feat: completed HW-1 for CS 2233
Nothing quite like cutting it this close 🤡
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Spring-2023/CS-2233/Assignment-1/Assignment.pdf
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Spring-2023/CS-2233/Assignment-1/Assignment.pdf
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Spring-2023/CS-2233/Assignment-1/Solution.typ
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Spring-2023/CS-2233/Assignment-1/Solution.typ
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#let m(math) = align(center)[$#math$]
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#let pgbreakmsg = align(center, text(blue, weight: "black", size: 1.5em)[See Next Page\ ↓])
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#let solve(work, solution) = align(
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center,
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)[
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#let solution = align(center, block(
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inset: 5pt,
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stroke: blue + .3pt,
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fill: rgb(0, 149, 255, 15%),
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radius: 4pt,
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)[#align(left)[#solution]])
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#if work == [] [
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#solution
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] else [
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#block(inset: 6pt, radius: 4pt, stroke: luma(50%) + .5pt, fill: luma(90%))[
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#align(left, text(font: "Liberation Sans", size: .85em, work))
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#solution
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]
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]
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]
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#let problem-header(number, points) = [== Problem #number. #text(weight: "regular")[[#points
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points]]]
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#let problem(number, points, body) = [
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== Problem #number. #text(weight: "regular")[[#points points]]
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#body
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]
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#set page(margin: (x: .4in, y: .4in))
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#set table(align: center)
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_*Price Hiller*_
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#v(-.8em)
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_*zfp106*_
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#v(-.8em)
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Homework Assignment 1
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#v(-.8em)
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CS 2233
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#v(-.8em)
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Section 001
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#align(
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center,
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block(
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inset: 6pt,
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radius: 4pt,
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stroke: luma(50%) + .5pt,
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fill: luma(90%),
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)[If you are interested in viewing the source code of this document, you can do so by clicking
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#text(
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blue,
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link(
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"https://gitlab.orion-technologies.io/philler/college/-/blob/Development/Spring-2023/CS-2233/Assignment-1/Solution.typ?ref_type=heads",
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"here",
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),
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). This document was written in Typst and a bit of infinite _withering_ pain in Neovim, a Vim
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derivative. Here's to hoping everything below is correct.],
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)
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= Problems
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#problem(
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1,
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10,
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)[
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- Complete all participation activities in zyBook sections: 1.1, 1.2, 1.3, 1.4,
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1.5.
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#solve[][Done]
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]
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#problem(
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2,
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15,
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)[
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Let $p$ denote "You passed CS 2233".
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Let $q$ denote "You passed CS 3333".
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Let $r$ denote "You can register for CS 3343".
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Let $s$ denote "You understand propsitional logic".
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Use $p$, $q$, $r$, and $s$, to create propositions representing the following
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statements.
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a. [5 points] You did not pass CS 2233, but you understand propositional logic.
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#solve[
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This can be alternatively expressed as
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#align(
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center,
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)["You did not pass CS 2233 and you understand propositional logic"]
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As the "but" in the statement is _not_ an exclusion, it is a conjunction in
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typical english language.
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][$¬p ∧ s$]
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b. [5 points] You cannot register for CS 3343 only if you have not passed both
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CS 2233 and CS 3333
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#solve[
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This can be alternatively expressed as
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#align(
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center,
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)["If you can register for CS 3343 then that implies you have passed CS 2233 and
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CS 3333"]
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Which can also be rewritten as
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#align(
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center,
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)["If you have passed CS 2233 and CS 3333 then you can register for CS 3343"]
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In the original statement, we're negating both sides, thus $¬r$ and $¬p ∧ ¬q$ which
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can be written as:
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][$¬r → ¬(p ∧ q)$]
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#pgbreakmsg
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#pagebreak()
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c. [5 points] If you can register for CS 3343, then you have passed CS 2233 and
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you understand propositional logic if you passed CS 2233\
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#solve[
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So this can be more easily understood by "solving" the latter half of the
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statement first.
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Expressing "You understand propositional logic if you passed CS 2233" logically
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would be:
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#m[s → q]
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Now slightly rewriting the statement with our logic embedded, "If you can
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register for CS 3343, then you have passed CS 2233 and $(s → q)$".
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Then further refining the statement: "If you can register for CS 3343 then $p ∧ (s → q)$.
