I need your help please, I already got the answer from you and there are some strange typing within the answer and i want you to help me to find out

I need your help please, I already got the answer from you and there are some strange typing within the answer and i want you to help me to find out what does that mean, also i need the idea behind the answer because it seems strange to me "if you don't mind",here you are the question and the answer i got as well:Show that the following identities hold for regular expressions over any alphabet:1. epsilon + R* R = R* pref(L) = {w: ∃x ∈ Σ* (wx ∈ L)}. Take a word w in Pref(L). By defnition, there exists a in_nite(??) word x over _ (??)such that wx belongs to Adh(L). Then for all n _ 0, wx[0; n 􀀀 (??) 1] belongs to Pref(L). Thus for all n _ 0, there exists a _nite (??) word y(n) 2 __ (??) such that w(n) := wx[0; n 􀀀(??) 1]y(n) belongs to L, and there are in_nitely many such wordsw(n). Conversely, let w be a pre_x of in_nitely many words in L. There exists a letter a(??)2 _ such that wa is a pre_x of in_nitely(??) many words in L. Iterating this argument, there exists a sequence (an)n_0 of letters in _ such that was _ _ _ anbelongs to Pref(L) for all n _ 0. This implies that was a1 _ _ _ belongs to Adh(L). Hence w belongs to Pref(L), which is closed. 2. (R *S* )* = (R + S)* suff(L) = {w: ∃x ∈ Σ* (xw ∈ L)}.Take a word w in Suff(L). By defnition, there exists a in_nite(??) word x over _ such that xw belongs to Adh(L). Then for all n _(??) 0, xw[0; n 􀀀(??) 1] belongs to Suff(L). Hence w belongs to Suff(L), which is closedAnother thing: what is the role of pref(L) in the first one, and also suff(L) in the second one?I would be grateful if you replied me