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+#let m(math) = align(center)[$#math$]
+#let pgbreakmsg = align(center, text(blue, weight: "black", size: 1.5em)[See Next Page\ ↓])
+
+#let solve(work, solution) = align(
+  center,
+)[
+  #let solution = align(center, block(
+    inset: 5pt,
+    stroke: blue + .3pt,
+    fill: rgb(0, 149, 255, 15%),
+    radius: 4pt,
+  )[#align(left)[#solution]])
+
+  #if work == [] [
+    #solution
+  ] else [
+    #block(inset: 6pt, radius: 4pt, stroke: luma(50%) + .5pt, fill: luma(90%))[
+      #align(left, text(font: "Liberation Sans", size: .85em, work))
+      #solution
+    ]
+  ]
+]
+
+#let problem-header(number, points) = [== Problem #number. #text(weight: "regular")[[#points
+    points]]]
+
+#let problem(number, points, body) = [
+  == Problem #number. #text(weight: "regular")[[#points points]]
+  #body
+]
+
+#set page(margin: (x: .4in, y: .4in))
+#set table(align: center)
+
+_*Price Hiller*_
+#v(-.8em)
+_*zfp106*_
+#v(-.8em)
+Homework Assignment 1
+#v(-.8em)
+CS 2233
+#v(-.8em)
+Section 001
+
+#align(
+  center,
+  block(
+    inset: 6pt,
+    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
+    #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.],
+)
+= Problems
+
+#problem(
+  1,
+  10,
+)[
+  - Complete all participation activities in zyBook sections: 1.1, 1.2, 1.3, 1.4,
+    1.5.
+  #solve[][Done]
+]
+
+#problem(
+  2,
+  15,
+)[
+  Let $p$ denote "You passed CS 2233".
+
+  Let $q$ denote "You passed CS 3333".
+
+  Let $r$ denote "You can register for CS 3343".
+
+  Let $s$ denote "You understand propsitional logic".
+
+  Use $p$, $q$, $r$, and $s$, to create propositions representing the following
+  statements.
+
+  a. [5 points] You did not pass CS 2233, but you understand propositional logic.
+
+  #solve[
+    This can be alternatively expressed as
+    #align(
+      center,
+    )["You did not pass CS 2233 and you understand propositional logic"]
+    As the "but" in the statement is _not_ an exclusion, it is a conjunction in
+    typical english language.
+  ][$¬p ∧ s$]
+
+  b. [5 points] You cannot register for CS 3343 only if you have not passed both
+  CS 2233 and CS 3333
+
+  #solve[
+    This can be alternatively expressed as
+    #align(
+      center,
+    )["If you can register for CS 3343 then that implies you have passed CS 2233 and
+      CS 3333"]
+    Which can also be rewritten as
+    #align(
+      center,
+    )["If you have passed CS 2233 and CS 3333 then you can register for CS 3343"]
+
+    In the original statement, we're negating both sides, thus $¬r$ and $¬p ∧ ¬q$ which
+    can be written as:
+  ][$¬r → ¬(p ∧ q)$]
+
+  #pgbreakmsg
+  #pagebreak()
+  c. [5 points] If you can register for CS 3343, then you have passed CS 2233 and
+  you understand propositional logic if you passed CS 2233\
+
+  #solve[
+    So this can be more easily understood by "solving" the latter half of the
+    statement first.
+
+    Expressing "You understand propositional logic if you passed CS 2233" logically
+    would be:
+    #m[s → q]
+
+    Now slightly rewriting the statement with our logic embedded, "If you can
+    register for CS 3343, then you have passed CS 2233 and $(s → q)$".
+
+    Then further refining the statement: "If you can register for CS 3343 then $p ∧ (s → q)$.