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And with a final refinement: $r → p ∧ (s → q)$
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Thus the final expression logically would be:
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][$r → p ∧ (s → p)$]
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]
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#problem(
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3,
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40,
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)[
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Show that $(¬q ∧ (p ∨ p)) → ¬q$ is a tautology, i.e. $(¬q ∧ (p ∨ p)) → ¬q ≡ T$
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a. [10 points] By creating a truth table
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// | q | p | ¬q | (p ∨ p) | (¬q ∧ (p ∨ p)) | (¬q ∧ (p ∨ p)) → ¬q |
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// |---|---|----|---------|----------------|---------------------|
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// | T | T | F | T | F | T |
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// | T | F | F | F | F | T |
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// | F | T | T | T | T | T |
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// | F | F | T | F | F | T |
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#solve[
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Notice that the furthest right column only has true values, thus the above
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proposition is a
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_tautology_ and is always _True_
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][
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#table(
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columns: 6,
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[q],
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[p],
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[¬q],
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[(p ∨ p)],
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[(¬q ∧ (p ∨ p))],
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[(¬q ∧ (p ∨ p)) → ¬q],
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[$T$],
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[$T$],
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[$F$],
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[$T$],
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[$F$],
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[$T$],
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[$T$],
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[$F$],
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[$F$],
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[$F$],
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[$F$],
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[$T$],
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[$F$],
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[$T$],
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[$T$],
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[$T$],
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[$T$],
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[$T$],
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[$F$],
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[$F$],
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[$T$],
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[$F$],
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[$F$],
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[$T$],
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)
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]
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b. [10 points] By creating a sequence of logical equivalences and annotating
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each step
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#solve[][
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#grid(columns: 2, row-gutter: .5em, column-gutter: 6em)[
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1. $¬q ∧ (p ∨ p) → ¬q$
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][Starting Proposition][
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2. $(¬q ∧ p) → ¬q ∵ (z ∨ z) ≡ p$
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][Idempotent Law][
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3. $¬(¬q ∧ p) ∨ ¬q ∵ y → z ≡ ¬y ∨ q$
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][Conditional Identitify][
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4. $(¬¬q ∨ ¬p) ∨ q ∵ ¬(z ∧ y) ≡ (¬z ∨ ¬q) $
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][De Morgan's laws][
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5. $(q ∨ ¬p) ∨ ¬q ∵ ¬¬z ≡ z$
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][Double Negation laws][
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6. $(q ∨ ¬q) ∨ ¬p ∵ (z ∨ y) ∨ x ≡ z ∨ (y ∨ x)$
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][Associative Laws][
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7. $T ∨ ¬p ∵ z ∨ ¬z ≡ T ∴ (q ∨ ¬q) ∨ ¬p ≡ T ∨ ¬p$
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][Complement Laws][
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8. $T ∵ T ∨ z ≡ T$
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][Domination Laws]
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]
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#pgbreakmsg
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#pagebreak(weak: true)
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Show that $¬q → (p ∧ r) ≡ (¬q → r) ∧ (q ∨ p)$
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c. [10 points] By creating a truth table
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// * ¬q → (p ∧ r)
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//
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// | $q$ | $p$ | $r$ | $¬q$ | $(p ∧ r)$ | $¬q → (p ∧ r)$ |
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// |-----|-----|-----|------|-----------|----------------|
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// | T | T | T | F | T | T |
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// | T | T | F | F | F | T |
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// | T | F | T | F | F | T |
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// | T | F | F | F | F | T |
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// | F | T | T | T | T | T |
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// | F | T | F | T | F | F |
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// | F | F | T | T | F | F |
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// | F | F | F | T | F | F |
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//
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//
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// * (¬q → r) ∧ (q ∨ p)
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//
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// | $q$ | $p$ | $r$ | $¬q$ | $¬q → r$ | $q ∨ p$ | $(¬q → r) ∧ (q ∨ p)$ |
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// |-----|-----|-----|------|----------|---------|----------------------|
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// | T | T | T | F | T | T | T |
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// | T | T | F | F | T | T | T |
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// | T | F | T | F | T | T | T |
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// | T | F | F | F | T | T | T |
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// | F | T | T | T | T | T | T |
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// | F | T | F | T | F | T | F |
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// | F | F | T | T | T | F | F |
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// | F | F | F | T | T | F | F |
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#solve[
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Notice that the furthest right column only has true values, thus the above
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statement is a
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_tautology_ and is always _True_
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][
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#table(columns: 2, stroke: none, [Truth table for *$¬q → (p ∧ r)$*
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#table(
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columns: 6,
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[$q$],
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[$p$],
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[$r$],
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[$¬q$],
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[$(p ∧ r)$],
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[$¬q → (p ∧ r)$],
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[$T$],
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[$T$],
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[$T$],
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[$F$],
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[$T$],
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[$T$],
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[$T$],
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[$T$],
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[$F$],
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[$F$],
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[$F$],
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[$T$],
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[$T$],
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[$F$],
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[$T$],
