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Hello. I'd like to check my answer for 1. (g)

$\forall$ x $\in$ A, P(x), $\forall$ y $\in$ A, C(y) $\wedge$ F(y), $\forall$ z $\in$A, C(z), T(x, y) $\implies$ T(x, z)

Is this correct?

Thank you

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What does the comma in your notation mean? –  dtldarek Oct 19 '13 at 21:11
    
To separate the domain and predicates? Are you not supposed to use commas? –  muros Oct 19 '13 at 21:19
    
Not the commas after $\in A$, but those before $\forall$-s, or the one between $C(z)$ and $T(x,y)$. –  dtldarek Oct 19 '13 at 22:05
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1 Answer

I'd suggest the use of parentheses to help disambiguate the scope of each quantified variable, replacing some of the commas. You also need a connective following $C(z)$. Here is my suggested translation:

"If any professor has taught every first-year course, then that professor has taught every course."

$$\forall x \in A, \Big(\{P(x) \land \forall y \in A, [(C(y)\land F(y)) \rightarrow T(x, y)]\}\rightarrow \forall z \in A, (C(z) \rightarrow T(x,z))\Big)$$

This essentially reads,

"for all x in the domain, IF [x is a professor such that, for all y in the domain, if (y is a first-year- (and) y is a course) then (x has taught y)), THEN, for all z in the domain, if z is a course, it follows that x has taught z.

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You have a lot of patience to read that very fine print! +1 –  Amzoti Oct 22 '13 at 1:12
    
@Amzoti The OP doesn't seem to be very responsive, so I don't know whether it's helping at all! –  amWhy Oct 22 '13 at 14:03
    
I have had that same issue with this OP. Hope you have a great day! –  Amzoti Oct 22 '13 at 14:20
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