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@ -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][