: luma(50%) + .5pt, stroke: luma(50%) + .5pt,
fill: luma(90%), 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( #text(
blue, blue,
link( link(
"https://gitlab.orion-technologies.io/philler/college/-/blob/Development/Spring-2023/CS-2233/Assignment-1/Solution.typ?ref_type=heads", "https://gitlab.orion-technologies.io/philler/college/-/blob/Development/Spring-2023/CS-2233/Assignment-1/Solution.typ?ref_type=heads",
"here", "here",
), ),
). This document was written in Typst and a bit of infinite _withering_ pain in Neovim, a Vim ). This document was written in Typst and a bit of infinite _withering_ pain in
derivative. Here's to hoping everything below is correct.], Neovim, a Vim derivative. Here's to hoping everything below is correct.],
) )
= Problems = Problems
@ -125,17 +126,17 @@ Section 001
Expressing "You understand propositional logic if you passed CS 2233" logically Expressing "You understand propositional logic if you passed CS 2233" logically
would be: would be:
#m[s → q] #m[s → p]
Now slightly rewriting the statement with our logic embedded, "If you can 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: Thus the final expression logically would be:
][$r → p ∧ (s → p)$] ][$r → (p ∧ (s → p))$]
] ]
#problem( #problem(
3, 3,
@ -379,8 +380,8 @@ Section 001
d. [10 points] By creating a sequence of logical equivalences and annotating d. [10 points] By creating a sequence of logical equivalences and annotating
each step each step
#solve[Notice by step three, both logical sequences are equivalent.][ #solve[Notice by step three, both logical sequences are equivalent.][
#table(columns: 2, [ #table(columns: 2, align: left, [
*$¬q → (p ∧ r)$* #align(center)[*$¬q → (p ∧ r)$*]
#grid(columns: 2, row-gutter: .5em, column-gutter: 1em)[ #grid(columns: 2, row-gutter: .5em, column-gutter: 1em)[
1. $¬¬q ∨ (p ∧ r)$ 1. $¬¬q ∨ (p ∧ r)$
][Conditional Identities][ ][Conditional Identities][
@ -388,8 +389,7 @@ Section 001
][Double Negation Law][ ][Double Negation Law][
3. $(q ∨ p) ∧ (q ∨ r)$ 3. $(q ∨ p) ∧ (q ∨ r)$
][Distributive Laws] ][Distributive Laws]
], [ ], [#align(center)[*$(¬q → r) ∧ (q ∨ p)$*]
#align(left)[*$(¬q → r) ∧ (q ∨ p)$*]
#grid(columns: 2, row-gutter: .5em, column-gutter: 1em)[ #grid(columns: 2, row-gutter: .5em, column-gutter: 1em)[
1. $(¬¬q ∨ r) ∧ (q ∨ p)$ 1. $(¬¬q ∨ r) ∧ (q ∨ p)$
][Conditional Identities][ ][Conditional Identities][