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Projectonework
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Projectonework#1
dmcnulty27 wants to merge 6 commits into
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Projectonework

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@dmcnulty27 dmcnulty27 assigned dmcnulty27 and unassigned dmcnulty27 Feb 5, 2023
@peihongx

peihongx commented Feb 5, 2023

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Hello,

Regarding this line:

(c) (¬A∧¬B∧C)∨(¬A∧¬C)??

I'm afraid that this is not a CNF. The most exterior logical connectives of a CNF should only be conjunctions ("∧") , but your answer uses a disjunction to connect two conjunction formulas, and thus you give a DNF (disjunctive normal form) here. I will find another time to read other parts of your project and add more comments!

Best,
Karl

@dmcnulty27

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Hello,

Regarding this line:

(c) (¬A∧¬B∧C)∨(¬A∧¬C)??

I'm afraid that this is not a CNF. The most exterior logical connectives of a CNF should only be conjunctions ("∧") , but your answer uses a disjunction to connect two conjunction formulas, and thus you give a DNF (disjunctive normal form) here. I will find another time to read other parts of your project and add more comments!

Best, Karl

Thank you Karl :)

@peihongx peihongx left a comment

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@dmcnulty27 Hi Dewbird, I've completed my review of your project!

Comment thread Project-1/README.md Outdated
Comment thread Project-1/README.md Outdated
Comment thread Project-1/README.md Outdated
Comment thread Project-1/README.md Outdated
Comment thread Project-1/README.md Outdated
(c) C that says that x and y are cousins
(d) O that says that x is an only child
(e) T that says that x has exactly two brothers
```

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I will give an example. (a) $B ≡ M\sqcap ∃parentOf^-.(∃parentOf\ge2)$. This is my definition. It means: x is a brother of y = x is a male & x is a child of some guy, who is a parent of at least 2 children. The key point is to introduce $parentOf^-$, which is the reverse of the parenthood, namely childhood.

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Karl I cannot get the coding to take in Github. So frustrating

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Karl I cannot get the coding to take in Github. So frustrating

There is no spacing between "$" and symbols next to it. Maybe you could search for "Markdown and Latex"? That is weird because I copy your code and paste to my comments, and then it works!

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image

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I will try again!

Comment thread Project-1/README.md Outdated
Comment thread Project-1/README.md Outdated
Comment thread Project-1/README.md Outdated
Male = {w, x}
Brother = {w, x}
Aunt = {y}
Parent_of = {(z,w), (z,x)}

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It seems you do not specify the value of "Remi", "Mary", "Frank" and "Donnie". Insetad, you assign a value to "William", "Xaivar", "Eunice", and "Zane". Maybe you can somehow change the latter to the former?

Comment thread Project-1/README.md
```(a) ∃x∀y says something like there is some x such that for all y
(a) ∀x∃y says for all x there is some y
(b) ∃x∀y∃z says there is some x such that for all y there is some z such that
(b) ∀x∃y∀z says for all x there is come y such that for all x

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Not "come" but "some"

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Woops thank you!

Comment thread Project-1/README.md

``` Here is my diagram of the universe: [label](file:///Universe.pub)
U: {0,1}
R: {(0,0), (0,1), (1,1)}

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It seems that your graph satisfy two formulas at the same time. In order to differ between them, maybe you need a graph including 3 distinct points.

@Finn1928

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For question 1, I get that a and c are tautologies, and b and c are contingent. It looks like you entered (c) twice, rather than doing (b).

For question 2, I either get errors when I type in (b) and (c), or I'm told they're tautologies. I've been told by other people that I can type something to the effect of "This is a tautology, so no CNF form is possible. p∨~p", but I don't really understand why.

Question 3 looks pretty good to me. I also included statements of non-identity in my answers, but it makes my answers pretty ugly. For instance, for 3(a) I say ∃x∃y∃z(P(z,y)∧P(z,x)∧¬Fx∧x≠y∧x≠z∧z≠y) This makes it clear we're talking about 3 different people (the parent isn't identical to any of the children, and the brothers aren't identical to each other. You can decide if you think these non-identity statements are important. I don't think Karl is including them. For me, 3(b) is ∃w∃x∃y∃z(P(z,y)∧P(w,z)∧P(w,x)∧Fx∧x≠y∧x≠z∧x≠w∧w≠y∧w≠z∧y≠z) 3(c) is ∃v∃w∃x∃y∃z(P(v,w)∧P(v,z)∧P(w,x)∧P(z,y)∧x≠v∧x≠y∧x≠z∧x≠w∧w≠y∧w≠z∧w≠v∧y≠z∧y≠v∧v≠z) 3(d) is ∃x∃y∀z((P(y,z)∧P(y,x)→z=x)∧y≠z∧y≠x) and 3(e) is ∃v∃w∃x∃y(P(v,x)∧P(v,w)∧¬Fw∧P(v,y)∧¬Fy∧∀z((P(v,z)∧¬Fz)→((z=x)∨(z=w)∨(z=y)))∧v≠w∧v≠x∧v≠y∧v≠z)

It looks like you have both your old and new answers to 4 and 5.

Besides that, this looks good to me!

Hopefully, these corrections go to the right place.
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3 participants