+
+    And with a final refinement: $r → p ∧ (s → q)$
+
+    Thus the final expression logically would be:
+  ][$r → p ∧ (s → p)$]
+]
+#problem(
+  3,
+  40,
+)[
+  Show that $(¬q ∧ (p ∨ p)) → ¬q$ is a tautology, i.e. $(¬q ∧ (p ∨ p)) → ¬q ≡ T$
+
+  a. [10 points] By creating a truth table
+
+  // | q | p | ¬q | (p ∨ p) | (¬q ∧ (p ∨ p)) | (¬q ∧ (p ∨ p)) → ¬q |
+  // |---|---|----|---------|----------------|---------------------|
+  // | T | T | F  | T       | F              | T                   |
+  // | T | F | F  | F       | F              | T                   |
+  // | F | T | T  | T       | T              | T                   |
+  // | F | F | T  | F       | F              | T                   |
+  #solve[
+    Notice that the furthest right column only has true values, thus the above
+    proposition is a
+    _tautology_ and is always _True_
+  ][
+    #table(
+      columns: 6,
+      [q],
+      [p],
+      [¬q],
+      [(p ∨ p)],
+      [(¬q ∧ (p ∨ p))],
+      [(¬q ∧ (p ∨ p)) → ¬q],
+      [$T$],
+      [$T$],
+      [$F$],
+      [$T$],
+      [$F$],
+      [$T$],
+      [$T$],
+      [$F$],
+      [$F$],
+      [$F$],
+      [$F$],
+      [$T$],
+      [$F$],
+      [$T$],
+      [$T$],
+      [$T$],
+      [$T$],
+      [$T$],
+      [$F$],
+      [$F$],
+      [$T$],
+      [$F$],
+      [$F$],
+      [$T$],
+    )
+  ]
+  b. [10 points] By creating a sequence of logical equivalences and annotating
+  each step
+
+  #solve[][
+    #grid(columns: 2, row-gutter: .5em, column-gutter: 6em)[
+      1. $¬q ∧ (p ∨ p) → ¬q$
+    ][Starting Proposition][
+      2. $(¬q ∧ p) → ¬q ∵ (z ∨ z) ≡ p$
+    ][Idempotent Law][
+      3. $¬(¬q ∧ p) ∨ ¬q ∵ y → z ≡ ¬y ∨ q$
+    ][Conditional Identitify][
+      4. $(¬¬q ∨ ¬p) ∨ q ∵ ¬(z ∧ y) ≡ (¬z ∨ ¬q) $
+    ][De Morgan's laws][
+      5. $(q ∨ ¬p) ∨ ¬q ∵ ¬¬z ≡ z$
+    ][Double Negation laws][
+      6. $(q ∨ ¬q) ∨ ¬p ∵ (z ∨ y) ∨ x ≡ z ∨ (y ∨ x)$
+    ][Associative Laws][
+      7. $T ∨ ¬p ∵ z ∨ ¬z ≡ T ∴ (q ∨ ¬q) ∨ ¬p ≡ T ∨ ¬p$
+    ][Complement Laws][
+      8. $T ∵ T ∨ z ≡ T$
+    ][Domination Laws]
+  ]
+  #pgbreakmsg
+  #pagebreak(weak: true)
+
+  Show that $¬q → (p ∧ r) ≡ (¬q → r) ∧ (q ∨ p)$
+
+  c. [10 points] By creating a truth table
+
+  // * ¬q → (p ∧ r)
+  //
+  //  | $q$ | $p$ | $r$ | $¬q$ | $(p ∧ r)$ | $¬q → (p ∧ r)$ |
+  //  |-----|-----|-----|------|-----------|----------------|
+  //  | T   | T   | T   | F    | T         | T              |
+  //  | T   | T   | F   | F    | F         | T              |
+  //  | T   | F   | T   | F    | F         | T              |
+  //  | T   | F   | F   | F    | F         | T              |
+  //  | F   | T   | T   | T    | T         | T              |
+  //  | F   | T   | F   | T    | F         | F              |
+  //  | F   | F   | T   | T    | F         | F              |
+  //  | F   | F   | F   | T    | F         | F              |
+  //
+  //
+  // * (¬q → r) ∧ (q ∨ p)
+  //
+  //  | $q$ | $p$ | $r$ | $¬q$ | $¬q → r$ | $q ∨ p$ | $(¬q → r) ∧ (q ∨ p)$ |
+  //  |-----|-----|-----|------|----------|---------|----------------------|
+  //  | T   | T   | T   | F    | T        | T       | T                    |
+  //  | T   | T   | F   | F    | T        | T       | T                    |
+  //  | T   | F   | T   | F    | T        | T       | T                    |
+  //  | T   | F   | F   | F    | T        | T       | T                    |
+  //  | F   | T   | T   | T    | T        | T       | T                    |
+  //  | F   | T   | F   | T    | F        | T       | F                    |
+  //  | F   | F   | T   | T    | T        | F       | F                    |
+  //  | F   | F   | F   | T    | T        | F       | F                    |
+  #solve[
