Given the following language definition: L = {w ϵ {a, b}* : x ϵ {a, b}+ (w = axa)} ||| Give a simple English description, two words it contains, and two it does not.
The words must all begin with 'a' and end with 'a'
can there be a string with zero length?
no
is this because of the plus sign?
but i don't understand fully what you are trying to say.
basically looking for strings which would be found in the language as defined in the question
\[ L = \{ w \in (a,b)^*:x\in (a,b) + (w=a \times a) \}\] how do you interpret this statement?
There is a language such that w is an element of L containing any combinations of the symbols 'a' and 'b'...
If X is in L, and contains at least one a and b, X must begin with a and end with a
this last part means that there cannot be a string with no b's unless if is an empty string
I have never seen those representation before. I think there is no word in English that begins with 'a' and ends with 'a' and only containing 'a' and 'b'. If a set does not have any elements then it's a null set.
try searching for few words online http://www.thefreedictionary.com/
English has nothing to do with it, except that I need to explain the fictional language in English
isn't this related to sets??
yes, the words which make up the language will be a set of words, the first being the empty set
sorry, word with no characters, not the empty set
then you can have infinite words. begin with 'a' then add up as much as you like, then again end with 'a'.
is there one exception, meaning the plus sign (one or more)?
x is an element of words with at least one a and one b if it begins with a and ends with a?
just get the two word that is contained in X and 2 word not contained in X. eg. aba, ababa <--- these two are contained in X eg. baba, abab <--- these two are not contained in X
L being superset of X, contains all those 4 elements, but X does not contain those last two elements of L, because of the criteria imposed on X for its members.
X appears to be a representation of a substring of strings which begin with a and end with a, not a subset of L
don't you need a word that X does not contain?
that L does not contain
a word that is contain in X but not contain in L?
X is merely a representation of a word within L it appears
so you need 2 words that L does not contain?
yes
any word containing 'a' and 'b' would be contained in L, either we construct a word from other characters or there is no such word.
bab would be a word not in L?
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