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My question is the exact order for this statement below. I'm definitely clueless. Even I have no idea where I should start from. Actually, what I wrote was "Every student has a computer or there is a friend who has a computer and is a friend with every student." I did this way $∀x(C(x))∨ (∀x∃y(C(y)∧(F(x,y)))$ Is it fine?

Define $C(x)$ : " $x$ has a computer" and $F(x,y)$: "$x$ and $y$ are friends" where $x$ and $y$ are over the domain of students. $∀x(C(x)∨∃y(C(y)∧(F(x,y)))$

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    Once we enter inside of the first parenthesis we are discussing only the specific student $x$ what we choose to inspect in closer detail. It is not necessary that there is a person with a computer that is friends with everyone, just that for whichever student we are inspecting that he has a friend with a computer regardless of whether or not that friend is friends with anyone else. – 2017-01-19
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    Are you trying to translate your written statement to a symbolic statement, or your symbolic statement to a written one? – 2017-01-19
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    deleted my original post for now it will be back later once I have edited in why it was wrong. – 2017-01-23

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Let's translate it literally first:

$\forall x \,(C(x) ∨ \exists y \,(C(y)∧(F(x,y)))$

For all students x, $(C(x) ∨ \exists y \,(C(y)∧(F(x,y)))$

For all students x, student x has a computer $∨ \, \exists y \,(C(y)∧(F(x,y)))$

For all students x, student x has a computer or there exists a student y such that $(C(y)∧(F(x,y)))$

For all students x, student x has a computer or there exists a student y such that student y has a computer $∧ \, (F(x,y))$

For all students x, student x has a computer or there exists a student y such that student y has a computer and students x and y are friends.

Or, more compactly: Every student has a computer or is a friends with a student who has a computer.

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    I actually disagree with your edit - I think the statement you translated would be $(\forall x C(x)) ∨ (\forall x \exists y (C(y)∧(F(x,y))))$. I do intend to say that for any student x, x has a computer, or there exists a student y such that y has a computer and x and y are friends. The symbolic statement written doesn't require that all students are of the same "type" - where the types are 1. having a computer or 2. having a friend with a computer. Am I misunderstanding? – 2017-01-20
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    Sorry I thought you were translating the first formula in the question. You're right that mine is for that and not the one you're translating. I got confused because the other answer translated the asker's attempt and I thought it was the same as the original. – 2017-01-20
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    I've edited back to your version but changed to "every", which I think is preferable as we won't have the strange plural "are friends with" that is ambiguous. Please feel free to rollback if you disagree with the edit. – 2017-01-20
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    No problem - that misunderstanding was why I commented with a clarifying question for the OP. And I agree with your change -thanks! – 2017-01-20
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So it turns out my answer was wrong. Instead of an answer this will now become an explanation as to why.

Here was my original answer:

A faithful translation of your statement written in English is indeed what you wrote in using logical symbols bellow it.

However; if you wished to produce a more compact statement that yields the same results; you could go with

∀x[∃y(C(y)∧(F(x,y))]

The only difference here is that the second statement does not double up on the case in which every student has a computer. The reason this occurs is because the $\exists$ operator permits the case in which every value is true.

And here is my explanation of what goes wrong.

Consider a world in which only one member exists: p.

In this world C(p) is true, and F(p,p) is false.

When we consider the first statement:

∀x(C(x))∨∀x(∃y(C(y)∧(F(x,y))))

Since there only exists one thing in W, this means that:

∀x(C(x)) in W

means C(p), which is true since C(p) is true.

Thus the whole first statement is true in W.

If we instead consider the second statement:

∀x(C(x)∨∃y(C(y)∧(F(x,y))))

we find that in W it means:

C(p)∨∃y(C(y)∧(F(p,y)))

which is true because C(p) is true.

Now if we were to consider my answer:

∀x[∃y(C(y)∧(F(x,y))]

we instead find that it means:

∃y(C(y)∧(F(p,y))

which is just C(p)∧F(p,p).

But F(p,p) is false. Thus the statement as a whole is false.

Therefore because there exists a world in which my statement is false and both statements in the OP are true; my statement is not equivalent.