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[$F$],
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[$F$],
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[$T$],
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[$T$],
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[$F$],
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[$F$],
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[$F$],
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[$F$],
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[$T$],
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[$F$],
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[$T$],
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[$T$],
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[$T$],
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[$T$],
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[$T$],
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[$F$],
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[$T$],
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[$F$],
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[$T$],
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[$F$],
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[$F$],
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[$F$],
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[$F$],
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[$T$],
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[$T$],
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[$F$],
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[$F$],
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[$F$],
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[$F$],
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[$F$],
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[$T$],
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[$F$],
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[$F$],
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)], [Truth table for *$(¬q → r) ∧ (q ∨ p)$*
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#table(
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columns: 7,
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[$q$],
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[$p$],
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[$r$],
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[$¬q$],
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[$¬q → r$],
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[$q ∨ p$],
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[$(¬q → r) ∧ (q ∨ p)$ ],
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[$T$],
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[$T$],
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[$T$],
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[$F$],
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[$T$],
|
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[$T$],
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[$T$],
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[$T$],
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[$T$],
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[$F$],
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[$F$],
|
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[$T$],
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[$T$],
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[$T$],
|
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[$T$],
|
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[$F$],
|
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[$T$],
|
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[$F$],
|
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[$T$],
|
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[$T$],
|
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[$T$],
|
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[$T$],
|
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[$F$],
|
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[$F$],
|
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[$F$],
|
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[$T$],
|
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[$T$],
|
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[$T$],
|
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[$F$],
|
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[$T$],
|
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[$T$],
|
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[$T$],
|
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[$T$],
|
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[$T$],
|
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[$T$],
|
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[$F$],
|
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[$T$],
|
||||
[$F$],
|
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[$T$],
|
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[$F$],
|
||||
[$T$],
|
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[$F$],
|
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[$F$],
|
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[$F$],
|
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[$T$],
|
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[$T$],
|
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[$T$],
|
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[$F$],
|
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[$F$],
|
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[$F$],
|
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[$F$],
|
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[$F$],
|
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[$T$],
|
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[$T$],
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[$F$],
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[$F$],
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)])
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]
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d. [10 points] By creating a sequence of logical equivalences and annotating
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each step
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#solve[Notice by step three, both logical sequences are equivalent.][
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#table(columns: 2, [
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*$¬q → (p ∧ r)$*
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#grid(columns: 2, row-gutter: .5em, column-gutter: 1em)[
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1. $¬¬q ∨ (p ∧ r)$
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][Conditional Identities][
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2. $q ∨ (p ∧ r)$
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][Double Negation Law][
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3. $(q ∨ p) ∧ (q ∨ r)$
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][Distributive Laws]
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], [
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#align(left)[*$(¬q → r) ∧ (q ∨ p)$*]
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#grid(columns: 2, row-gutter: .5em, column-gutter: 1em)[
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1. $(¬¬q ∨ r) ∧ (q ∨ p)$
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][Conditional Identities][
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2. $(q ∨ r) ∧ (q ∨ p)$
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][Double Negation Law][
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3. $(q ∨ p) ∧ (q ∨ r)$
|
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][Commutative Laws]
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])
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Thus $¬q → (p ∧ r) ≡ (¬q → r) ∧ (q ∨ p)$
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]
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]
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#pgbreakmsg
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#pagebreak()
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#problem(
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4,
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20,
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)[
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a. [10 points] Show that the $∨$ operator is associative by creating a truth
|
||||
table showing that $p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r$
|
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#solve[Notice that the furthest right columns of both tables are equivalent, therefore
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the $∨$ operator is associative][
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#table(columns: 2, stroke: none, [Truth table for *$p ∨ (q ∨ r)$*
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#table(
|
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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,
|
||||
[$q$],
|
||||
[$p$],
|
||||
[$r$],
|
||||
[$(p ↓ q)$],
|
||||
[$(p ↓ q) ↓ r$],
|
||||
[$T$],
|
||||
[$T$],
|
||||
[$T$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$T$],
|
||||
[$T$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$T$],
|
||||
[$T$],
|
||||
[$F$],
|
||||
[$T$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$T$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$T$],
|
||||
[$F$],
|
||||
[$T$],
|
||||
[$T$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$T$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$T$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$T$],
|
||||
[$T$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$F$],
|
||||
[$T$],
|
||||
[$F$],
|
||||
)])]
|
||||
]
|
||||
Loading…
Reference in New Issue
Block a user