+    Notice that the furthest right column only has true values, thus the above
+    statement is a
+    _tautology_ and is always _True_
+  ][
+    #table(columns: 2, stroke: none, [Truth table for *$¬q → (p ∧ r)$*
+      #table(
+        columns: 6,
+        [$q$],
+        [$p$],
+        [$r$],
+        [$¬q$],
+        [$(p ∧ r)$],
+        [$¬q → (p ∧ r)$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+      )], [Truth table for *$(¬q → r) ∧ (q ∨ p)$*
+      #table(
+        columns: 7,
+        [$q$],
+        [$p$],
+        [$r$],
+        [$¬q$],
+        [$¬q → r$],
+        [$q ∨ p$],
+        [$(¬q → r) ∧ (q ∨ p)$ ],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+      )])
+  ]
+
+  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, [
+      *$¬q → (p ∧ r)$*
+      #grid(columns: 2, row-gutter: .5em, column-gutter: 1em)[
+        1. $¬¬q ∨ (p ∧ r)$
+      ][Conditional Identities][
+        2. $q ∨ (p ∧ r)$
+      ][Double Negation Law][
+        3. $(q ∨ p) ∧ (q ∨ r)$
+      ][Distributive Laws]
+    ], [
+      #align(left)[*$(¬q → r) ∧ (q ∨ p)$*]
+      #grid(columns: 2, row-gutter: .5em, column-gutter: 1em)[
+        1. $(¬¬q ∨ r) ∧ (q ∨ p)$
+      ][Conditional Identities][
+        2. $(q ∨ r) ∧ (q ∨ p)$
+      ][Double Negation Law][
+        3. $(q ∨ p) ∧ (q ∨ r)$
+      ][Commutative Laws]
+    ])
+    Thus $¬q → (p ∧ r) ≡ (¬q → r) ∧ (q ∨ p)$
+  ]
+]
+
+#pgbreakmsg
+#pagebreak()
+#problem(
+  4,
+  20,
+)[
+  a. [10 points] Show that the $∨$ operator is associative by creating a truth
+  table showing that $p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r$
+
+  #solve[Notice that the furthest right columns of both tables are equivalent, therefore
+    the $∨$ operator is associative][
+    #table(columns: 2, stroke: none, [Truth table for *$p ∨ (q ∨ r)$*
+      #table(
+        columns: 5,
+        [$q$],
+        [$p$],
+        [$r$],
+        [$(q ∨ r)$],
+        [$p ∨ (q ∨ r)$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$F$],
+      )], [Truth table for *$(p ∨ q) ∨ r$*
+      #table(
+        columns: 5,
+        [$q$],
+        [$p$],
+        [$r$],
+        [$(p ∨ q)$],
+        [$(p ∨ q) ∨ r$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$F$],
+      )])
+  ]
+
+  b. [10 points] The NOR operator $↓$ is the negation of a disjunction: $p ↓ q ≡ ¬(p ∨ q)$.
+  Its truth table is:
+
+  #table(
+    columns: 3,
+    [$p$],
+    [$q$],
+    [$p ↓ q$],
+    [$T$],
+    [$T$],
+    [$F$],
+    [$T$],
+    [$F$],
+    [$F$],
+    [$F$],
+    [$T$],
+    [$F$],
+    [$F$],
+    [$F$],
+    [$T$],
+  )
+
+  Show that The NOR operator is not associative by creating a truth table showing
+  that it is not the case that *$p ↓ (q ↓ r) ≡ (p ↓ q) ↓ r$*. In other words,
+  create a truth table showing that *$(p ↓ (q ↓ r)) ↔ ((p ↓ q) ↓ r)$* is not a
+  tautology.
+
+  #solve[Notice that the two tables' outputs are different in the furthest righthand
+    column. If the NOR were associative, the furthest right columns of both tables
+    would be identical. Since this is not the case, the NOR operator isn't
+    associative.][#table(columns: 2, stroke: none, [Truth table for *$p ↓ (q ↓ r)$*
+      #table(
+        columns: 5,
+        [$q$],
+        [$p$],
+        [$r$],
+        [$(q ↓ r)$],
+        [$p ↓ (q ↓ r)$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+      )], [Truth table for *$(p ↓ q) ↓ r$*
+      #table(
+        columns: 5,
+        [$q$],
+        [$p$],
+        [$r$],
+        [$(p ↓ q)$],
+        [$(p ↓ q) ↓ r$],
+        [$T$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$T$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$F$],
+        [$T$],
+        [$F$],
+      )])]
+]