**Due date:**Friday, February 3, 2023 10:00 pm (Eastern Standard Time)
<aside> đź”– (a)
$$ \begin{aligned}&p\rarr q\\ &p\\\hline\therefore &q\end{aligned} \text{ by Modus Ponens} $$
$$ \begin{aligned}&r\rarr q\\ &\neg q\\\hline\therefore &\neg r\end{aligned} \text{ by Modus Tollens}\\ \hspace{2in}Q.E.D $$
(b) Yes, “the waterslide is not running” can be concluded by the information.
We can assume that:
$p:$ There is a lifeguard shortage.
$q:$ The pools are open. then $\neg q$: The pools are closed.
$r:$ The waterslide is running.
from $\mathbf{(i)}$ we have $p$;
from $\mathbf{(ii)}$we have $p\rarr\neg q$;
from $\mathbf{(iii)}: r \rarr q$
So we can conclude $\neg r:$ “The waterslide is not running” by the conclusion of part(a)
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<aside> đź”– (a)
$$ \begin{aligned} &2n^3+3n^3+n =(2n+1)(n+1)(n)\\ &n=3k\text{ is a multiple of 3}\\ \hline\\\therefore&2n^3+3n^3+n \\&= (2n+1)(n+1)(3k)\\&= 3[(2n+1)(n+1)k] \text{ is a multiple of 3}\end{aligned} $$
(b) similarly,
$$ \begin{aligned} &2n^3+3n^3+n =(2n+1)(n+1)(n)\\ &n=3k+2\\ \hline\\\therefore&2n^3+3n^3+n \\&= (2n+1)(3k+3)(n)\\&= 3[(2n+1)(k+1)n] \text{ is a multiple of 3}\end{aligned} $$
(c) similarly,
$$ \begin{aligned} &2n^3+3n^3+n =(2n+1)(n+1)(n)\\ &n=3k+1\\ \hline\\\therefore&2n^3+3n^3+n \\&= (6k+3)(n+1)(n)\\&= 3[(2k+1)(n+1)n] \text{ is a multiple of 3}\end{aligned} $$
(d)
first, if $p\rarr r:$
$$ \begin{aligned} &p: \exist k_3(m=3k_3),m\text{ is a multiple of 3}\\&r: \exist k_6(m=6k_6 =3(2k_6)),m\text{ is a multiple of 6}\\ \hline \\&p\rarr r\ and\ k_3 =2k_6\end{aligned} $$
then if $q\rarr r:$
$$ \begin{aligned} &q: \exist k_2(m=2k_2),m\text{ is a multiple of 2}\\&r: \exist k_6(m=6k_6 =2(3k_6)),m\text{ is a multiple of 6}\\ \hline \\&q\rarr r\ and\ k_2 =3k_6\end{aligned} $$
so $(p\rarr r)\vee (q\rarr r)\equiv (p\wedge q)\rarr r$
If an integer m is a multiple of $3$, then there exists an integer k such that $m = 3k$. If m is also an even number, then $3k$ is even. We saw in class that if $pq$ is even, then either $p$ is even or $q$ is even. So since $3$ is not even, $k$ must be even. This means that $k = 2â„“$ for some integer $â„“$, but then $m = 3(2â„“) = 6â„“$,so m is a multiple of $6$.
In case you had forgotten the claim “if $pq$ is even, then $p$ is even or $q$ is even”, this is the contra positive of “if $p$ is odd and $q$ is odd, then $pq$ is odd”, which can be proved directly from the observation $(2k + 1)(2ℓ + 1) = 2(2kℓ + k + ℓ) + 1$.
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<aside> đź”– Answer:
(a)
(i)
(ii)Yes: Logical Equivalences Involving Conditional Statements and Logical Equivalences Involving Biconditional Statements.
(b)
(i) $(p_1\vee p_2\vee p_3)\rarr q \equiv (p_1\vee (p_2\vee p_3))\rarr q \equiv(p_1\vee r)\rarr q$
(ii) $(p_1\vee r)\rarr q\equiv (p_1\rarr q)\wedge(r\rarr q)$
(iii) $(p_1\rarr q)\wedge(r\rarr q)\equiv(p_1\rarr q)\wedge((p_2\vee p_3)\rarr q)\equiv(p_1\rarr q)\wedge[(p_2\rarr q)\wedge (p_3\rarr q)]$
(iv)$(p_1\rarr q)\wedge[(p_2\rarr q)\wedge (p_3\rarr q)]\equiv[(p_1\rarr q)\wedge(p_2\rarr q)\wedge(p_3\rarr q)]$
(c) I’d like to use Mathematical Induction MI to prove it: $n=2$ proved before and then prove for every $n$, if the statement holds for $n$, then it holds for $n+1$.
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<aside> đź”– (a)
$$ \neg p \equiv\neg\big(\mathcal{E}(m)\wedge\mathcal{E}(n)\wedge\mathcal{E}(o)\big)\equiv\neg\mathcal{E}(m)\vee\neg\mathcal{E}(n)\vee\neg\mathcal{E}(o). $$
(b) $\{x|x=2k,k\text{ is } \Z\}$.
(c) $\exist x(\mathcal{O}(x))$
(d)
(i) $\neg\bigg(\Big(\exist k(m=2k)\Big)\wedge\Big(\exist k(n=2k)\Big)\wedge\Big(\exist k(o=2k)\Big)\bigg)\\\equiv\bigg(\Big(\forall k(m\ne 2k)\Big)\vee\Big((\forall k(n\ne 2k)\Big)\vee\Big((\forall k(o\ne 2k)\Big)\bigg)$
(ii) $\neg\bigg(\forall x \exist k(x=2k)\bigg)\equiv \exist x\forall k(x\ne 2k)$
(e) Yes, they are same. And I prefer the (c) because it is much more universal and shorter.
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