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    Can someone please allude to what mistake (if any) I have made in this answer? I have asked in chat but no one has responded. I also cannot ask a formal question about it here or on meta without some information on what is wrong. – 2017-01-19
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    Your statement says that all students have a friend who has a computer- but that doesn't match the original statement. The C(X) part is necessary, because if every student has a computer but no one is friends with anyone, your statement would be false. Also, your quantifiers should be in the opposite order. – 2017-01-19
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    My quantifiers are in the same order as the OP's and your own. Also, my statement says that all students have a friend who has a computer. This means that there is not a case in which no one is friends with anyone. So my statement could not be false in that manner. Y is a student and X is a student. X may not have a computer and any chosen Y does. However if all of Y could be picked (including choosing yourself) as a friend to share with, and all of Y has a computer, then all of Y makes up the entire domain of students. Hence everyone has a computer and C(X) is not necessary. – 2017-01-20
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    Your statement says "for all students x, there exists a student y such that y has a computer and x is friends with y". This is neither a translation of the symbolic statement the OP presented nor the written statement he provided. Graham above has broken down the translation of the written statement to a symbolic one, and mentions the order of the quantifiers. As for the C(X) part - I suppose it boils down to an existential question - are you friends with yourself? For the record, I didn't downvote you, but I am explaining why one might have since you asked. – 2017-01-20
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    OK thanks for the feedback. I suppose in the future I should not try and answer more than the questions asks. Thanks again – 2017-01-20
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    @emma Turns out it was in fact wrong. I have edited it into a response as to why it was wrong instead of an answer now. – 2017-01-23
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    There were some typographical errors and I also tidied up the reasoning. And here's an upvote for you! – 2017-01-23
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Almost.

  • Every student has a computer or there is a friend who has a computer and is a friend with every student.

  • $\big(\forall x{\in}\mathrm S~C(x)\big)$ or there is a friend who has a computer and is a friend with every student.

  • $\Big(\forall x{\in}\mathrm S~C(x)\Big) \lor \Big(\exists y{\in}\mathrm F~\big($ "who has a computer and is a friend with every student.$\big)\Big)$

  • $\Big(\forall x{\in}\mathrm S~C(x)\Big) \lor \Big(\exists y{\in}\mathrm F~\big(C(y)$ "and is a friend with every student.$\big)\Big)$

  • $\Big(\forall x{\in}\mathrm S~C(x)\Big) \lor \Big(\exists y{\in}\mathrm F~\big(C(y) \land (\forall x{\in}\mathrm S~F(y,x))\big)\Big)$

  • $\Big(\forall x{\in}\mathrm S~C(x)\Big) \lor \Big(\exists y{\in}\mathrm F~\forall x{\in}\mathrm S~\big( C(y)\land F(y,x)\big)\Big)$

Key point: the order of the existential and universal quantifiers cannot be swapped.

The statement $∀x(C(x))∨(∀x∃y(C(y)∧(F(x,y)))$, if we assume the domain of Students, translates as

"Either every student has a computer, or every student is friends with a student who has a computer."

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    The phrase "either...or" is in many cases taught as being the analogue to the exclusive or. Recommend removal of the offending word "either." – 2017-01-19
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    Recommend that it should not be taught so. – 2017-01-19
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    Its a bit unavoidable at this point. Many dictionaries explicitly define the phrase as such. A short google search yielded [Cambridge Dictionary](http://dictionary.cambridge.org/us/dictionary/english/either-or) and [Merriam-Webster](https://www.merriam-webster.com/dictionary/either%E2%80%93or) both define it as the exclusive or. – 2017-01-19
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Just for completeness, I shall prove that the two statements you have in your question are not equivalent (I also assumed that both are missing a bracket at the end):

(1) $∀x(C(x))∨(∀x∃y(C(y)∧(F(x,y))))$.

(2) $∀x(C(x)∨∃y(C(y)∧(F(x,y))))$.

Let $W$ be a world with exactly $2$ objects $p,q$ such that the following hold:

  • $C(p)$, $\neg C(q)$.

  • $\neg F(p,p)$, $F(q,p)$.

Then (1) is false in $W$ because both "$∀x(C(x))$" and "$∀x∃y(C(y)∧(F(x,y)))$" are false in $W$. To see that that second half is false in $W$, note that it in particular asserts that "$( \exists y\ ( C(y) \land F(p,y) )$" is true, but that is false because neither $p$ nor $q$ is a valid witness ($p$ fails since "$F(p,p)$" is false, and $q$ fails since "$C(q)$" is false).

But (2) is true in $W$ because it is true when applied to both $p$ and $q$ (for $p$ since $C(p)$, and for $q$ since $C(p)$ and $F(q,p)$).

Therefore (1) and (2) are not equivalent because they have different truth values in some world.