parsing - Explanation on this FIRST function -

ll(1) grammar:

(1) var -> id dimlist    (2) dimlist -> ε dimlist'   (3) dimlist' -> dim dimlist' (4) dimlist' -> ε   (5) dim -> [ const ] 

and, in script reading, says function first(ε dimlist') gives {#, [}. but, how?

my guess since right side of (2) begins ε, skips epsilon , takes first(dimlist') is, considering (3) , (5), equal {[}, also, because of (4), takes follow(dimlist') {#}.

other way that, since (2) begins ε skips epsilon , takes first(dimlist') takes follow(dimlist) (2)...

first 1 makes more sense me, though i'm still in process of learning basics of ll(1) grammars appreciate if takes time make clear, thank you.

edit: and, of course, neither of these true.

the usual definition of first function result in first(dimlist) (or, if like, first(ε dimlist') being {ε, [}. ε in first(ε dimlist') because both ε , dimlist' nullable. [ element because first symbol in derivation of ε dimlist, same saying first symbol in derivation of dimlist'.

another way of saying that:

first(ε dimlist' #) = {#, [}

we define function predict:

predict(ω) = first(ω follow(ω))

and can see

predict(dimlist) = first(dimlist follow(dimlist)) = {#, [}

here, first(ω) set of strings of terminals (of length ≤ 1) appear @ beginning of derivation of ω, while predict(ω) set of strings of terminals (of length ≤ 1) present in input when derivation of ω possible.

it's not uncommon confuse first , predict, it's better keep difference straight.

note of these functions can generalized strings of maximum length k, written firstk, followk , predictk, , definition of predictk similar above:

predictk(ω) = firstk(ω followk(ω))